UPPSC AE Electrical 13 Dec 2020 Official Paper I (Previous Year Paper)

दिए गए विकल्पों में से ‘तेल’ शब्द तद्भव शब्द है। अत: विकल्प 2 ‘तेल’ सही उत्तर है। अन्य विकल्प असंगत हैं।

स्पष्टीकरण:

‘तेल’ का अर्थ है – वनस्पतियों आदि से निकलनेवाला स्निग्ध तरल पदार्थ। तैल तत्सम शब्द है। जिसका तद्भव शब्द तेल है।

Key Pointsअन्य शब्द:

Additional Information

विशेष

उपरोक्त विकल्पों में से 'ज्योतिषी', 'सारंग' का कोई अर्थ नही है, दोनों के मध्य कोई सम्बन्ध नही है। अत: सही विकल्प 4 'ज्योतिषी' है।

• सारंग का अर्थ:- सुहावना, सुंदर, मन भावन होता है।
• भौंरा, कामदेव, तलवार, सभी सारंग के अनेकार्थक शब्द है।

सारंग के एक से अधिक अर्थ – हाथी, कोयल, कामदेव, सिंह, धनुष भौंरा, मृग, मयूर, स्त्री, नानावर्ण, सुन्दर, सरस, बादल, वृक्ष, छाता, वस्त्र, बाल, शंख, शिव, कपूर, चन्दन, आभूषण, स्वर्ण मधुमक्खी, कमल, तलवार।

Confusion Points

• अनेकार्थक शब्द में एक ही शब्द के भिन्न भिन्न अर्थ होते है।
• जैसे:- कनक का मतलब 'सोना (स्वर्ण)' , गेंहू, और धतूरा तीनो होता है।

Additional Information

‘अंदर-अंदर कड़ाही में गुड पगना’ मुहावरे का सही अर्थ है – गुप्त मंत्रणा होना। अन्य विकल्प असंगत हैं। 

Key Points

स्पष्टीकरण:-

मुहावरा का शाब्दिक अर्थ ‘अभ्यास’ है। 

मुहावरा शब्द अरबी भाषा का शब्द है।

हिन्दी में ऐसे वाक्यांशों को मुहावरा कहा जाता है, 

जो अपने साधारण अर्थ को छोडकर विशेष 

अर्थ को व्यक्त करते हैं।

Additional Informatio

अन्य विकल्प:-

दिए गए विकल्पों में 'सहस्र' शब्द की वर्तनी शुद्ध है।

सहस्र का अर्थ - हजार होता है।

Key Points अन्य प्रमुख शुद्ध वर्तनी:

Additional Information 

"मरणासन्न – अव्ययीभाव समास" यह समास - युग्म बिलकुल गलत है।  अत: प्रश्नानुसार सही विकल्प 4 'मरणासन्न – अव्ययीभाव समास' है।

• मरणासन्न का समास विग्रह :- जो मृत्यु के आसन पर हो अर्थात जो मृत्यु के निकट हो
• अत: यहाँ उत्तर पद 'आसन्न' प्रधान पद है, इसलिए यह तत्पुरुष समास होगा।
• जबकि "जिस समास में पहला पद प्रधान हो और समस्त शब्द अव्यय का काम करे।" वह 'अव्ययी भाव समास' होता है।

Key Points

• सतसई – द्विगु समास ---> "सात का समाहार अथवा समूह"
• तुलसीकृत – तत्पुरुष समास ---> "तुलसी द्वारा कृत अथवा तुलसी द्वारा लिखा"
• मंदोदरी – बहुव्रीहि समास  ---> "मंद उदर हो जिसका अर्थात मन्दोदरी"

Additional Information

अलग होने के अर्थ में 'से' कारक-चिह्न अपादान कारक चिन्ह का प्रयोग होता है।  

Key Points 

Additional Information 

• कारक के आठ भेद होते हैं- कर्म कारक, कर्ता कारक, सम्बन्ध कारक, अधिकरण कारक, संबोधन कारक, आपदान कारक, सम्प्रदान कारक, करण कारक।
• जो शब्द क्रिया के साथ सम्बन्ध प्रकट करते हैं वे कारक कहलाते अथवा कारक चिन्हों को परसर्ग या विभक्ति कहते हैं।

‘पवन’ शब्द का सन्धि-विच्छेद 'पो + अन' होगा। अतः सही विकल्प 'पो + अन' है।

Key Points

• 'पवन' में 'अयादि स्वर संधि' है।
• 'पवन' का विच्छेद करने पर 'पो + अन = पवन' होगा और नियम 'ओ + अ = अव्' है।

Additional Information

इसका सही उत्तर विकल्प 1 है। अन्य विकल्प असंगत हैं।

• उपरोक्त विकल्पों में से शुद्ध वर्तनी- निर्झरिणी है।
• अन्य विकल्प वर्तनीगत अशुद्ध हैं।
• निर्झरिणी शब्द का अर्थ नदी होता है।

Additional Information

नयन शब्द में उपसर्ग का प्रयोग नही हुआ है। 

नयन शब्द में 'नी' मूल शब्द और अन् प्रत्यय लगा हुआ है, अतः यहाँ पर उपसर्ग का प्रयोग नही हुआ है। 

अन्य विकल्प:

• सहज - इसमें सह उपसर्ग है। - सह + ज
• अनुभव - इसमें 'अनु' उपसर्ग है। अनु + भव
• संचार - इसमें सम् उपसर्ग है। सम् + चार

Additional Information 

दिए गए विकल्पों में सही उत्तर विकल्प 2 'अघ' होगा। अन्य विकल्प इसके अनुचित उत्तर हैं। 

Key Points

• 'अनघ' शब्द का उचित विलोम शब्द 'अघ' है।
• अघ का अर्थ जिसने अघ या पाप किया हो।
• अनघ का अर्थ जिसने पाप नही किया हो।

अन्य विकल्प: 

Additional Information

'पिशुन' शब्द पक्षी का पर्यायवाची नहीं है। अन्य विकल्प 'पक्षी' के पर्यायवाची शब्द हैं।

Key Points 

• 'पिशुन' के अन्य पर्यायवाची शब्द 'काक, करठ, कौआ, काग, वायस' आदि हैं। अतः सही विकल्प 'पिशुन' है।
• 'पक्षी' शब्द के अन्य पर्यायवाची 'खग, नभचर, शकुनि, विहग, पखेरू, अण्डज, द्विज।' आदि हैं।

Additional Information 

"उत्तराधिकार में प्राप्त सम्पत्ति – धरोहर" यह विकल्प युग्म गलत है। अत: प्रश्नानुसार सही विकल्प 1 'उत्तराधिकार में प्राप्त सम्पत्ति – धरोहर' है।  

1. "उत्तराधिकार में प्राप्त सम्पत्ति" के लिए सही वाक्यांश :-  "रिक्थ" कहलाता है।
2. साथ ही इसे "पैतृक धन" भी कहा जाता है।

Key Points

• धरोहर : किसी की मानत/जो वस्तु दूसरे के यहाँ रखी हो

Additional Information

महीसुर शब्द का अर्थ 'ब्राह्मण' होता है।

महीसुर के पर्यायवाची शब्द -  ब्राह्मण, महीदेव, वाडव, द्विज, भूमिसुर, द्विजाति, विप्र, भूसुर, अग्रजन्मा, भूदेव, भूमिदेव।

अन्य विकल्प:

• पृथ्वी का रक्षक - ओजोन परत को कहा जाता है।
• महिषासुर - एक असुर।
• राक्षस - दैत्य, निशाचर।

दिए गए विकल्पों में सही उत्तर विकल्प 1 'विवाह’  होगा। अन्य विकल्प इसके अनुचित उत्तर हैं। 

Key Points

• दिए गए शब्दों में 'विवाह' शब्द तत्सम है जिसका तद्भव रूप 'ब्याह' होगा।
• ऐसे शब्द जो संस्कृत से ज्यों के त्यों लिए गए, तत्सम होते हैं।
• अन्य सभी शब्द तद्भव हैं।

अन्य विकल्प: 

Additional Information

‘नयन’ शब्द में ‘अयादि संधि’ है तथा इसका संधि विच्छेद ‘ने + अन = नयन’ है। अतः सही विकल्प ‘अयादि संधि’ है।

Key Points 

स्पष्टीकरण

Additional Information

अन्य  विकल्प

इसका सही उत्तर विकल्प 1 है। अन्य विकल्प असंगत हैं।

• उपरोक्त विकल्पों में से शुद्ध वर्तनी उज्ज्वल है।
• अन्य विकल्प वर्तनीगत अशुद्ध हैं।

अन्य विकल्प:

Additional Information

'बुद्धिहीन' शब्द विशेषण का उदाहरण है।  

यह गुणवाचक विशेषण का उदाहरण है।  

Key Points

• ‘बुद्धिहीन’ शब्द ‘गुणवाचक विशेषण’ है।

Additional Information

• किसी जाति, द्रव्य, गुण, भाव, व्यक्ति, स्थान और क्रिया आदि के नाम को संज्ञा कहते हैं।
• जैसे - पशु (जाति), सुंदरता (गुण), व्यथा (भाव), मोहन (व्यक्ति), दिल्ली (स्थान), मारना (क्रिया)।
• सर्वनाम - संज्ञा के स्थान पर प्रयुक्त होने वाले शब्दों को सर्वनाम कहते हैं। जैसे - मैं, वह, वे, उन्हें, अपने तुम, हम, मैं, तू, आप, यह, वह, जो, सो, कोई, कुछ, कौन, क्या’
• सर्वनाम के छः भेद हैं –निश्चयवाचक, पुरुषवाचक (व्यक्तिवाचक), निजवाचक, प्रश्नवाचक, अनिश्चयवाचक, संबंधवाचक
• जिस शब्द अथवा शब्द -समूह के द्वारा किसी कार्य के होने अथवा करने का बोध हो, वह क्रिया कहलाता है।
• जैसे - देवेश दूध पिता है; अर्चना काम कर रही है।

त, थ, द, ध, आदि का उच्चारण स्थान है दन्त्य।

Key Points 

वर्णों का उच्चारण करते समय जिह्वा जिस स्थान को स्पर्श करती है अथवा जिस स्थान से ध्वनि निकलती है उसे उच्चारण स्थान कहते हैं। मुख्य रूप से 6 उच्चारण स्थान होते हैं -

Additional Information

दो स्थानों को मिलाकर भी कुछ वर्णों के उच्चारण स्थान होते हैं-

इन वाक्यों में 'महादेवी वर्मा छायावाद की प्रसिद्ध कवयित्री है।' यह व्याकरण की दृष्टि से शुद्ध वाक्य है 

दिए गए विकल्पों में से सही विकल्प 4 'महादेवी वर्मा छायावाद की प्रसिद्ध कवयित्री हैं।' शुद्ध वाक्य है। अन्य विकल्प सही उत्तर नही हैं। 

Key Points

• 'महादेवी वर्मा छायावाद की प्रसिद्ध कवयित्री हैं।' वाक्य शुद्ध है  क्योंकि इसमें कोई त्रुटि नहीं है।
• अन्य विकल्पों में पुनरुक्ति की अशुद्धियां हैं।

Additional Information

वाक्य सम्प्रेषण की सबसे महत्वपूर्ण और सार्थक इकाई होती है। अतः वाक्यगत अशुद्धियाँ को शुद्ध रूप में लिखना सम्प्रेषण को अधिक सरल बनाता है। वाक्य में, संज्ञा, सर्वनाम, लिंग, वचन, क्रिया-विशेषण, क्रिया, विशेषण आदि संबंधी अशुद्धियाँ  हो सकती हैं।

दिए गए विकल्पों में 'उद्धार' शब्द व्यंजन संधि आधारित शब्द है। अतिरिक्त विकल्प स्वर तथा विसर्ग संधि हैं। अतः सही विकल्प 2 'उद्धार' है।

उद्धार का संधि विच्छेद = उत् + हार।

Key Points

• व्यंजन संधि में त् + ह आये तो त् का द् तथा ह का ध में रूपांतरण होगा। जैसे – उत् + हरण = उद्धरण, पद् + हती = पद्धति।

अन्य विकल्प

Additional Information

‘चौराहा’ शब्द में 'द्विगु' समास है क्योंकि इसका पहला पद संख्यावाचक है। शेष विकल्प असंगत हैं। अतः सही विकल्प 'द्विगु' है।

‘चौराहा’ का समास विग्रह :-  चार राहो का समूह होता है। 

Key Points

स्पष्टीकरण 

निवृत्ति का विलोम 'प्रवृति' होगा। अतः विकल्प (3) सही उत्तर है।

• निवृत्ति अर्थ होता है वापस आना, लौटना।
• प्रवृति का अर्थ होता है मन का झुकाव।

अन्य विकल्प

उपरोक्त विकल्पों में 'शारदा' का पर्यायवाची शब्द 'गिरा' है। अत: विकल्प 'गिरा' इसका सही उत्तर है।

• शारदा के पर्यायवाची शब्द -  सरस्वती, वीणापाणि, वागीश, भारती, वागेश्वरी, विमला, शारदा, गिरा।

Key Points

Additional Information

"मृत्यु के इच्छुक" वाक्यांश के लिए शब्द 'मुमूर्षु' होगा। अन्य विकल्प असंगत हैं।

Key Points

अन्य  विकल्प :

Additional Informationविशेष :-

• हिंदी भाषा में जब कई शब्दों के स्थान पर एक शब्द का प्रयोग किया जाये तो उसे वाक्यांश के लिए एक शब्द कहा जाता है। इसे ‘अनेक शब्दों के लिए एक शब्द’ भी कहा जाता है। वाक्यांश को संक्षेप में सामासिक पद का भी रूप दिया जाता है। कुछ ऐसे लाक्षणिक पद या शब्द भी है, जो अपने में पूरे एक वाक्य या वाक्यांश का अर्थ रखते है।

'चीनांशुक शब्द का अर्थ है - 'रेशम' है। 

चीन से आनेवाला रेशमी वस्त्र चीनांशुक कहलाता है।

Key Pointsअन्य विकल्प:

• तंतु - तागा
• रेणु - धूल, कण
• चीनी मिट्टी - इसका उपयोग बर्तन, प्याले, कटोरी, थाली, अस्पताल में काम में लाए जानेवाले सामान, बिजली के पृथक्कारी बनाने में आता है।

Operating torque:

In moving iron instruments, the deflecting torque is given by

${T_d} = \frac{1}{2}{I^2}\frac{{dL}}{{d\theta }}$

I is the operating current in Ampere

L is the inductance in Henry

θ is the deflection angle in radians

As the operating torque is directly proportional to the square if the operating current. Hence, this instrument possesses high operating torque.

Applicable to both DC and AC circuits:

• Whatever the direction of the current in the coil of the instrument, the iron vanes are so magnetized that there is always a force of attraction in the attraction type and repulsion in the repulsion type of instruments.
• Thus, moving iron instruments are unpolarised instruments i.e. they are independent of the direction in which the current passes. Therefore, the instruments can be used on both ac and dc.

Power consumption in these instruments is higher for low voltage range.

Important Points

Concept:

The following is a procedure to measure the frequency of a Lissajous pattern obtained from CRO.

• For a given Lissajous figure, draw both horizontal and vertical lines passing through the Lissajous figure (Never draw either horizontal line (or) vertical line via pre-existing intersection in the Lissajous figure)
• Count the number of cuts made by both lines
• The unknown to known frequency ratio can be measured as follows.

$\frac{{{f_v}}}{{{f_h}}} = \frac{{{n_h}}}{{{n_v}}}$

Where

fv = unknown vertical frequency

fh = known horizontal frequency

nh = no. of cuts as made by horizontal

nv = no. of cuts as made by a vertical line

Calculation:

From the given Lissajous figure,

Vertical tangencies (NV) = 6

Horizontal tangencies (NH) = 5

The frequency ratio of the vertical signal to the horizontal signal is

$\frac{{{f_V}}}{{{f_H}}} = \frac{{{N_H}}}{{{N_V}}} = \frac{5}{6}$

⇒ f_H = 1250 Hz

Multiplier setting = 20 kΩ

Meter resistance = x

Total resistance = 20 kΩ + x = R₁

Voltmeter reading (V₁) = 440 V

$\begin{array}{l} {V_1}\alpha {I_1}x\ {I_1}\alpha \frac{1}{{{R_1}}} \end{array}$

For multiplier setting = 80 kΩ

Meter resistance = x

Total resistance = 80 KΩ + x = R₂

Voltmeter reading (V₂) = 352 V

$\begin{array}{l} {V_2}\alpha {I_2}x\ {I_2}\alpha \frac{1}{{{R_2}}}\ \frac{{{V_2}}}{{{V_1}}} = \frac{{{I_2}}}{{{I_1}}} = \frac{{{R_1}}}{{{R_2}}}\ \frac{{352}}{{440}} = \frac{{20 + x}}{{80 + x}} \Rightarrow x = 220\ k{\rm{\Omega }} \end{array}$

For multiplier setting = 40 KΩ

R₃ = 260 kΩ

$\frac{{{V_3}}}{{{V_1}}} = \frac{{{I_3}}}{{{I_1}}} = \frac{{{R_1}}}{{{R_3}}} \Rightarrow \frac{{{V_3}}}{{440}} = \frac{{240}}{{260}} \Rightarrow {V_3} = 406\ V$

Explanation:

If a pointer is having less weight, it will be having high torque/weight ratio so that even for a very small deflection pointer starts moving and indicates measured value. PMMC instruments have a more torque/weight ratio. Hence PMMC instruments are more accurate.

Important Points

PMMC instruments:

PMMC instrument also called D'Arsonval meter.

These are used to measure the current and voltage in the DC circuit.

They are used eddy current type damping

The deflection angle can be calculated as

$\theta = \frac {BANI}{K_c}$

B = Flux density in Wb / sq.m 

A = Area of the frame

N = No. of turns

I = Current flow through the coil

Kc  = Spring control coefficient

Advantages:

• The scale is uniform (Linear scale)
• High torque to weight ratio compared to all the other type of instruments
• Frequency error or reactance error is absent
• High sensitivity
• Hysteresis loss is lesser
• No internal heating problem
• High accuracy
• Very less power consumption

Disadvantages:

• It works only for Dc not used for AC
• Cost is very high

Explanation:

• The one-wattmeter method is used for the measurement of three-phase power in case balanced loads only
• It is used for both star and delta connected loads
• Two wattmeter method and three wattmeter methods are used for both balanced and unbalanced loads
• It is also used for both star and delta connected loads
• Two wattmeter method is best suitable for three-phase three-wire system and three wattmeter method is suitable for the three-phase four-wire system

Additional Information

Measurement of polyphase power: 3 - ϕ

1. For 3ϕ – 4 wire → neutral is present

2. As per Blondel’s theorem, if neutral is not present for ‘n’ wire system, then ‘n-1’ number of wattmeter’s are required for ‘n’ wire system.

When a loss-free overhead transmission line is terminated through a resistance equal to surge Impedance of the line, there will be no reflection of travelling wave hence the value of R is 400 Ω.

Important Points:

The surge impedance or characteristic impedance of a long transmission line is given by,

${Z_C} = \sqrt {\frac{Z}{Y}}$

Z is series impedance per unit length per phase

Y is shunt admittance per unit length per phase

Surge Impedance for the transmission line is about 400 ohms it is around 40 ohms for underground cables.

Surge impedance loading (SIL):

Surge impedance loading is defined as the power load in which the total reactive power of the lines becomes zero. The reactive power generated by the shunt capacitance is consumed by the series inductance of the line.

SIL $= \frac{{{V^2}}}{{{Z_C}}};MW$

In a HVDC system, 12 pulse bridge converters can eliminate 5^th and 7^th harmonics because the order of the harmonics for 6 – Pulse Bridge and 12 – pulse bridge converters are

(6 k ± 1) respectively.

Concept:

The core to neutral capacitance is,

C_N = C_e + 3 C_c

Where core-core and core-earth capacitances of the cable are C_c and C_e respectively.

${C_e} = \frac{{{C_1}}}{3}$

${C_c} = \frac{{{C_2} - {C_e}}}{2}$

Where C₁ is the capacitance between 3-cores bunched together and the sheath

C₂ is the capacitance between a conductor and the other two connected together to the sheath

Calculation:

Given that, C₁ = 15 μF and C₂ = 10 μF

C_e = 15/3 = 5 μF

${C_c} = \frac{{10 - 5}}{2} = 2.5;\mu F$

The core to neutral capacitance, C_N = 5 + 3(2.5) = 12.5 μF

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 a plant is that all the unit is must operate at the same incremental fuel cost.

Calculation:

Given that

P1 + P2 = 200

$\begin{array}{l} \frac{{d{C_1}}}{{d{P_1}}} = 2 + 0.01{P_1}\ \frac{{d{C_2}}}{{d{P_2}}} = 1.6 + 0.02{P_2} \end{array}$

For economical operations, $\frac{{d{C_1}}}{{d{P_1}}} = \frac{{d{C_2}}}{{d{P_2}}}$

2 + 0.01 P₁ = 1.6 + 0.02 P2

2P2 – P1 = 40

3P2 = 240

P2 = 80 MW

P1 = 120 MW

Concept:

• In extra-high voltage (EHV) transmission lines and power systems, switching surges is an important factor that affects the design of insulation.
• All transmission lines rated for 220 kV and above incorporate switching surge spark overvoltage for their insulation levels.
• A switching surge is a short duration transient voltage produced in the system due to a sudden opening or closing of a switch or circuit breaker or due to an arcing at a fault in the system.
• The waveform is not unique. The transient voltage may be an oscillatory wave or a damped oscillatory wave of frequency ranging from a few hundred hertz to few kilohertz.
• It may also be considered as a slow rising impulse having a wavefront time of 0.1 to 10 ms, and a tail time of one to several ms. Switching surges contain larger energy than the lightning impulse voltages.

Explanation:

The insulation of modern EHV lines is designed based on switching voltage​.

When a transmission line is loaded with impedance it is represented as follows:

V = Incident voltage

V’= Reflected voltage

V’’= Refracted or transmitted voltage.

Transmission line impedance is surge impedance Zs.

Load impedance is ZL.

Short circuited is line is considered when ZL = 0

Coefficient of reflection is given by the expression ${V_{reflection}} = \frac{{V'}}{V}$

${V_{reflection}} = \frac{{{Z_L} - {Z_s}}}{{{Z_L} + {Z_S}}}$

Calculations: 

${{\rm{V}}_{{\rm{reflection}}}} = \frac{{0 - {Z_S}}}{{0 + {Z_S}}}$

V reflection = -1

Therefore coefficient of reflection of voltage for a short circuited line is -1.

Additional Information

Formulas for the Coefficients of reflection and refraction are as follows

${v_{refraction}} = \frac{{2{Z_L}}}{{{Z_L} + {Z_S}}}$   

 ${v_{reflection}} = \frac{{{Z_L} - {Z_S}}}{{{Z_L} + {Z_S}}}$

${i_{refraction}} = \frac{{2{Z_S}}}{{{Z_L} + {Z_S}}}$   

 ${i_{reflection}} = \frac{{{Z_S} - {Z_L}}}{{{Z_L} + {Z_S}}} = - {v_{reflection}}$​

Line to Line fault:

The boundary conditions for the figure shown above are

I_a = 0

I_b + I_c = 0

V_b = V_c

${I_{a0}} = \frac{1}{3}\left( {{I_a} + {I_b} + {I_c}} \right) = 0$

${I_{a1}} = \frac{1}{3}\left( {{I_a} + \lambda {I_b} + {\lambda ^2}{I_c}} \right) = \frac{1}{3}\left( {\lambda - {\lambda ^2}} \right){I_b}$

${I_{a1}} = \frac{1}{3}\left( {{I_a} + {\lambda ^2}{I_b} + \lambda {I_c}} \right) = \frac{1}{3}\left( {{\lambda ^2} - \lambda } \right){I_b}$

From the above relations, for a line-to-line fault, the zero-sequence component of current is absent and the positive-sequence component of current is equal in magnitude but opposite in phase to negative sequence component of current, i.e.

I_a1 = -I_a2

Important Points

The following circuit shown is a system with line to ground fault.

Here, Va = 0, Ib = 0, Ic = 0

Now, ${I_{a1}} = \frac{1}{3}\left( {{I_a} + k{I_b} + {k^2}{I_c}} \right)$

${I_{a2}} = \frac{1}{3}\left( {{I_a} + {k^2}{I_b} + k{I_c}} \right)$

${I_{a0}} = \frac{1}{3}\left( {{I_a} + {I_b} + {I_c}} \right)$

By substituting Ib and Ic values,

${I_{a1}} = {I_{a2}} = {I_{a0}} = \frac{{{I_a}}}{3}$

Hence in LG faults, all of the three components la0, la1 and la2 are equal.

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 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:

In the given circuit diagram, both the shunt switches are closed. Therefore, both the primary and secondary are delta-connected.

Concept:

Load factor: The ratio of average load to the maximum demand during a given period is known as the load factor.

Load factor = average load/maximum demand

If the plant is in the operation of T hours

$Load;factor = \frac{{Avearge;load \times T}}{{Maximum;demand \times T}}$

$= \frac{{Units;generated;in;T;hours}}{{Maximum;demand \times T}}$

• The load factor may be daily load factor, monthly or annually if the period considered is a day or month or year
• Load factor is always less than 1 because the average load is smaller than the maximum demand
• It plays a key role in determining the overall cost per unit generated
• Higher the load factor of the power station, lesser will be the cost per unit generated, it is because higher load factor means lesser maximum demand
• The station capacity is so selected that it must meet the maximum demand
• Now, lower maximum demand means a lower capacity of the plant which reduces the cost of the plant

Calculation:

Given that, maximum demand = 2 kW

Daily energy consumption = 30 Units.

Load factor $= \frac{{30}}{{24 \times 2}} = 0.625$

Concept:

Voltage regulation of transmission line is defined as the ratio of the difference between sending and receiving end voltage to receiving end voltage of a transmission line between conditions of no-load and full load.

It can be expressed in percentage as

$\% VR = \frac{{{V_s} - {V_R}}}{{{V_R}}} \times 100$

Where Vs is the sending end voltage per phase and VR is the receiving end voltage per phase.

Percent regulation, $= \frac{{IRcos{\phi _r} \pm IXsin{\phi _r}}}{{{V_r}}} \times 100;$

Application:

Maximum regulation occurs at lagging power factor only

$\frac{d}{{d\phi }}\left( {R\cos \phi + X\sin \phi } \right) = 0$

$\Rightarrow \tan \phi = \frac{X}{R}$

Additional Information

Zero regulation occurs at leading power factor only

$\Rightarrow \frac{{IRcos{\phi _r} - IXsin{\phi _r}}}{{{V_r}}} \times 100 = 0;$

$\Rightarrow IRcos{\phi _r} = IXsin{\phi _r}$

$\Rightarrow {\phi _r} = {\tan ^{ - 1}}\left( {\frac{R}{X}} \right)=\frac{\pi}{2}-\theta ;$

Concept:

In G(s) plane, the Nyquist plot of G(S) passes through the negative real axis at the point (-a, j0)

a = magnitude of G(s) at ω = ωpc

ωpc is phase cross over frequency.

at ${\rm{\omega}} = {{\rm{\omega }}_{{\rm{pc}}}},\angle G\left( {j\omega } \right) = - 180^\circ$

Application:

The -180° phase line of Bode diagram indicates the phase angle of the function is -180°.

At this phase, the Nyquist plot of G(S) passes through the negative real axis.

Important Points

The slope of magnitude plot of a Bode plot at any frequency is given by

Slope = (Z – P) × 20 dB/decade

Where, Z is the number of zeros at that frequency

P is the number of poles at that frequency

If a closed-loop pole of the system is on the jω axis, then the magnitude equals 1 and the phase is 180 degrees at that frequency. If that occurs, the Nyquist plot will pass through the -1 point at that frequency. It indicates that there are equal number of poles and zeros.

Bode plot is logarithmic scale plot, so we take log value there.

Therefore, the unit circle of the Nyquist plot transforms into 0 dB line of the amplitude plot of the Bode diagram at any frequency

Concept:

State space representation:

ẋ(t) = A(t) x(t) + B(t) u(t)

y(t) = C(t) x(t) + D(t) u(t)

y(t) is output

u(t) is input

x(t) is a state vector

A is a system matrix

This representation is continuous time-variant.

Controllability:

A system is said to be controllable if it is possible to transfer the system state from any initial state x(t0) to any desired state x(t) in a specified finite time interval by a control vector u(t)

Kalman’s test for controllability:

ẋ = Ax + Bu

Qc = {B AB A2B … An-1 B]

Qc = controllability matrix

If |Qc| = 0, system is not controllable

If |Qc|≠ 0, system is controllable

Observability:

A system is said to be observable if every state x(t0) can be completely identified by measurement of output y(t) over a finite time interval.

Kalman’s test for observability:

Q0 = [CT ATCT (AT)2CT …. (AT)n-1 CT]

Q0 = observability testing matrix

If |Q0| = 0, system is not observable

If |Q0| ≠ 0, system is observable.

Calculation:

\(A = \left[ {\begin{array}{{20}{c}} 1&2\ 0&2 \end{array}} \right],;B = \left[ {\begin{array}{{20}{c}} 0\ 1 \end{array}} \right],;C = \left[ {\begin{array}{*{20}{c}} b&0 \end{array}} \right]\)

Observability:

\(CA = \left[ {\begin{array}{{20}{c}} b&0 \end{array}} \right]\left[ {\begin{array}{{20}{c}} 1&2\ 0&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} b&2b \end{array}} \right]\)

If |Q0| ≠ 0, system is observable.

Observable for all non-zero values of b.

Calculation:

The power required is:

A desktop computer of 400 W

An electric geyser of 1 kW i.e. 1000 W

A toaster of 500 W

Total power required to generate = 400 + 1000 + 500 = 1900 W = 1900 J per unit time

The efficiency of PV = 25% = 0.25

Therefore, to get the required output, the input must be

${P_i} = \frac{{{P_o}}}{\eta } = \frac{{1900}}{{0.25}} = 7600;W$

Solar radiation of 200 J/m2 and per unit time

The required area $= \frac{{7600}}{{200}} = 38;{m^2}$

Lead sheath: In order to protect the cable from moisture, gases or other damaging liquids (acids or alkalis) in the soil and atmosphere, a metallic sheath of lead or aluminium is provided over the insulation as shown in the figure. It has minimum dielectric stress in a cable.

The extruded Lead sheath also serves as Metallic Screen. Copper Tape/Copper sheath can also be used but Copper is costly and would rather be used as overhead conductors with less mechanical stresses.

Bedding: Over the metallic sheath is applied a layer of bedding which consists of fibrous material like jute or hessian tape. It is to protect the metallic sheath against corrosion and from mechanical injury due to armouring.

Armouring: Over the bedding, armouring is provided which consists of one or two layers of galvanized steel wire or steel tape. Its purpose is to protect the cable from mechanical injuries while laying it or handling it.

Explanation:

Area control error (ACE) that is defined as

ACE = (P_tie – P_sch) + B_fΔf = ΔP_tie + B_f Δf

Where P_tie and P_sch are tie-line power and scheduled power through tie-line respectively and the constant B_f is called the frequency bias constant.

A negative ACE means that the area is not generating enough power to send the desired amount of the area. There is a deficiency in net power output.

Note:

• Without frequency bias, the indicated deficiency would be less because there would be no positive offset added when actual frequency is less than scheduled frequency and the ACE would be less.
• The area would produce sufficient generation to supply its own load and the prearranged interchange but would not provide the additional output to assist neighbouring interconnected area to raise the frequency.

Concept:

Neutral current in 3-ϕ, star connected system represents the unbalance of the 3-ϕ conductors.

Neutral current is the sum of all phase current

In = IR + IY + IB

\(\left[ {\begin{array}{{20}{c}} {{I_R}}\ {{I_Y}}\ {{I_B}} \end{array}} \right] =\left[ {\begin{array}{{20}{c}} 1&1&1\ 1&{{\alpha ^2}}&\alpha \ 1&\alpha &{{\alpha ^2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{I_{R0}}}\ {{I_{R1}}}\ {{I_{R2}}} \end{array}} \right]\)

\(\left[ {\begin{array}{{20}{c}} {{I_{R0}}}\ {{I_{R1}}}\ {{I_{R2}}} \end{array}} \right]=\dfrac13\left[ {\begin{array}{{20}{c}} 1&1&1\ 1&{{\alpha}}&\alpha^2 \ 1&\alpha^2 &{{\alpha }} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{I_R}}\ {{I_Y}}\ {{I_B}} \end{array}} \right] \) ...(1)

IR0, IR1, and IR2 are the zero, positive and negative sequence currents.

Application:

From (1)

I_R0 = (1/3)(I_R + I_Y + I_B)

IR + IY + IB = 3 IR0

⇒ In = 3 IR0

Classification of Buses:

The power system is nothing but the interconnection of the various bus.

Each of these buses is associated with four electrical parameters namely voltage with magnitude and phase angle, active power, and reactive power.

Slack Bus:

• Slack Bus is also known as Swing or Reference Bus.
• One of the generator buses is considered as a slack bus.
• The slack bus does not exist in real rather it is assumed for the consideration of losses occurring during power transmission.
• Actually, there exist only two buses in the power system, Load Bus and Generator Bus for which active power is specified.
• Since active power delivered by Generator Bus and consumed by Load Bus differ
• This means that a power loss equal to the difference between Generator Bus P and Load Bus P is occurring.
• This loss can only be calculated after the solution of Load Flow.
• Therefore to supply power loss, an extra generator bus is considered for which bus magnitude and voltage are specified, and active power and reactive power are to be calculated.
• This active power of slack bus is the equivalent power loss occurring in different systems.
• Generally, the phase angle of the slack bus is taken for reference for the entire load flow solution. Therefore this bus is also called Reference Bus.​

Load Bus:

• For load bus, real power P and reactive power Q are known but the magnitude and phase angle of bus voltage is unknown.
• It is desired to find the bus voltage using load flow analysis. At load bus, voltage is allowed to vary within some specified limit.

Generator Bus:

• Generator Bus is also called a voltage-controlled bus.
• The generator is connected to this bus.
• Therefore the bus voltage corresponding to generation voltage and active power generation corresponding to generator rating is specified for this bus.

The quantities specified for different type of buses shown in the below table.

$\% \eta = \frac{{{P_r}}}{{{P_r} + losses}} \times 100$

P_r = Receiving end power

$= \frac{{3{V_R}{I_R}\cos {\phi _R}}}{{3{V_R}{I_R}\cos {\phi _R} + 3{I^2}R}} \times 100$

As the power factor increases, efficiency also increases.

Efficiency also increases when we increase the transmission voltage as I²R losses decreases.

Explanation:

Feedback refers to the comparison of the final output to the desired output at respective input so as to get accurate and error free result and in the system improves the transient response of the system. Feedback system is basically a low pass filter.

Low pass filter is mainly as integral controller and it is used as the controller in the system so as to increase the accuracy by reducing or proper eliminating the steady state error of the control system.

Important Points

The name of the filter is decided by the type of frequency it passes:

Low pass filter:

This filter passes the low frequency and blocks or attenuates the high frequency.

High Pass Filter:

This filter passes the high frequency and blocks the low frequency.

Band Pass Filter:

This filter passes a certain band of frequencies and blocks low and high frequencies.

Band Stop Filter:

This filter blocks a certain band of frequencies and passes low and high frequencies.

Concept:

Gain margin (GM): The gain margin of the system defines by how much the system gain can be increased so that the system moves on the edge of stability.

It is determined from the gain at the phase cross over frequency.

$GM = \frac{1}{{{{\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|}_{\omega = {\omega _{pc}}}}}}$

Phase crossover frequency (ωpc): It is the frequency at which phase angle of G(s) H(s) is -180°.

$\angle G\left( {j\omega } \right)H\left( {j\omega } \right){|_{\omega = {\omega _{pc}}}} = - 180^\circ$

Phase margin (PM): The phase margin of the system defines by how much the phase of the system can increase to make the system unstable.

$PM = 180^\circ + \angle G\left( {j\omega } \right)H\left( {j\omega } \right){|_{\omega = {\omega _{gc}}}} = - 180^\circ$

It is determined from the phase at the gain cross over frequency.

Gain crossover frequency (ωgc): It is the frequency at which the magnitude of G(s) H(s) is unity.

${\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|_{\omega = {\omega _{gc}}}} = 1$

In the Nyquist plot, the gain at phase cross over frequency is the gain at which the plot cuts the negative real axis.

Calculation:

Given open loop transfer function is, ${\rm{T}}\left( {\rm{s}} \right) = \frac{{\rm{K}}}{{{{\left( {{\rm{s}} + 1} \right)}^3}}}$

The magnitude $\left| {T\left( {j\omega } \right)} \right| = \frac{K}{{{{\left( {\sqrt {{\omega ^2} + 1} } \right)}^3}}}$

At gain crossover frequency, the magnitude of the open loop transfer function must be unity.

$\Rightarrow \frac{K}{{{{\left( {\sqrt {{{\left( {\sqrt 3 } \right)}^2} + 1} } \right)}^3}}} = 1$

⇒ K = 8

Concept:

According to Mason’s gain formula, the transfer function is given by

TF = \frac{{\mathop \sum \nolimits_{k = 1}^n {M_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}

Where, n = no of forward paths

Mk = kth forward path gain

Δk = the value of Δ which is not touching the kth forward path

Δ = 1 – (sum of the loop gains) + (sum of the gain product of two non-touching loops) – (sum of the gain product of three non-touching loops)

Application:

From the given SFG, the loops are

L1 = -bc, L2 = -fg, L3 = cjgi

Two non-touching loops: L1L2

Determinant, Δ = 1 – (L1 + L2 +L₃) + L1L2

= 1 + bc + fg – cigj + bcfg

Concept:

The transfer function of the standard second-order system is:

$TF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\omega _n^2}}{{{s^2} + 2\zeta {\omega _n}s + \omega _n^2}}$

ζ is the damping ratio

ωn is the natural frequency

Characteristic equation: ${s^2} + 2\zeta {\omega _n} + \omega _n^2$

Roots of the characteristic equation are: $- \zeta {\omega _n} + j{\omega _n}\sqrt {1 - {\zeta ^2}} = - \alpha \pm j{\omega _d}$

α is the damping factor

Calculation:

The given differential equation is:

$4\frac{{{d^2}c\left( t \right)}}{{d{t^2}}} + 8\frac{{dc\left( t \right)}}{{dt}} + 16c\left( t \right) = 16r\left( t \right)$

By applying the Laplace transform,

(4s2 + 8s + 16) C(s) = 16 R(s)

$\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{16}}{{4{s^2} + 8s + 16}} = \frac{4}{{{s^2} + 2s + 4}}$

By comparing the above transfer function with the standard second-order system,

$\omega _n^2 = 4 \Rightarrow {\omega _n} = 2$

⇒ 2ζωn = 2 ⇒ ζ = 0.5

Damping ratio = 0.5

Natural frequency = 2 rad/sec

Concept:

1. Every branch of a root locus diagram starts at a pole (K = 0) and terminates at a zero (K = ∞) of the open-loop transfer function.
2. Root locus diagram is symmetrical with respect to the real axis. 3. Number of branches of the root locus diagram are:

N = P if P ≥ Z

= Z, if P ≤ Z

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

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

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: ${\theta _l} = \frac{{\left( {2l + 1} \right)\pi }}{{P - Z}}$

l = 0, 1, 2, … |P – Z| – 1

7. On the real axis to the right side of any section, if the sum of total number of poles and zeros are odd, root locus diagram exists in that section. 8. Break in/away points: These exist when there are multiple roots on the root locus diagram.

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

So, the roots of $\frac{{dK}}{{ds}}$ are the break points.

Calculation:

Given loop transfer function is, $G\left( s \right) = \frac{K}{{s{{\left( {s + 3} \right)}^2}}}$

For breakaway points, $\frac{{dK}}{{ds}} = 0$

$\Rightarrow \frac{d}{{ds}}\left[ {s{{\left( {s + 3} \right)}^2}} \right] = 0$

$\Rightarrow \frac{d}{{ds}}\left[ {{s^3} + 6{s^2} + 9s} \right] = 0$

⇒ 3s2 + 12s + 9 = 0

⇒ s2 + 4s + 3 = 0

⇒ (s + 3) (s + 1) = 0

⇒ s = -3, s = -1

s = -1 is a valid breakaway point.

Concept:

1. Every branch of a root locus diagram starts at a pole (K = 0) and terminates at a zero (K = ∞) of the open-loop transfer function.
2. Root locus diagram is symmetrical with respect to the real axis. 3. Number of branches of the root locus diagram are:

N = P if P ≥ Z

= Z, if P ≤ Z

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

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

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: ${\theta _l} = \frac{{\left( {2l + 1} \right)\pi }}{{P - Z}}$

l = 0, 1, 2, … |P – Z| – 1

7. On the real axis to the right side of any section, if the sum of total number of poles and zeros are odd, root locus diagram exists in that section. 8. Break in/away points: These exist when there are multiple roots on the root locus diagram.

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

So, the roots of $\frac{{dK}}{{ds}}$ are the break points.

Application:

The loop transfer function of a system is given by: $G\left( s \right)H\left( s \right) = \frac{{K\left( {s + 6} \right)}}{{\left( {s + 3} \right)\left( {s + 5} \right)}}$

Number of open loop poles = 2

Number of open loop zeros = 1

Number of Asymptotes = |2 – 1| = 1

Centroid, $\sigma = \frac{{\left( {0 - 3 - 5} \right) - \left( { - 6} \right)}}{1} = - 2$

Concept:

First-order system:

The transfer function of the standard first-order system is given by

$TF = \frac{{k}}{{\left( {1 + τ s} \right)}}$

Where,

τ = time constant of the system

The time constant can be defined as the negative reciprocal of the pole of the system.

Calculation:

From the given figure,

$\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\frac{{3K}}{{2s + 1}}}}{{1 + \frac{{3K}}{{2s + 1}}}}$

$= \frac{{3K}}{{2s + 3K + 1}}$

$= \frac{{3K}}{{3K + 1}} \times \frac{1}{{1 + \frac{{2s}}{{3K + 1}}}}$

From the above transfer function, the time constant is

$T = \frac{2}{{3K + 1}}$

The time constant of the system is less than 0.2

$\Rightarrow \frac{2}{{3K + 1}} < 0.2;$

⇒ K > 3

State space representation:

ẋ(t) = A(t)x(t) + B(t)u(t)

y(t) = C(t)x(t) + D(t)u(t)

y(t) is output

u(t) is input

x(t) is a state vector

A is a system matrix

This representation is continuous time-variant.

Controllability:

A system is said to be controllable if it is possible to transfer the system state from any initial state x(t0) to any desired state x(t) in a specified finite time interval by a control vector u(t)

Kalman’s test for controllability:

ẋ = Ax + Bu

Qc = {B AB A2B … An-1 B]

Qc = controllability matrix

If |Qc| = 0, system is not controllable

If |Qc|≠ 0, the system is controllable

Observability:

A system is said to be observable if every state x(t0) can be completely identified by measurement of output y(t) over a finite time interval.

Kalman’s test for observability:

Q0 = [CT ATCT (AT)2CT …. (AT)n-1 CT]

Q0 = observability testing matrix

If |Q0| = 0, system is not observable

If |Q0| ≠ 0, system is observable.

Duality property of controllability and observability:

• If the pair (A, B) is controllable then the pair (AT, BT) is observable.
• If the pair of (A, C) is observable then the pair of (AT, CT) is controllable.

If there are no pole-zero cancellations in the transfer function then the system is completely controllable and observable.

Concept:

According to Mason’s gain formula, the transfer function is given by

TF = \frac{{\mathop \sum \nolimits_{k = 1}^n {M_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}

Where, n = no of forward paths

Mk = kth forward path gain

Δk = the value of Δ which is not touching the kth forward path

Δ = 1 – (sum of the loop gains) + (sum of the gain product of two non-touching loops) – (sum of the gain product of three non-touching loops)

Calculation:

From the given signal flow graph, the forward paths are

P1 = rsu, P2 = efh

The loops are

L1 = st, L2 = fg

Two non-touching loops = L1L2 = stfg

Δ = 1 – st – fg + stfg

= 1 – st – fg (1 – st) = (1 – st) (1 – fg)

Δ1 = 1 – fg

Δ2 = 1 – st

Transmittance between X1 and X2 is,

$TF = \frac{{rsu\left( {1 - fg} \right) + efh\left( {1 - st} \right)}}{{\left( {1 - st} \right)\left( {1 - fg} \right)}}$

$= \frac{{rsu}}{{1 - st}} + \frac{{efh}}{{1 - fg}}$