GATE EE 2014 Official Paper: Shift 1 (Previous Year Paper)

The phrasal verb ‘cope with’ means to ‘adjust to’. The best meaning of the phrase is captured by ‘adapt to’.

The context of the sentence requires a word that will express the contradiction in the speaker’s mind. Since he did not think the performance to be impressive, that is why he could not understand the reason for her getting the first prize. Thus, the words ‘superb’ and ‘exhilarating’ can be ignored. Between ‘medium’ and ‘mediocre’, the latter is a better fit to describe a performance.

The meaning of the phrase ‘the buck stops here’ means that the minister would be responsible for all his decisions and with him lies the final power of decision making and thus option C is the correct meaning of the phrase.

Calculation:

\(\begin{array}{l} {\left( {z + \frac{1}{z}} \right)^2} = 98\ {Z^2} + \frac{1}{{{z^2}}} + 2 \times z \times \frac{1}{z} = 98\ {Z^2} + \frac{1}{{{z^2}}} = 98 - 2 = 96 \end{array}\)

Given the quadratic equations,

\(\begin{array}{l}
a{x^2}; + bx + c = 0\
a{x^2} + b\left| x \right| + c = 0
\end{array}\)

\(\left| x \right|\) has two values \(+ x;and; - x\) . So, it may be resolved into two equations

\(a{x^2}; + bx + c = 0\)  with 2 roots

And \(a{x^2} - bx + c = 0\) with 2 roots

Hence, there are total 4 real roots.

The inference that can be drawn from the passage is that the low lying terrain of the gap is responsible for the weather patterns in Tamil Nadu and Kerala. We do not know the reasons of formation of the gap or the nature of the regions near the Palghat Gap thus eliminating options A and B. No correlation between temperatures and rainfall can be inferred from the passage, thus option D is not the correct inference.

Since the genetic roots are being confirmed for psychiatric illnesses, the assumption that can be made is that these diseases have a genetic basis. Thus option B is the correct assumption.

Total round trip fare for group of 5 tourist without discount

\(= 5 \times 200 = Rs.1000\)

(i) Discount for round trip = 10%of total fare \(= \frac{{10}}{{100}} \times 1000 = Rs.;100\)

 (ii) Discount for having a group of 5 tourist = 10%of total fare \(= \frac{5}{{100}} \times 100 = Rs.;500\)

Total discount = Discount for round + Discount for having a group of 5 tourist.

\(= 100 + 50 = Rs.150\)

Thus, the net round trip fare for group of 5 tourist after discount is

Net fare = total fare-total discount

\(\begin{array}{l}
= Rs.1000 - Rs.150\
= Rs.;850
\end{array}\)

Total respondents = 300 (150 men + 150 women)

Total number of men who owns car = 100

Total number of men who owns scooter = 90

Total number of men who do not own a scooter = 40(Car) + 20(Nothing) = 60

Total number of women who owns car = 80

Total number of women who owns scooter = 66

Total number of women who do not own a scooter = 34(Car) + 50(Nothing) = 84

Percentage of respondents who do not own a scooter \(= \frac{{60 + 84}}{{300}} \times 100 = 48%\)

We draw a tetrahedron structure ABCD as

Now, P is a point inside the tetrahedron.

The point P is connected to each corners A, B, C and D of the tetrahedron.

So, we have the internal planes as ABP, APC, BPC, DPC, DPB and DPA.

I.e. the total number of internal planes = 6.

Concept:

The system Ax = B, with coefficient matrix A and augmented matrix [A|B], the number of unknowns is n.

Below, we summarize the possibilities for solving the system.

i. Ax = B is inconsistent (i.e., no solution exists) if and only if rank[A] < rank[A|B].

ii. Ax = B has a unique solution if and only if rank[A] = rank[A|B] = n.

iii. Ax = B has infinitely many solutions if and only if rank[A] = rank[A|B] < n

Calculation:

\(x; + ;2y; + ;2z; = ;{b_1},;;;;;5x; + ;y; + ;3z; = ;{b_2}\) 

Above given equation is in the form of Non - Homogeneous equation.

Here, number of variables (n) = 3. 

Now,

AX = B 

\(\begin{array}{l} \left[ {A;:B} \right] = \left[ {\begin{array}{*{20}{c}} 1&2&2&{{b_1}}\ 5&1&3&{{b_2}} \end{array}} \right]\ {R_2} \Rightarrow {R_2} - 5{R_1}\ \left[ {\begin{array}{*{20}{c}} 1&2&2&{{b_1}}\ 0&{ - 9}&{ - 7}&{{b_2} - 5{b_1}} \end{array}} \right] \end{array}\)

There exist two non - zero rows irrespective of values of b₁ and b₂.

Rank of \(A; = ;2; =\) Rank of \(\left[ {A;:;B} \right]\) is less than the number of variable. 

∴ The system has infinite number of solutions.

Note:

Homogeneous equations are always consistent

Non - homogeneous equations has either no solution, one solution or infinite solutions.

Concept:

Maxima and Minima of a function:

Let f(x) is supposed to be continuous for all values of x in the neighbourhood of x = a. Then f(a) is called as maximum or minimum of f(x) according to as f(a+h) is less than or greater than for all sufficiently small independent value of h, positive or negative, provided both of them are not equal to zero.

Necessary conditions for a point to be maxima or minima: The necessary conditions for f(x) to have maxima or minima at x = a is that

\({\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}}_{\left( {{\rm{x}} = {\rm{a}}} \right)}} = 0{\rm{;}}\)

Sufficient conditions for a point to be maxima and minima:

\({\frac{{{{\rm{d}}^2}{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}}_{\left( {{\rm{x}} = {\rm{a}}} \right)}} > 0 \to {\rm{minima}}\)

\({\frac{{{{\rm{d}}^2}{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}}_{\left( {{\rm{x}} = {\rm{a}}} \right)}} < 0 \to {\rm{maxima}}\)

Calculation:

Given, 

\(f\left( x \right) = x{e^{ - x}}\)

On differentiating both sides,

\(f'\left( x \right) = - x{e^{ - x}} + {e^{ - x}}\)

The stationary points are calculated as,

For \(f'\left( x \right) = 0,;x = 1\)

At x = 1, f" (x) <

∴ Maximum exists at \(x = 1\) and is equal to \(f\left( 1 \right) = {e^{ - 1}}\)

Given the differential equation,

\(\frac{{{d^2}x}}{{d{t^2}}} = - 9x\)

Taking the Laplace transform,

s²X(s) – sX(0) – X^’(0) = -9X(s)

or s²X(s) – s(1) – (1) = - 9X(s)

or X(s) (s²+9) = s + 1

\(X\left( s \right) = \frac{{s + 1}}{{{s^2} + 9}} = \frac{s}{{{s^2} + 9}} + \frac{1}{{{s^2} + 9}}\)

\(x\left( t \right) = cos\ 3t + \frac{1}{3}\sin 3t\) (Taking  inverse Laplace.)

Concept:

\(\rm x(0^+) = \displaystyle\lim_{x \rightarrow \infty} sX(s)\)

Calculation:

Using the initial value theorem,

\(\rm x(0^+) = \displaystyle\lim_{x \rightarrow \infty} sX(s)\)

\(\Rightarrow \rm x(0^+) = \displaystyle\lim_{s \rightarrow \infty} s. \frac{(3s + 5)}{(s^2 + 10s + 21)}\)

\(\Rightarrow \rm x(0^+) = \displaystyle\lim_{s \rightarrow \infty} \frac{s^2( 3 + \frac{5}{s})}{s^2 \left( 1 + \frac{10}{S} + \frac{21}{S^2} \right)}\)

\(\Rightarrow \rm x(0^+) = \displaystyle\lim_{s \rightarrow \infty} \frac{3 + \frac{5}{S}}{1 + \frac{10}{S} + \frac{21}{S^2}}\)

∴ x(0+) = 3

Alternate method:

using Inverse Laplace Transform method,

we have,

\(\rm X(s) = \frac{3S + 5}{S^2 + 10S + 21} = \frac{3S + 5}{S^2 + 10 S + 21 + 4 - 4}\)

\(\rm X(s) = \frac{3S + 5}{(S + 5)^2 - 2^2} = \frac{3S + 15 - 10}{(S + 5)^2 - 2^2}\)

\(\rm X(S) = \frac{3(S + 5)}{(S + 5)^2 - 2^2} - \frac{10}{(S + 5)^2 - 2^2}\)

Taking inverse Laplace transform

x(t) = 3e-5t cosh2t - 5e-5t sinh2t

x(t) = e-5t(3 cosh2t - 5 sinh2t)

At t = 0+,

x(0+) = e0(3cosh 0 - 5 sinh0)

x(0+) = 1(3 - 0)

x(0+) = 3

\(\begin{array}{l} \omega = u + iv\ f\left( z \right) = z;{z^*} = \left( {x + iy} \right)\left( {x - iy} \right)\ = {x^2} + {y^2}; = 1;;;;;;\left( {\because\left| Z \right| = 1} \right)\ = 1 + 0i\ \therefore ;u = 1,v = 0 \end{array}\)

Hence all the points on ‘S’ are mapped to the point \(\left( {1,;0} \right)\) in w – plane.

Concept:

suppose z = p + jQ is a complex number, then the conjugate of z can be written as 

z^* = P - jQ

• z - z^* = 2 Re{z}
• z + z^* = 2j Img{z}
• z^* = z; when z is purely real.
• z + z^* = 0; when z is purely imaginary
• zz^* = |z|²

Calculation:

we have,

function, f(z) = zz^* & |z| = 1

Let, z = x + jy

then z^* = x - jy

so, zz^* = (x + jy) (x - jy) = x² + y² = |z|²

⇒ x² + y² = 1

above equation denotes unit circle on the z-plane having origin center & radius = 1

Therefore, f(z) maps 's' to the point (1, 0) in the complex plane.

Concept:

The sign convention for power calculation is given below.

When current entering the positive terminal, the power absorbed is positive and the power delivered is negative.

Power absorbed (P) = VI

Power delivered or supplied (P) = -VI

When current entering the negative terminal, the power delivered is positive and the power absorbed is negative.

Power delivered or supplied (P) = VI

Power absorbed (P) = -VI

In any element power absorb is equals to negative times of power delivered or supplying to element.

Calculation:

By KCL current through 15 V source is 2 A. When current is entering in to +ve terminal of a battery means, it is absorbing the power.

When current is entering in to –ve terminal means, delivering the power.

100 V source is absorbing the power

⇒ 10(100) = 1000 watts

80 V source is delivering the power

⇒ 8(80) = 640 watts

15 V source is delivering the power

⇒ 2(15) = 30 watts

Total power absorbed by the circuit elements

= 1000 – (640 + 30) = 330 watts

CONCEPT:

Capacitor:

• The capacitor is a device in which electrical energy can be stored.
• In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite sign and separated by an insulating medium.
• The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​

​Parallel plate capacitor:

• A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.
• The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​
• The electric field intensity at the outer region of the parallel plate capacitor is always zero whatever be the charge on the plate.
• The electric field intensity in the inner region between the plates of a parallel plate capacitor remains the same at every point.
• When the dielectric medium is filled in the space between the plates of the parallel plate capacitor, its capacitance increases.
• The electric field intensity in the inner region between the plates of a parallel plate capacitor is given as,

• The potential difference between the plates is given as,

• The capacitance C of the parallel plate capacitor is given as,

Where A = area of the plates, d = distance between the plates, Q = charge on the plates, σ = surface charge density, and K = dielectric constant

CALCULATION:

C = C₁ + C₂

_{\(C = \frac{{{A_1}{\varepsilon _1}}}{d} + \frac{{{A_2}{\varepsilon _2}}}{d}\)}

_{\(= \frac{{A{\varepsilon _0}}}{{2d}} + \frac{{A{\varepsilon _r}{\varepsilon _0}}}{2d}\)}

\(= \frac{{A{\varepsilon _0}}}{{2d}}\left( {1 + {\varepsilon _r}} \right)\)

\(C = \frac{{{C_0}}}{2}\left( {1 + {\varepsilon _r}} \right)\)

Concept:

Transients are present in the network when the network is subject to changes either by changing source magnitude or by changing any circuit elements

• the provided circuit should consist of any energy storage elements.
• No transients occur in the resistive element.
• Inductor doesn’t allow sudden change in current i.e. I_L(0^-) = I_L(0⁺).
• Capacitor doesn’t allow sudden change in voltage i.e. V_C(0^-) = V_C(0⁺)

Case 1: t < 0 , Under steady-state,

Case 2: t > 0

When the switch is opened:

By using Laplace transform approach,

\({V_s}\left( s \right) = \left( {\frac{I}{s}} \right)\left( {\frac{1}{{sC}}} \right) + \left( {\frac{I}{s}} \right)\left( {R} \right) + \left( {\frac{{ - 2}}{s}} \right)\)

\({V_s}\left( s \right) = \left( \frac{1}{s} \right) \left( IR-2+\frac{I}{sC} \right)\)

\({V_s}\left( s \right) = \left( \frac{1}{s} \right) \left( 20\times 10^{-3}\times 100-2+\frac{20\times 10^{-3}}{s\times 1 \times 10^{-6}} \right)\)

\({V_s}\left( s \right) = \left( \frac{1}{s} \right) \left( 2-2+\frac{20;\times ;10^3}{s} \right)\)

\({V_s}\left( s \right) = \frac{{20 ;\times ;{{10}^3}}}{{{s^2}}}\)

\(\Rightarrow {V_s}\left( s \right) = \frac{{2; \times ;{{10}^4}}}{{{s^2}}}\)

⇒ V_S(t) = 2 × 10⁴ t u(t)

The equation obtained is the equation of a straight line passing through origin.

Concept:

Z(t) = x(t) * y(t)

If two signal are convoluted then the resulting convoluted signal has following range. 

sum of lower limits < t < sum of upper limits.

Calculation:

we have:

x(t) is Non-zero for T_x < t <T_x'

y(t) is Non-zero for T_y < t < T_y'

z(t) = x(t) * y(t)

Lower limit of z(t) is sum of lower limits of x(t) & y(t)

Upper limit of z(t) is sum of upper limits of x(t) & y(t).

Hence,

limits of z(t) is,

T_x + T_y < t < T_x' + T_y'

Therefore,

z(t) is zero outside of (T_x + T_y) < t < (T_x' + T_y')

Example:

A Periodic square wave is shown and defined as

\(x\left( t \right) = \left{ {\begin{array}{*{20}{c}} 1&{\left| t \right| < T/4}\ 0&{;\frac{T}{4} < \left| t \right| < T/2} \end{array}} \right.\)

The Various Fourier series coefficients

\({a_0} = \frac{1}{T}\mathop \smallint \limits_{ - {T_1}}^{{T_1}} dT = \frac{{2T}}{T}\)

\(= \frac{2}{T}\frac{T}{4} = \frac{1}{2}\)

\({a_k} = \frac{1}{T}\mathop \smallint \limits_{ - T/4}^{ + T/4} {e^{ - jk{\omega _0}t}}dt = \frac{{ - 1}}{{{j^{k{\omega _o}T}}}}\left. {{e^{ - jk{\omega _o}T}}} \right|_{ - T/4}^{T/4}\)

\({a_k} = \frac{2}{{k{\omega _0}T}}\left[ {\frac{{{e^{ - j\frac{{k{\omega _0}T}}{4}}} - {e^{ - j\frac{{k{\omega _0}T}}{4}}}}}{{2j}}} \right]\)

\({a_k} = \frac{{sin\left( {\pi ;K/2} \right)}}{{\pi k}}\)

ak = 0 for k = even

\(\left. {\begin{array}{*{20}{c}} {{a_1} = {a_{ - 1}} = \frac{1}{\pi }}\ {{a_3} = a - 3 = \frac{{ - 1}}{{3\pi }}} \end{array}} \right]Alternate;Positive;and;Negative;\)

Hence, The square wave, we obverse that the Fourier series coefficients exist and the reconstruction converges at the most points.

Concept:

The frequency of rotor current or rotor induced EMF in an induction motor is,

                   f_r = sf

where,

          s is slip 

          f is supply frequency

Explanation:

For the given induction motor, we have \(P; = ;8\), \(f; = ;50Hz\), \(N; = ;700\) rpm

So, we obtain \(Ns = \frac{{120f}}{P} = \frac{{120 \times 50}}{8} = 750rpm\).

Therefore,

                 the slip \(s = \frac{{{N_s} - {N_r}}}{{{N_s}}} = \frac{{750 - 700}}{{750}} = \frac{1}{{15}}\)

Thus, the frequency of rotor current of the motor is obtained as

Frequency of rotor current = s × f \(= \frac{1}{{15}} \times 50 = 3.33Hz\)

Concept:

EMF equation of Transformer is given by \(E = \sqrt 2 \pi fϕ N\)

Where ϕ  is the flux , given as \(ϕ = B \times A\)

Explanation:

we know that, 

An induced EMF in a transformer can be expressed as,

E = 4.44 fNϕ

\(∵ \frac{E}{f} = \rm constant\)

⇒ Nϕ = constant

⇒ ϕ ∝ \(\frac{1}{N}\) ....eq (1)

Now, 

flux (ϕ) = \(\rm \frac{MMF}{Reluctance} = \frac{NI_m}{R} = \frac{NI_m}{l/\mu A}\)

⇒ \(\rm \phi = \frac{NI_m \mu A}{l}\)

From eq (1)

\(\rm ⇒ \frac{\mu I_m NA}{l} \propto \frac{1}{N}\)

⇒ \(\rm \frac{\mu N^2 AI_m}{l}\) = constant

⇒ N2ImA = constant

{∵ I0 (No-load current) = constant}

⇒ N2A = Constant

⇒ \(\rm N_1^2 A_1 = N_2^2 A_2\)

\(\Rightarrow \left( \frac{N_2}{N_1} \right)^2 = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{2r_2}{r_2} \right)^2\)

\(\left( \frac{N_2}{N_1} \right)^2 = 4\)

\(\frac{N_2}{N_1} = 2 \) i.e., Number of turns become double.

For the open circuit test, we have line – to – line voltage at I_f = 2.3 A as \({V_{OCL - L}}; = ;400;V\)

So, the open circuit phase voltage is obtained as \({V_{O{C_{Ph}}}} = {\left. {\frac{{400}}{{\sqrt 3 }}V} \right|_{{I_f} = 2.3A}};\)

Again, for short circuit test, we have phase rms current at If = 1.5A as \({I_{S{C_{Ph}}}} = {\left. {10A} \right|_{{I_f} = 1.5A}};\)

Since ISC and If are linearly related, so we get the short circuit phase current at I_f = 2.3 A as \(I_{S{C_{Ph}}}' = {\left. {\frac{{10}}{{1.5}} \times 2.3 = \frac{{46}}{3}A} \right|_{{I_f} = 2.3A}};\)

Thus, per phase synchronous impedance is \({Z_{Ph}} = {\left. {\frac{{{V_{o{c_{Ph}}}}}}{{I_{S{C_{Ph}}}'}}} \right|_{{I_f} = 2.3A}} = \frac{{\frac{{400}}{{\sqrt 3 }}}}{{\frac{{46}}{3}}} = \frac{{400 \times \sqrt 3 }}{{46}} = 15.06{\text{Ω }};\)

The electrical insulating material should have the following properties.

• High insulation resistance i.e. high resistivity.
• High dielectric strength
• Low permittivity
• High mechanical strength
• High thermal conductivity
• Non-hygroscopic i.e., it should not absorb moisture from air or soil
• Non-inflammable
• Unaffected by acids and alkalis
• The electrical and chemical properties of the material should not be affected by the temperature.

When fault takes place at F₁

Both generator 1 and generator 2 are supplying the fault current.

The voltage at bus A is due to generator 2 the angle of the generator is zero so that the voltage at A is negative.

Hence V_F1 lags I_F1 

I_x fault current will be I_x∠90° 

I_F1 = - I_x = (1∠180°) I_x∠-90° 

I_F1 = I_x∠90° 

Therefore V_F1 lags I_F1

When fault takes place at F₂

Both generator 1 and generator 2 are supplying the fault current.

The voltage at bus A is due to generator 1 the angle of the generator is δ and it is positive so that the voltage angle at bus A is also positive

Hence V_F2 leads I_F2

Concept:

Zero sequence networks for different 3-phase loads are shown below.

The zero sequence equivalent circuits of 3-phase transformers can be drawn by using the following general circuit.

Z is the zero-sequence impedance of the windings of the transformer. These are two series and two shunt switches. One series and one shunt switches are for both sides(primary and secondary) separately.

The series switch of a particular side is closed if it is star grounded and the shunt switch is closed if that side is delta connected, otherwise, they are left open.

Application:

Given two bus system,

The zero-sequence network of the corresponding transformers will be

The Routh Stability Criterion is used to test the stability of an LTI system. It is necessary and sufficient that each term of the first column of the Routh Array of the Characteristic Equation is positive for the system to be stable.

Cases of Routh Stability Criterion Breakdown:

• If the first term in any row of Routh Array is zero while rest of the row has at least one non-zero term. Because of this term, the terms in the next row will become infinite.
• When all elements in any row of the Routh array are zero. This condition indicates that there are symmetrical/imaginary roots in the s-plane.

Concept:

The root locus branch is present where the number of (poles+ zeroes) on right is even.

If the number of pole + zeros is odd on the right side then it is inverse root locus.

Calculation:

Given root locus diagram:

Here there is a zero on the right side of s plane i.e. we have an odd number of pole + zero

This implies the root locus plot is inverse root locus 

The inverse root locus is similar to the system having Positive feedback 

For Positive feedback and gain varying from -∞ to 0. 

Transfer Function is given by 

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{G(s)}{{( 1 - G(s)H(s))}}\)......(1)

\(G\left( s \right) = \frac{K}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\)

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{K}{{\left( {s + 1} \right)\left( {s + 2} \right) - K}}\)will give the root locus given the diagram.

Additional Information

Construction rules of Inverse Root locus:

1. Every branch of an Inverse root locus diagram starts at the open-loop zero (K = -∞) and terminates at a pole (K = 0) of the open-loop transfer function.
2. Inverse root locus diagram is symmetrical with respect to the real axis. If the pole-zero plot is symmetrical with respect to any point on the real axis, then the IRLD also symmetrical with respect to that point, provided that the point does not have any poles or zeros.
3. Number of branches of the inverse root locus diagram are:

N = P if P ≥ Z

= Z, if P ≤ Z 

⇒ Number of branches of the root locus diagram is equal to the order of the system.

Where P & Z are the number of finite poles and zeros of G(s)H(s)

4. Number of asymptotes in an inverse root locus diagram = |P – Z| 5. Centroid: It is the intersection of the asymptotes and always lies on the real axis i.e., the centroid is real. It is denoted by σ.

Centroid may or may not be the part of Inverse Root locus diagram.

\(\sigma = \frac{{\sum {P_i} - \sum {Z_i}}}{{\left| {P - Z} \right|}}\)

ΣPi is the sum of real parts of finite poles of G(s)H(s)

ΣZi is the sum of real parts of finite zeros of G(s)H(s)

6. Angle of asymptotes: \({\theta _l} = \frac{{\left( {2l } \right)\pi }}{{P - Z}}\)

l = 0, 1, 2, … |P – Z| – 1

7. A segment of the real axis exists on the Inverse root locus diagram if the sum of a total number of poles and zeros of G(s)H(s) is even. 8. Break-in/away points: These exist when there are multiple roots on the inverse root locus diagram.

At the breakpoints gain K is either maximum and/or minimum.

So, the roots of\(\frac{{dK}}{{ds}}\) are the breakpoints.

Concept:

In a two-wattmeter method, for lagging load

The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit is:

\(Q=\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)\)

Power factor = cos ϕ

\(\phi =\text{ta}{{\text{n}}^{-1}}\left( \frac{\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)}{{{W}_{1}}+{{W}_{2}}} \right)\)

Calculation:

Given:

Let W1 = W

W2 = 2 W

W1 – W2 = - W

W1 + W2 = 3 W

\(tan;\phi=\dfrac{\sqrt{3}\times W}{3W}=\dfrac{1}{\sqrt 3}\)

ϕ = 30° or π/6

Power factor = cos 30  = 0.866

Additional Information

In a two watt-meter method used to measure three phase power. Readings of both watt-meters are

\({W_1} = \sqrt 3 {V_p}{I_p}\cos \left( {30 - \phi } \right)\)

\({W_2} = \sqrt 3 {V_p}{I_p}\cos \left( {30 + \phi } \right)\)

In a two watt-meter method, Power factor can be calculated as

\(\cos \phi = \cos \left[ {{{\tan }^{ - 1}}\sqrt 3 \left( {\frac{{{W_1} - {W_2}}}{{{W_1} + {W_2}}}} \right)} \right]\)

In a two watt-meter method, total power can be calculated as

P = W1 + W2

Where, Vp = Voltage across pressure coil

Ip = Current through the current coil

ϕ = Phase angle

Case 1^st

• For ϕ = 0°,
• cos ϕ = p.f = 1
• \({W_1} = \sqrt 3 {V_p}{I_p}\cos \left( {30} \right) = \frac{3}{2}{V_p}{I_p}\)
• \({W_2} = \sqrt 3 {V_p}{Z_p}\cos \left( {30} \right) = \frac{3}{2}{V_p}{I_p}\)
• ∴ W1 = W2
• Hence two watt-meters used to measure three phase power reads equal reading at unity power factor.

Case 2^nd

• For ϕ = 30°,
• cos ϕ = p.f. = 0.867
• \({W_1} = \sqrt 3 {V_p}{I_p}\cos \left( 0 \right) = \sqrt 3 {V_p}{I_p}\)
• \({W_2} = \sqrt 3 {V_p}{I_p}\cos \left( {60^\circ } \right) = \frac{{\sqrt 3 }}{2}{V_p}{I_p}\)
• ∴ W1 = 2 W2
• Reading of one watt-meter is half of other

Case 3^rd

• For ϕ = 60°,
• cos ϕ = p.f. = 0.5
• \({W_1} = \sqrt 3 {V_p}{I_p}\cos \left( { - 30^\circ } \right) = \frac{{\sqrt 3 }}{2}{V_p}{I_p}\)
• \({W_2} = \sqrt 3 {V_p}{I_p}\cos \left( {90^\circ } \right) = 0\)
• One of the wattmeters have zero reading and the other reads complete power​

Note:

• For angle more than 60° or cosϕ < 0.5, wattmeter will give negative reading.
• To take negative reading either potential coil or current coil connection should be reversed.

Case 4^th

• For ϕ = 90°,
• cos ϕ = p.f. = 0
• \({W_1} = \sqrt 3 {V_p}{I_p}\cos \left( {30^\circ } \right) = \frac{{\sqrt 3 }}{2}{V_p}{I_p}\)
• \({W_2} = \sqrt 3 {V_p}{I_p}\cos \left( {120^\circ } \right) = \frac{{ - \sqrt 3 }}{2}{V_p}{I_p}\)
• ∴ W1 = -W2  (zero p.f.)
• One of the wattmeter have negative reading

In an oscilloscope screen, a Linear sweep signal is applied at the horizontal axis.

Additional Information

The CRO stands for a cathode ray oscilloscope.it consists of a set of blocks which are, shown in the following block diagram.

Subparts of Cathode Ray Tube are

• Electron gun: It emits electrons and forms them into a beam with the help of a heater, cathode, grid, and pre-accelerating, accelerating, and focusing anode.
• Vertical deflection plates: VDP is applied with a test signal. Whose waveform has to be observed on the screen VDP is kept in a horizontal position.
• Horizontal deflection plates:  HDP is used for shifting the electron beam horizontally i.e. horizontal time scale is adjusted by changing HDP potential manually.
• Screen: A standard screen size is 8 cm by 10 cm. The screen is coated with a phosphor that emits light when struck by the electron beam

Hence, the Time-base generator is a part of CRO but not a part of CRT.

Uses of CRO:

• Used to measure the voltage, current, frequency, inductance, admittance, resistance, and power factor.
• used to monitor the signal properties as well as characteristics and also
• Used to control the analog signals.

Important Points

• Vertical deflection plates are mounted horizontally.
• Horizontal plates are mounted vertically as shown in the block diagram.

Concept:

Overall modulus of a cascaded system is the product of their individual modules.

For example:

A counter of modulus p, modulus q, and modulus r is cascaded, then the modulus of the overall counter is p × q × r

Calculation:

Given, 

The modulus of all three counters is 5

Then the overall modulus = 5 × 5 × 5 = 125

To determine the balancing condition, we reconstruct the wein bridge as

For balance, we have

\(Z_1Z_4=Z_2Z_3\)

\(\begin{array}{l} \left( {{R_1} - \frac{J}{{\omega {C_1}}}} \right){R_4} = \left( {\frac{{{R_2}}}{{1 + J\omega {C_2}{R_2}}}} \right){R_3}\ or\ \frac{{{R_3}}}{{{R_4}}} = \frac{{{R_1}}}{{{R_2}}} + \frac{{{C_2}}}{{{C_1}}} + J\left( {\omega {C_2}{R_1} - \frac{1}{{\omega {C_1}{R_3}}}} \right) \end{array}\)

Comparing real and imaginary parts, we get

\(\frac{{{R_3}}}{{{R_4}}} = \frac{{{R_1}}}{{{R_2}}} + \frac{{{C_2}}}{{{C_1}}}\)

And \(\omega {C_2}{R_1} - \frac{1}{{\omega {C_2}{R_1}}} = 0\)

\(\Rightarrow \omega = \frac{1}{{\sqrt {{R_1}{C_1}{R_2}{C_2}} }}\)

Concept:

AC analysis:

• Find AC equivalent circuit by replacing all capacitors by short circuits, inductors by open circuits, dc voltage sources by ground connections and dc current sources by open circuits
• Replace transistor by its small-signal model
• Use small-signal ac equivalent to analyze ac characteristics of amplifier
• Combine end results of dc and ac analysis to yield total voltages and currents in the network

Explanation:

To draw an A.C equivalent circuit:

• All Capacitor are short circuited.
• All D.C voltage sources are replaced by short circuit.

Internal voltage gain \(\rm \left( \frac{V_0}{V_i} \right) = \frac{-h_{fe}R_L}{h_{ie}}\)     ---(1)

Overall voltage gain \(\rm \left( \frac{V_0}{V_i} \right) = \frac{V_0}{V_{i_1}} \times \frac{V_{i_1}}{V_i}\)     ---(2)

using voltage Division Rule:

\(\rm V_{i_1} = \frac{h_{ie}}{R_B + h_{ie}} \times V_i\)

\(\rm \frac{V_{i_1}}{V_i} = \frac{h_{ie}}{R_B + h_{ie}}\)     ---(3)

From equation (1), (2) and (3)

\(\rm \frac{V_0}{V_i} = A_V = - \frac{h_{fe} R_L}{h_{ie}} \times \frac{h_{ie}}{R_B + h_{ie}} = \frac{-h_{fe}{R_L}}{R_B + h_{ie}}\)

here, ∵ h_ie is not given,

so,

R_B + h_ie ≈  R_B

Therefore,

\(\frac{V_0}{V_i} = A_V ≈ \frac{-h_{fe}R_L}{R_B} = \frac{-100 \times 10}{10} = -100\)

|A_V| = 100

Concept:

RMS value of load current = RMS value of source current

Explanation:

RMS value of load current (I_or) = Average value of load current (I₀) = 10 A

During positive half cycle:

                           D₁, D₂ : ON  

                             i_s = I₀

During Negative half cycle:

                          D₃, D₄ : ON

                             i_s = -I₀

Average value of source current = 0 A

RMS value of source current = 10 A

Equivalent circuit diagram is given below,

RMS output current,

\(\rm \Rightarrow I_{or} = \frac{V{or}}{R} = \frac{V_m/\sqrt 2}{R} = \frac{\frac{230\sqrt 2}{\sqrt 2}}{R}\)

\(\Rightarrow \rm I_{or} = \frac{230}{R}\)

\(\Rightarrow 10 = \frac{230}{R}\)

∴ R = 23 Ω

Concept:

Thyristor can block both forward and reverse voltage in  off state.

Explanation:

Switch (i):

when V_a > V_b

     Diode : OFF, Transistor : ON

when V_a < V_b

             Diode : ON, transistor : OFF

Therefore, switch (i) can not block voltages of either polarity when the diode or transistor is in the OFF state.

Switch (ii):

when Va > Vb

           Diode : ON, Transistor : ON

when Va < Vb

           Diode : OFF, transistor : OFF

Therefore, the switch (ii) can block voltages of either polarity when the diode & transistor are in OFF condition.

Switch (iii):

when Va > Vb

              SCR : ON, 

when Va < Vb

              SCR : OFF

Therefore, switch can block voltages of either polarity when the SCR in OFF state.

Switch (iv)

when Va > Vb

             Diode : OFF

when Va < Vb

             SCR : OFF

Therefore, switch (iv) can Not block voltages of either polarity when SCR or diode is in OFF state.

Concept:

Greatest Integer Function: It is also known as floor function.

It is written as f(x) = [x]

The value of [x] is the largest integer that is less than or equal to x.

In Mathematical notation,

[x] = max{m ∈ z | m ≤ x}

Here notation "m ∈ z" means 'm' is an integer.

Fractional function:

x - [x] = {x} ⇒ fractional part of a integer

It's value always between 0 & 1 & {x} can also be zero.

0 ≤ {x} < 1

Also the constant term in Fourier analysis is defined as:

\(\rm a_0 = \frac{1}{T} \displaystyle\int _0^T g(x) dx\)

Calculation:

g(x) = x - [x] = {x}

It is periodic with a period (T) = 1

As we know,

∵ \(\rm a_0 = \frac{1}{T} \displaystyle\int _0^T g(x) dx\)

\(\rm a_0 = \frac{1}{1} \displaystyle\int _0^1 x dx\) (∵ x - [x] = x in (0, 1)

\(a_0 = \left. \left[ \frac{x^2}{2} \right] \right|_0^1\)

\(a_0 = \frac{1}{2}\)

∴ Constant term = 1/2

For ‘n’ number of coin tosses

for \(n\) head, 0 tails; \(n - 1\) heads ,1  tails; \(n-2\) heads, 2 tails; \(n-3\) heads,3  tails; …………..; 1 heads, \(n - 1\) tails; 0 heads, \(n\) tails.

The difference between the number of heads and tails is \(n,;n - 2,;n - 4,; \ldots ..,;2,;0.\)

It is not \(\left( {n - 1} \right)\), \(;\left( {n - 3} \right),;\left( {n - 5} \right),; \ldots \ldots ;,\) etc.

∴ The probability of difference being ‘n – 3’ = 0.

Concept:

Green's theorem:

\(\oint \left( {{\rm{Mdx}} + {\rm{Ndy}}} \right) = \int!!!\int \left( {\frac{{\partial {\rm{N}}}}{{\partial {\rm{x}}}} - \frac{{\partial {\rm{M}}}}{{\partial {\rm{y}}}}} \right){\rm{dxdy}}\)

Calculation:

\(\begin{array}{l} F = yz\hat i\ \mathop \smallint \limits_c^{} F.dr = \mathop \smallint \limits_c^{} \left( {yz\hat i} \right)\left( {\hat idx + \hat jdy + \hat kdz} \right)\ = \mathop \smallint \limits_c^{} yzdx\ = \mathop \smallint \limits_c^{} ydx;\left( {\therefore z = 1;given} \right)\ = \mathop \smallint \limits_c^{} \left( {ydx - ody} \right) \end{array}\)

From Green’s Theorem

\(\begin{array}{l} \mathop \oint \limits_c^{} Pdx + Qdy = \mathop \int!!!\int \limits_R^{} \left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y;;}}} \right)dx;dy\ \therefore \mathop \smallint \limits_c^{} ydx - 0dy = \mathop \int!!!\int \limits_{{x^2} + {y^2} = 1}^{} \left( {0 - 1} \right)dx;dy\ \left( {P{\rm{ }} = {\rm{ }}y,{\rm{ }}Q{\rm{ }} = {\rm{ }}0} \right)\ = \mathop \int!!!\int \limits_{{x^2} + {y^2} = 1}^{} - dxdy = - \mathop \int!!!\int \limits_{{x^2} + {y^2} = 1}^{} dxdy \end{array}\)

= - Area of circle (x² + y² = 1)

= - π × (1)²

= - π

\(\mathop \smallint \limits_c^; \vec F.d\vec r = - \pi\)

NOTE - 

Stoke's theorem:

\(\mathop \oint \limits_C^; \vec A.d\vec l = \mathop \int!!!\int \limits_S^; \left( {\vec \nabla \times \vec A} \right).d\vec s = ;\mathop \int!!!\int \limits_S^; \vec V.d\vec s\)

Gauss Divergence theorem:

\(\oint A.ds=\iiint{\left( \nabla .A \right)dv}\)

\(\oint \overrightarrow{F.}\hat{n}ds=\iiint{\left( \nabla .F \right)dv}\)

CONCEPT:

• The property of any conductor that opposes the flow of electric current through it is called resistance. It is denoted by R.
• The resistance of any conductor changes by changing the temperature of the conductor.

The relationship between resistance and temperature is given by:

R = R0 (1 + α Δ T)

Where R0 is resistance at a reference temperature, α is the temperature-coefficient of the conductor, and Δ T is the change in temperature.

Working resistance:

Generally, the conductor has the highest resistance when it is working at rated voltage and rated power rating.

As temperature increases due to the heating effect of current, resistance increases.

Power rating \({{P}_{rated}}=~\frac{V_{rated}^{2}}{R}\)

Calculation:

Given, 

Output power P = 40 W

Rated voltage of lamp = 240 V

\({{P}_{rated}}=~\frac{V_{rated}^{2}}{R}\)

\({R_{lamp}} = \frac{{{V^2}}}{P} = \frac{{{{240}^2}}}{{40}} = 1440Ω\)

The working resistance of the lamp is R = 1440 Ω 

At temperature t = 26°, resistance R₀ is 120 Ω, temperature co-efficient = 4.5 × 10^-3/°C

From the formula R = R0 (1 + α Δ T)

⇒ 1440 = 120 (1 + 4.5 × 10-3 (t₂ - 26°C))

⇒ 12 = 1 + 4.5 × 10-3 (t2 - 26°C)

⇒ 11 = 4.5 × 10-3 (t2 - 26°C)

⇒ 2444.45°C = t2 - 26°C

⇒ t₂ = 2470.45°C

Therefore, incandescent lamp ‘ON’ state filament temperature in °C is approximately 2470.45°C

Given that, V = 300 volts

\(R = \left( {25 + \frac{I}{2}} \right)\)                 ........(1)

According to Ohm's law

V = IR                            .........(2)

From equation (1) & (2), we get

\(300 = I\left[ {25 + \frac{I}{2}} \right]\)

⇒ I² + 50 I – 600 = 0

⇒ I = 10 A or I = -60 A

I = 10 A is correct based on the given direction

Concept:

Fortescue’s Theorem:

A unbalance set of ‘n’ phasors may be resolved into (n - 1) balance n-phase system of different phase sequence and one zero phase sequence system.

A zero-phase sequence system is one in which all phasors are of equal magnitude and angle.

Considered three phasors are represented by a, b, c in such a way that their phase sequence is (a b c).

The positive phase sequence will be (a b c) and the negative phase sequence will be (a c b).

Assumed that subscript 0, 1, 2 refer to zero sequences, positive sequence, negative sequence respectively.

Current Ia, Ib, Ic represented an unbalance set of current phasor as shown,

Each of the original unbalance phasor is the sum of its component and it can be written as,

Ia = Ia0 + Ia1 + Ia2

Ib = Ib0 + Ib1 + Ib2

Ic = Ic0 + Ic1 + Ic2

For a balance position phase sequence (a b c) we can write the following relation,

Ia0 = Ib0 = Ic0

Ib1 = α2 Ia1

Ic1 = α Ia1

Ib2 = α Ia2

Ic2 =α2 Ia2

From the above equation Ia, Ib, Ic can be written in terms of phase sequence component,

Ia = Ia0 + Ia1 + Ia2

Ib = Ib0 +  α2 Ia1 + α Ia2

Ic = Ia0 + α Ia1 + α2 Ia2

The above equation can be written in form of Matrix as shown,

\(\left[ {\begin{array}{*{20}{c}} {{I_a}}\ {{I_b}}\ {{I_c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&1&1\ 1&{{α ^2}}&α \ 1&α &{{α ^2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{I_{a0}}}\ {{I_{a1}}}\ {{I_{a2}}} \end{array}} \right]\)

Calculation:

I_a = 1∠-90° P.u

I_b2 = 4∠-150° P.u

I_c0 = 3∠90° P.u

I_a = Ia1 + Ia2 + Ia0 

I_b2 = α I_a2

⇒ 4∠-150° = (1∠120°) I_a2

⇒ I_a2 = 4∠-270° 

I_a0 = Ib0 = Ic0 = 3∠90° 

I_a1 = 8∠90° 

I_b1 = α² I_a1 = 8∠150° 

I_b = Ib0 + Ib1 + Ib2 

= 3∠90°  + 8∠150° + 4∠-150° 

= 11.53∠154.4°  

Therefore the magnitude of phase current Ib in pu is 11.53∠154.4°

Concept:

The magnetic field forms a closed loop, i.e. the amount of field leaving a point equals the amount entering. i.e. Magnetic monopoles do not exist.

Since the divergence of a field gives the net outflow of a field and is calculated as \(∇ .\vec F\)

\(So,;∇ .B = 0\)

B = Magnetic flux density

Since B is related to H (magnetic field intensity) as:

B = μH

∇.H = 0 as well.

Calculation:

For magnetic fields ∇⋅B = 0 by verification

a)  ∇⋅B = 0

b)  ∇⋅B = 0

c)  ∇⋅B = 2x + 2y + 2z ≠ 0

Maxwell’s Equations in Differential Form are as shown:

\(∇ \times E = - \frac{{\partial B}}{{\partial t}}\)

\(∇ \times H = J + \frac{{\partial D}}{{\partial t}}\)

\(∇ .D = {\rho _v}\)

\(∇ .B = 0\)

Concept:

unit step signal:

\(\rm u(t) = \left { \begin{matrix} 1; & t>0 \\ 0; & \rm otherwise \end{matrix} \right.\)

ramp signal:

\(\rm r(t) = \left { \begin{matrix} t; & t\ge 0 \\ 0; & \rm otherwise \end{matrix} \right.\)

• Magnitude of step signal represent magnitude
• Magnitude of ramp signal represent slope

Explanation:

⇒ X(t) = u(t) - u(t - T) + \(\frac{1}{T}\) r(t - T) - \(\frac{1}{T}\)r(t - 2T)                      {∵ r(t) = tu(t)}

⇒ x(t) = u(t) - u(t - T) + \(\frac{1}{T}\)(t - T) u(t - T) - \(\frac{1}{T}\)(t - 2T) u(t - 2T)

Alternative method:

Concept:

Basic signal can be expressed as step signal as   (final value-initial value) u(t - critical point)

ramp signal as (final slope value - initial slope value) r(t - critical point) 

so, 

          x(t) = (1 - 0)u(t) + (0 - 1) u(t - T) + (1 - 1) u(t - 2T)+ (0 - 0) r(t) + \(\left( \frac{1}{T} - 0 \right)\)  r(t - T)+ \(\left(0- \frac{1}{T} \right)\) r(t - 2T)

⇒ X(t) = u(t) - u(t - T) + \(\frac{1}{T}\) r(t - T) - \(\frac{1}{T}\)r(t - 2T)

⇒ x(t) = u(t) - u(t - T) + \(\frac{1}{T}\)(t - T) u(t - T) - \(\frac{1}{T}\)(t - 2T) u(t - 2T)

Using long division method

we have:

\(X\left( Z \right) = \frac{1}{{1 - {Z^{ - 3}}}}\)

since a causal sequence given, so numerator & denominator polynomial arrange in descending powers of z.

\(\Rightarrow X(z) = \frac{Z^3}{Z^3 - 1}\)

x(Z) = 1 + Z^-3 + Z^-6 + ....

taking inverse Z-transform,

x[n] = δ[n] + δ[n - 3] + δ[n - 6] + ......

At n = 2,

x[2] = δ[2] + δ[2 - 3] + δ[2 - 6] + ......

x[2] = 0 + 0 + 0 + ....

x[2] = 0

At n = 3,

x[3] = δ[3] + δ[3 - 3] + δ[3 - 6] + ....

x[3] = 0 + 1 + 0 + .....

x[3] = 1

Given the Fourier transform pair,

\(\begin{array}{l} F\left( \omega \right) = \mathop \smallint \limits_{ - \infty }^\infty f\left( t \right){e^{ - j\omega t}}dt\ g\left( t \right) = \mathop \smallint \limits_{ - \infty }^\infty F\left( u \right){e^{ - j\omega t}}du \end{array}\)

The Inverse Fourier transform of F (ω) is given as

\(f\left( t \right) = \frac{1}{2π}\mathop \smallint \limits_{ - \infty }^\infty F\left( \omega \right){e^{j\omega t}}d\omega\)

Replacing t by ‘ –t ‘ and ω by u in the above expression, we have,

\(f\left( { - t} \right) = \frac{1}{2π}\mathop \smallint \limits_{ - \infty }^\infty F\left( u \right){e^{-ju t}}du\)

g(t) =  2π f (-t)

If f (t) is even, then f (-t) = f (t)

∴ g (t) = 2π f (t)

⇒ g (t) will be proportional to f (t), if f (t) is an even function.

Concept:

Hysteresis losses: These are due to the reversal of magnetization in the transformer core whenever it is subjected to alternating nature of magnetizing force.

\({W_h} = \eta B_{max}^xfv\)

\({B_{max}} \propto \frac{V}{f}\)

Where

x is the Steinmetz constant

Bm = maximum flux density

f = frequency of magnetization or supply frequency

v = volume of the core

At a constant V/f ratio, hysteresis losses are directly proportional to the frequency.

Wh ∝ f

Eddy current losses: Eddy current loss in the transformer is I2R loss present in the core due to the production of eddy current.

\({W_e} = K{f^2}B_m^2{t^2}V\)

\({B_{max}} \propto \frac{V}{f}\)

Where,

K - coefficient of eddy current. Its value depends upon the nature of magnetic material

Bm - Maximum value of flux density in Wb/m2

t - Thickness of lamination in meters

f - Frequency of reversal of the magnetic field in Hz

V - Volume of magnetic material in m3

At a constant V/f ratio, eddy current losses are directly proportional to the square of the frequency.

We ∝ f2

Iron losses or core losses or constant losses are the sum of both hysteresis and eddy current losses.

Wi = Wh­ + We

At constant V/f ratio, Wi = Af + Bf2

Explanation:

For V = 230 V and F = 50 Hz, we have core loss = 1050 W   or Iron loss + Hysteresis loss = 1050 W

Since, iron loss and hysteresis loss are given by Iron loss \(= {P_i}; = {k_i};{V^2}\) and Hysteresis loss \(= ;{P_H} = {K_H};{V^{1.6}};{f^{ - 0.6}}\)

Substituting it in equation (i) we get  \({K_i}{\left( {230} \right)^2} + {K_H}\frac{{{{\left( {230} \right)}^{1.6}}}}{{{{\left( {50} \right)}^{0.6}}}} = 1050\) or

\({K_i}\left( {52900} \right) + {K_H}\left( {574.62} \right) = 1050\)

Again for \(V; = ;138;V\) and \(f; = ;30;Hz\), we have

Core loss = 500 W   or Iron loss + Hysteresis loss = 500 W

or \({K_P}{\left( {138} \right)^2} + {K_H}\frac{{{{\left( {138} \right)}^{1.6}}}}{{{{\left( {30} \right)}^{0.6}}}} = 500\) or K_P (19044) + K_H (344.77) = 500

Solving equations (i) and (ii), we get

\({K_P}; = ;0.01027;;;;{K_H}; = ;0.855\)

And, For 230 V, 50Hz we have

Hysteresis loss = \({K_H};{V^{1.6}};{f^{ - 0.6}}\) \(= 0.855 \times {\left( {230} \right)^{1.6}} \times {\left( {50} \right)^{ - 0.6}}; = ;508.53;W\)

Hence, the eddy loss is    Eddy loss = ­\({K_{I{V^2}}} = {\text{}}0.01024 \times 529000 = 541.69;W\)

For the given data, we draw the circuit as

So the current through field resistance is \({I_f} = \frac{V}{{{R_f}}} = \frac{{230}}{{230}} = 1A\)

Hence, we have the armature current \({I_a} = I - {I_f}\) = 5 – 1 = 4A

From the given circuit, we have \(V; = ;{E_{f1}}; + ;{I_a}; \times ;{R_s}\)  Or, \(230; = ;{E_{f1}}; + ;4; \times ;0.4\)  Or, \({E_{f1}}\) = 228.4 V

At full load  \(I; = ;70;A\)

So, \({I_a}; = ;70;-;1; = ;69;A\)  and \({E_{f1}}; = ;230;-;69; \times ;0.4;;; = ;202.4;V\)

as emf \( \propto\) flux \(\times\) speed     ,    for DC shunt motor flux (\(\phi\)) is constant. So,

\({E_f} \propto \omega\)  or \(\frac{{{E_{f1}}}}{{{E_{f2}}}} = \frac{{{\omega _1}}}{{{\omega _2}}}\)  or \(\frac{{228.4}}{{202.4}} = \frac{{1400}}{{{\omega_2}}}\)

Thus, \({\omega ^2} = \frac{{202.4}}{{228.4}} \times 1400 = 1241.1rpm\)

Concept:

For the induction motor, we have \({T_f}\left( {full;load;torque} \right) \propto \frac{{{s_f}{R_2}}}{{R_2^2 + {{\left( {{s_f}{X_2}} \right)}^2}}}\) or \({T_{max}}\left( {\mathop {\max }\ torque} \right) \propto \frac{1}{{2{X_2}}}\)

So, we get,

 \(\frac{{{T_f}}}{{{T_{max}}}} = \frac{{2{s_f}{R_2}{X_2}}}{{R_2^2 + {{\left( {{s_f}{X_2}} \right)}^2}}}\)  or \(\frac{{{T_f}}}{{{T_{max}}}} = \frac{2}{{\frac{{{S_{max}}}}{{{s_f}}} + \frac{{{S_f}}}{{{S_{max}}}}}}\)                       ……..(i)

\(\frac{{{T_f}}}{{{T_{max}}}} = \frac{2s_{max} s_f}{{{{{S_{max}^2}}} + {{{S_f}^2}}{{}}}}\)

Where,

s_f = Full load slip

s_max = (R/X) = Maximum slip

Calculation:

Now, the slip at full load is 5%, i.e   s_f = 0.05  and

 \({s_{max}} = \frac{{{R_2}}}{{{X_2}}} = \frac{{0.1}}{{0.5}} = 0.2\)

Substituting these values in equation (i), we get

 \(\frac{{{T_f}}}{{{T_{max}}}} = \frac{2}{{\frac{{0.2}}{{0.05}} + \frac{{0.05}}{{0.2}}}}\)  or

 \(\frac{{{T_{max}}}}{{{T_f}}} = \frac{{4 + 0.25}}{2}\)

 \(\frac{{{T_{max}}}}{{{T_f}}} = 2.125\) 

Hence, the ratio of maximum to full load torque is 2.125

If all the bulbs glow dark simultaneously, then all the gear and infinite bus are in same phase sequence. But, in the given problem, they are coming is sequence(not simultaneously) \({L_a}\) then \({L_b}\), and then \({L_c}\). So phase sequence of the generator is opposite to infinite bus system and frequency is high (as dark period is for more)

Let I be current supplied by source -1.

‘r’ be resistance / length

400 = (400 r) I + 200 r (I – 200) + (200 r) (I – 300) + (200 r) (I – 500) + 400

0 = 400r I + 200r I – 40000r + 200r I – 60000 + 200r I – 100000r 

⇒ I = 200 A

Current in branch – B is I – 200 = 200 – 200 = 0

For point P, source - 1 supplies OA current source 2 supplies 100 A current.

\(\begin{array}{l}
\left[ {\begin{array}{*{20}{c}}
A&B\
C&D
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&2\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{j0.1}\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{0.1\angle 90^\circ }\
0&1
\end{array}} \right]\
\left[ {\begin{array}{*{20}{c}}
{{V_S}}\
{{I_S}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
A&B\
C&D
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_r}}\
{{I_r}}
\end{array}} \right]
\end{array}\)

V_S = V₁ ∠0, B = 0.1 ∠90

V_r = 1∠δ₂, A=1∠0

V_S = AV_r + BI_r

\({I_r} = \frac{{{V_S}}}{B} - \frac{A}{B}{V_r} = \frac{{{V_1}\angle 0}}{{0.1\angle 90}}\left( {1\angle {\delta _2}} \right)\)

= 10V₁ ∠-90 -10∠δ₂ – 90

\(\begin{array}{l}
I_r^ \star = 10{V_1}\angle 90 - 10\angle 90 - {\delta _2}\
{S_r} = {P_r} + j{Q_r} = {V_r}I_r^ \star = \left( {1\angle {\delta _2}} \right)\left( {10{V_1}\angle 90 - {\delta _2}} \right)
\end{array}\)

= 10 V₁ ∠90 + δ₂ – 10∠90  

= 10 [V₁ cos (90 + δ₂) + jV₁ sin (90 + δ₂)] – j10

= 10 [-V₁ sin δ₂ + jV₁ cos δ₂]-j10

= [-10V₁ sin δ₂] + j[10V₁ cos δ₂ - 10]  

Given S_r = 1 + j0.5

-10V₁ sin δ₂ = +1 ⇒ 10V₁ sin δ₂ = -1

10V₁ cos δ₂ – 10 = 0.5 ⇒ 10V₁ cos δ₂ = 10.5

\( \Rightarrow \tan {\delta _2} = \frac{{ - 1}}{{10.5}}\)

⇒ δ₂ = -5.44

10V₁ sin δ₂ = -1

⇒ V₁ = 1.054

Concept:

The condition for optimum operation and economic dispatch is

\(\frac{{d{F_1}}}{{d{P_1}}} = \frac{{d{F_2}}}{{d{P_2}}} = \frac{{d{F_3}}}{{d{P_3}}} \ldots \ldots \ldots \ldots = \frac{{d{F_n}}}{{d{P_n}}} = \lambda \)

Where,

\(\frac{{dF}}{{dP}}\) is incremental fuel costs in Rs/MWh for a power plant.

The above equation shows that the criterion for the most economical division of load between plants is that all the unit is must operate at the same incremental fuel cost.

Calculation:

Given fuel cost functions of two power plants are

Plant \(P1:{C_1} = 0.05Pg_1^2 + AP{g_1} + B\)

Plant \({P_2}:{C_2} = 0.10Pg_2^2 + 3AP{g_2} + 2B\)

The incremental fuel cost of the two plants will be

For plant 1

\(\frac{{d{C_1}}}{{dP{g_1}}} = 2 \times 0.05{P_{{g_1}}} + A = 100\)

\(\frac{{d{C_1}}}{{dP{g_1}}} = 0.1{P_{{g_1}}} + A = 100\)

For plant 2

\(\frac{{d{C_2}}}{{dP{g_2}}} = 2 \times 0.1{P_{{g_2}}} + 3A = 100\)

\(\frac{{d{C_2}}}{{dP{g_2}}} = 0.2{P_{{g_2}}} + 3A = 100\)

By solving both the incremental fuel cost equations

P_g1 = 800 MW

P_g2 = 200 MW

\(\Rightarrow \frac{{{P_{{g_1}}}}}{{{P_{{g_2}}}}} = \frac{4}{1}\)

Concept:

Plug setting multiplier is the ratio of fault current and the product of plug setting and CT ratio.

Plug setting multiplier \(= \frac{{fault;current}}{{plug;setting \times CT;ratio}}\)

Plug setting = relay setting × secondary CT current

Relay current (I relay) = fault current / CT ratio

Calculation:

\({t_{op}} = \frac{{0.14 \times T.M.S}}{{{{\left( {P.S.M} \right)}^{0.02}} - 1}}\)

For relay B

T.M.S. = 0.1

P.S.M. = 5A

I_f(max) = 2000 A

I_f(min) = 500 A

Maximum load current at C = 100 A

For setting current of 1A, plug setting is 100%.

\(\begin{array}{l} PSM = \frac{{Max\ fault\ current}}{{CT\ ratio \times Current\ Setting}} = \frac{{2000}}{{100 \times 1}} = 20\ {t_{op}} = \frac{{0.14 \times T.M.S.}}{{{{\left( {P.S.M.} \right)}^{0.02}} - 1}} = \frac{{0.14 \times 0.1}}{{{{\left( {20} \right)}^{0.02}} - 1}} = 0.23sec \end{array}\)

Concept:

According to RH stability criteria there are two necessary conditions:

First is there should not be any sign changes in the first column of the tabulation and Second Is there should not be any missing coefficients in the equation 1 + G(s) H(s).

The RH criteria can be solved as following:

\({s^3} - - - {\rm{;;;}}{{\rm{a}}_0} \searrow {\rm{;;;;;}} \swarrow {{\rm{a}}_2}{\rm{;;;;;;}}{{\rm{a}}_4}\)

\({s^2} - - - {\rm{;;;}}{{\rm{a}}_1}{\rm{;}} \nearrow {\rm{;;;}} \nwarrow {{\rm{a}}_3}{\rm{;;;;;;;}}0\)

\({s^1} - - - {\rm{;;;}}\frac{{{{\rm{a}}_1}{{\rm{a}}_2} - {{\rm{a}}_0}{{\rm{a}}_3}}}{{{{\rm{a}}_1}}}{\rm{;}} = {\rm{A;;;;;;;;;;}}0{\rm{;}}\)

\({s^0} - - - {\rm{;;;}}\frac{{{\rm{A}}{{\rm{a}}_3}}}{{\rm{A}}} = {{\rm{a}}_3}\)

Calculation:

The characteristic equation is

\(\eqalign{ & 1 + G\left( s \right) = 0 \cr & 1 + \frac{{\left( {s + α } \right)}}{{{s^3} + \left( {1 + α } \right){s^2} + \left( {α - 1} \right)s + \left( {1 - α } \right)}} = 0 \cr & \Rightarrow {s^3} + \left( {1 + α } \right){s^2} + α s + 1 = 0; \cr} \)

\(\begin{array}{*{20}{c}} {{s^3}}\ {{s^2}}\ {{s^1}}\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} 1&α\ 1+α&{1}\ \frac{{α \left( {1 + α } \right) - 1}}{1+α }&0\ 1&{} \end{array}} \right.\)

By R-H criteria, \(\left( {1 + α } \right)α > 1\)

\(\eqalign{ & \Rightarrow {α ^2} + α -1 > 0 \cr & \Rightarrow α = 0.618;and; - 1.618 \cr}\)

As for stability, all the coefficients should be +ve

For α  = -1.618, the coefficient comes out to be negative.

For α  = +0.618, all the coefficient comes out to be positive.

∴  For a stable system α should be 0.618.

Concept:

Bode plot transfer function is represented in standard time constant form as 

\(T\left( s \right) = \frac{{k\left( {\frac{s}{{{\omega _{{c_1}}}}} + 1} \right) \ldots }}{{\left( {\frac{s}{{{\omega _{{c_2}}}}} + 1} \right)\left( {\frac{s}{{{\omega _{{c_3}}}}} + 1} \right) \ldots }}\)

ωc1, ωc2, … are corner frequencies.

In a Bode magnitude plot,

• For a pole at the origin, the initial slope is -20 dB/decade
• For a zero at the origin, the initial slope is 20 dB/decade
• The slope of magnitude plot changes at each corner frequency
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The final slope of Bode magnitude plot = (Z – P) × 20 dB/decade

Where Z is the number of zeros and P is the number of poles

Calculation:

By given magnitude plot

\(G\left( s \right) = \frac{{K\left[ {1 + \frac{s}{2}} \right]\left[ {1 + \frac{s}{4}} \right]}}{{s\left[ {1 + \frac{s}{8}} \right]\left[ {1 + \frac{s}{{24}}} \right]\left[ {1 + \frac{s}{{26}}} \right]}}\)

By comparing with the given transfer function,

\(G\left( s \right) = \frac{{K\left( {1 + 0.5s} \right)\left( {1 + as} \right)}}{{s\left( {1 + \frac{s}{8}} \right)\left( {1 + bs} \right)\left( {1 + \frac{s}{{36}}} \right)}}\)

\(a = \frac{1}{4};b = \frac{1}{{24}};\)

For finding K:

\(K = {\left( {{\omega _1}} \right)^n}\) ; where n is no. of poles from the given plot

\(K = {\left( 8 \right)^1}\)

\(\Rightarrow K = 8\)

so, \(\frac{a}{{bK}} = \frac{{\frac{1}{4}}}{{\frac{1}{{24}} \cdot 8}} = 0.75\)

Concept:

Eigenvalues are a special set of scalars associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots/characteristic values and are denoted by λ.

Let A be a square matrix of order 'n' and 'λ' be a scalar.

|A - λI| = 0 is called characteristic equation of A.

The roots of the characteristic equation are called eigenvalues of A.

Properties of Eigenvalues:

• The sum of Eigenvalues of a matrix A is equal to the trace of that matrix A
• The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A
• If λ is an eigenvalue of a matrix A, then λn will be an eigenvalue of a matrix An.
• If λ is an eigenvalue of a matrix A, then kλ will be an eigenvalue of a matrix kA where k is a scalar

Calculation:

Characteristic equation is \(\left| {A - \lambda I} \right| = 0\)

\(\Rightarrow \left| {\begin{array}{*{20}{c}} { - \lambda }&1&{ - 1}\ { - 6}&{ - 11 - \lambda }&6\ { - 6}&{ - 11}&{5 - \lambda } \end{array}} \right| = 0\)

\(\Rightarrow {\lambda ^3} + 6{\lambda ^2} + 11\lambda + 6 = 0\)

⇒ λ = -1, -2, -3 are the Eigen values of A.

\(\begin{array}{l} {\lambda _{max}} = - 1;and;{\lambda _{min}} = - 3\ \left| {\frac{{{\lambda _{max}}}}{{{\lambda _{min}}}}} \right| = \left| {\frac{{ - 1}}{{ - 3}}} \right| = \frac{1}{3} \end{array}\)

Important Points

• If λ is an eigenvalue of a non-singular matrix A then |A| / λ is the eigenvalue of adj(A).
• If λ is an eigenvalue of a non-singular matrix A then 1 / λ is the eigenvalue of A-1.
• If λ is an eigenvalue of an orthogonal matrix then 1/λ is also another eigenvalue of the same matrix.

Given,

Impedance in both the parallel branches are zero

Therefore the current from the voltage source will be

\(i\left( t \right) = \frac{{100\sin \left( {ω t} \right)}}{{0.5}}\)

i(t) = 200 sin ωt

The impedance on both the parallel branches are equal, so the same amount of current will be flowing through each branch

⇒ I₁ = I₂ = 100 sin ωt

The voltage at node A is

V_A = (- j1) 100 sin ωt

The voltage at node B is

V_B = (j1) 100 sin ωt

Voltage measured by the voltmeter will be 

V = V_A - V_B = - 200j sin ωt

\({V_{RMS}} = \frac{{{V_m}}}{{\sqrt 2 }} = 141.42\ volts \)

Concept:

PMMC  Instrument:

In PMMC instruments, the deflecting torque is given by

Td = BINA

Where Td = Deflecting torque in N-m

B = Flux density, wb/m2

I = Current in the Coil, amperes

N = number of turns of the coil

A = Effective coil area m2

Dynamometer Ammeter:

The deflection torque of the dynamometer ammeter is given by

\({T_d} = {I^2}\frac{{dM}}{{dθ }}\)

Where,

I = fixed or current coil current

dM = change in mutual inductance

dθ = change in deflection

Calculation:

Both the PMMC and Electrodynammter type ammeters are in series therefore they measure the same amount of current and have the same deflecting torques (same spring constants and deflection in both the instruments)

Deflecting torque in PMMC = Deflecting torque in Electrodynamomter type instrument

\(BINA = I^2 \frac{dm}{dθ}\)

0.2 I (80 x 10^-6) x 100 = I² x 0.5 x 10^-3

I = 3.2 A

Concept: 

Instrumentation amplifier:

\({{V}_{o}}=\left( {{V}_{2}}-{{V}_{1}} \right)\left( 1+\frac{2{{R}_{1}}}{{{R}_{g}}} \right)\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)\)

The output voltage may be positive or negative based on the inputs.

Virtual ground -

The concept of the virtual ground is stated as if anyone of the i/p terminals is grounded physically the other i/p terminal will also be at ground potential even though, it is not grounded physically.

• One key feature of an Op-Amp is the differential input, and when put together in a circuit, this can form a virtual ground.
• The virtual ground concept is helpful for the analysis of Op Amps. This concept makes Op-Amp circuit analysis much easier.

Calculation:

\(\begin{array}{l} \frac{{{V_2} - {V_1}}}{{2R}} + \frac{{{V_2} - {V_{02}}}}{R} = 0\ \Rightarrow {V_{02}} = \frac{{3{V_2} - {V_1}}}{2}\ \frac{{{V_1} - {V_2}}}{{2R}} + \frac{{{V_1} - {V_{01}}}}{R} = 0\ \Rightarrow {V_{01}} = \frac{{3{V_1} - {V_2}}}{2}\ {I_1} = {I_f}\ \frac{{{V_{02}} - \frac{{{V_{01}}}}{2}}}{R} = \frac{{{V_{01}} - {V_0}}}{R}\ \Rightarrow {V_0} = {V_{01}} - {V_{02}}\ = \left( {\frac{{3{V_1} - {V_2}}}{2}} \right) - \left( {\frac{{3{V_2} - {V_1}}}{2}} \right)\ \Rightarrow {V_0} = 2\left( {{V_1} - {V_2}} \right) \end{array}\)

Concept:

3 variable K-maps:

• For a 3-variable Boolean function, there is a possibility of 8 output minterms.
• The general representation of all the minterms using 3-variables is shown below.

Calculation:

\(F = \bar A \bar C + BC\)

From the options:

Option 1:

Output is A̅.B.C

Option 2:

Output is A̅ + B.C

Option 3:

Output is A̅.C̅ + B.C

Option 4:

Output is A̅.C + B.C

The given circuit is LPF (law pass filter)

\(\begin{array}{l}
\frac{{{V_o}}}{{{V_{in}}}} = \frac{1}{{1 + SCR}} = \frac{1}{{1 + \left( {j\omega } \right)\left( {{{10}^3}} \right)\left( {{{10}^{ - 6}}} \right)}}\
= \frac{1}{{1 + j\omega \left( {{{10}^{ - 3}}} \right)}} = \frac{{1000}}{{1000 + j\omega }}\
= \frac{1}{{1 + \frac{s}{{1000}}}}
\end{array}\)

Corner frequency is at \(\omega = 1000\)

Low frequency gain is unity at \(\omega = 0\)

To enable 3L × 8L decoder, there enable lines E₁ (which is connected as an output of AND gate) should be high and E̅₂ and E̅₃ should be active low

Which means \(\frac{{{I_0}}}{{̅ m}}\) should be active low which is indicating that it is memory-mapped \(\frac{I}{0}\) interfacing.

So, the address of the device will be in 16 – bits.

To select output port through decoder 2^nd line the status of A₁₅(I₂) A₁₄(I₁) A₁₃(I₀) = 010 and to enable decoder through E₁ enable line A₁₂ = 1 and A₁₁ = 0 and by default as a starting address lines (A₁₀ ……A₀) should be zero. So, the overall port address is 

\(\begin{array}{*{20}{c}} {{A_{15}}}&{{A_{14}}}&{{A_{13}}}&{{A_{12}}}&{{A_{11}}}&{{A_{10}}}&{{A_9}}&{{A_8}}&{{A_7}}&{{A_6}}&{{A_5}}&{{A_4}}&{{A_3}}&{{A_2}}&{{A_1}}&{{A_0}}\ 0&1&0&1&0&0&0&0&0&0&0&0&0&0&0&0 \end{array} = {\left( {5000} \right)_H}\)

Concept:

• The freewheeling diode prevents the load voltage from becoming negative. Whenever load voltage tends to go negative, FD comes into play.
• As a result, the load current is transferred from the main thyristor to FD, allowing the thyristor to regain its forward blocking capability.
• A freewheeling diode placed across the inductive load will provide a path for the release of energy stored in the inductor while the load voltage drops to zero.

Explanation:

Given information,

                             V_o=70 V

                          RL Load:  R =10 Ω, L=0.05H

The diode present in parallel to the load acts as a freewheeling diode. Therefore, the output voltage will not be negative as shown below.

Average load voltage is given as,

\(\begin{array}{l} {V_o} = \frac{{{V_m}}}{{2\pi }}\left( {1 + cos\alpha } \right)\ 70 = \frac{{325}}{{2\pi }}\left( {1 + Cos\alpha } \right)\ \Rightarrow \alpha = 69.3^\circ \end{array}\)

Concept:

The circuit diagram of the buck converter is shown below.

The waveforms of current flows through the resistor, inductor, and capacitor are shown below.

The waveforms of ripple current and voltage are shown below.

Calculation:

For buck converter, peak to peak ripple current is given by,

\(\rm Δ I = \frac{V_s D(1 - D)}{fL}\)

where,

\(D = \frac{T_{ON}}{T} = \frac{0.05}{0.10} = 0.5 \) (∵ T = T_ON + T_OFF)

V_s = 100 V

\(\rm f = \frac{1}{T} = \frac{1}{0.1 × 10^{-3}}\)

L = 1 × 10^-3

\(Δ I = \frac{100 \times 0.5 \times 0.58 \times 0.1 \times 10^{-3}}{1 \times 10^{-3}}\)

Δ I = 2.5 A

Concept:

The Fourier Series representation of a wave function x_T(t) is

\({x_T}(t) = {a_0} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos \left( {n{\omega _0}t} \right) + {b_n}\sin \left( {n{\omega _0}t} \right)} \right)}\)

Where for a time period of (-T/2 to T/2)

\({a_0} = {2 \over T}\int\limits_{{-{T} \over 2}}^{{T \over 2}} {{x_T}\left( t \right)dt}\)

\({a_n} = {2 \over T}\int\limits_{{-{T} \over 2}}^{{T \over 2}} {{x_T}\left( t \right)\cos \left( {n{\omega _0}t} \right)dt}\)

\({b_n} = {2 \over T}\int\limits_{{-{T} \over 2}}^{{T \over 2}} {{x_T}\left( t \right)\sin \left( {n{\omega _0}t} \right)dt}\)

Even symmetrical wave: a₀ ≠ 0, a_n ≠ 0, b_n = 0

Odd symmetrical wave: a₀ = 0, a_n = 0, b_n ≠ 0

Half wave symmetry: 

a₀ = 0, an = 0 for n = even n and bn = 0 for all n.

an ≠ 0 and bn ≠ 0  for n = odd.

Quarter-Wave Symmetry:

If the periodic function were to be made even, then

a₀ = 0, bn = 0 for all n and an = 0 for n = even

If a quarter-wave symmetric periodic function is made odd,

a₀ = 0, an = 0 for all n and bn = 0 for n = even

Calculation:

Given output voltage waveform have odd quarter wave symmetry

⇒ a₀ = a_n = 0, b_n ≠ 0

\(b_{n}=\frac{1}{\pi}\int_{0}^{2\pi}f(t)\sin n\omega_{0}tdt\)

\(b_{n}=\frac{2}{\pi}\int_{0}^{\pi}f(t)\sin n\omega_{0}tdt\)

For fundamental n =1

\(\begin{array}{l} {b_1} = \frac{2}{\pi }\left[ {\int_0^\alpha {{V_{dc}}\sin \omega t} dt - \int_\alpha ^{\pi - \alpha } {{V_{dc}}\sin \omega t} dt + \int_{\pi - \alpha }^\pi {{V_{dc}}\sin \omega t} dt} \right]\ {b_1} = \frac{{2{V_{dc}}}}{\pi }\left[ {\left| {\cos \omega t} \right|_0^{^\alpha } + \left| {\cos \omega t} \right|_\alpha ^{\pi - \alpha } - \left| {\cos \omega t} \right|_{\pi - \alpha }^\pi } \right]\ {b_1} = \frac{{2{V_{dc}}}}{\pi }\left[ {2 - 4\cos \alpha } \right]\ {b_1} = \frac{{4{V_{dc}}}}{\pi }\left[ {1 - 2\cos \alpha } \right] \end{array}\)

RMS value of \({V_{01}} = \frac{{4{V_{dc}}}}{{\pi \sqrt 2 }}\left( {1 - 2cos\alpha } \right)\)

\(⇒ 50 = \frac{{400}}{{\pi \sqrt 2 }}\left( {1 - 2cos\alpha } \right) ⇒ \alpha = 77.15^\circ \)