May 2019 Solutions

Q.1. Attempt any five of the following :- [20 M]

Ans. i) A transformer is a static device by means of which electric power in one circuit is transformed into electric power of same frequency in another circuit without changing frequency.

ii) It can raise or lower the voltage in a circuit but with corresponding decrease or increase in current.

iii) The physical basic of transformer is mutual induction between two circuits linked by a common magnetic flux . In its simples form, it consists of two inductive coils which are electrically separated but magnetically linked through a path of low reluctance as shown in fig.

iv) The two coil posses high mutual inductance. If one coil is connected to a source of alternating voltage, an alternating flux is set up in the laminated core, most of which is linked with other coil in which it produces mutually induced emf. If the second coil circuit is closed, a current flows in it and so electric energy is transferred from first coil to second coil.

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Ans.

Consider a star connected and delta connected circuit .

Let resistance be R₁, R₂ and R₃ of the star connection and Let R₁₂, R₂₃, R₃₁ are the resistances of the delta circuit.

In star connection , The resistance between terminals 1 and 2 is (R1+R2). Similarly the other terminals are also gives as inthe same way as the first one.

R_{1}+R_{2}=\frac{R_{12}\times (R_{23}+R_{31})}{R_{12}+R_{23}+R_{31}} —-(1)

R_{2}+R_{3}=\frac{R_{23}\times (R_{31}+R_{12})}{R_{12}+R_{23}+R_{31}} —-(2)

R_{3}+R_{1}=\frac{R_{31}\times (R_{12}+R_{23})}{R_{12}+R_{23}+R_{31}} —-(3)

By using this three equations and multiplying eq.(1) and (ii) , (ii) and (iii) , (iii) and (i) and adding them together and simplifying them we get

R_{12}=\frac{R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{3}}=R_{1}+R_{2}+\frac{R_{1}R_{2}}{R_{3}} R_{12}=\frac{R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{1}}=R_{2}+R_{3}+\frac{R_{2}R_{3}}{R_{1}} R_{31}=\frac{R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{2}}=R_{1}+R_{3}+\frac{R_{3}R_{1}}{R_{2}}

Hence, the equivalent delta resistance between any two terminals is ngiven by the sum of stra resistances between those terminals plus the product of these two star resistances divide by the third star resistances.

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Ans.

Working principle of DC motor :- The working of DC motor is based on the principle that Whenever a current carrying conductor is placed in a magnetic field it experiences a force. An electric motor is a machine that converts an electric energy into mechanical energy.

The principle of operation of DC motor can be explained with the help of Fleming’s left hand rule .

Fleming left hand rule can be stated as , ”The direction of force produced on a current carrying conductor which is placed in a magnetic field can be determined by Fleming’s left hand rule. If the three fingers namely the fore fingers, the middle finger and the thumbs of left hand are held mutually perpendicular to each other such that if the forefinger indicates direction of magnetic flux, middle finger indicates direction of current, then thumb will indicate direction of force experienced by the conductor.

This is the principle operation of dc motor .

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Ans.

A series circuit is at electrical resonance when its net reactance is zero. The frequency fr at which resonance occurs is called resonant frequency(f_r).

Expression for Resonant frequency

Consider the circuit where, Resistance, capacitance and inductor is connected in series with the voltage across it.

The impedance of resonance is the resistance i.e. Z=R

As per the series resonance definition as net reactance is zero.

So X_L=X_C

X_L= ωL

X_{C}=\frac{1}{\omega C}

Putting value of X_L and X_C in the condition of resonance

ωL= \frac{1}{\omega C}

ω²LC=1

ω=2πf_r

4π²f_r²LC=1

f_{r}^{2}=\frac{1}{4\pi ^{2}LC} f_{r}=\frac{1}{2\pi \sqrt{LC}}

This is the expression for resonant frequency.

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Ans.

Given :-

To Find :- R=?

Solution :-

Step-1 :- Converting the 100 V voltage source into the current source

V= IR

100 = I x 10

I= 10 A

Step-2 :- By converting the voltage source into current source the resistance , 10 Ω, 14Ω and R are connected in parallel.

So, the equivalent resistance is given by,

R_{eq}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}

We will take R₁= 10, R₂ = 14, R₃ = R

R_{eq}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}=\frac{1}{10}+\frac{1}{14}+\frac{1}{R}=0.1714+\frac{1}{R} R_{eq}=\frac{0.1714R+1}{R}

As we know the ohm’s law

V= IR_eq

Putting value of R_eq in the ohm’s law

I=\frac{V}{R_{eq}} I=\frac{V}{R_{eq}}=\frac{100}{\left ( \frac{0.1714R+1}{R} \right )}

3\times \frac{R}{0.1714R+1}=100

3R = 100(0.1714R+1)

3R=17.14R

-100 = 17.14R-3R+100

-100=14.14 R

R=-100/14.14

R=-7.07 Ω

(- sign indicates that resistance is in the reverse direction )

So, as resistance can’t be negative

R= 7.07 Ω

Method-II

Solution :-

Step-I :- As 10 and 14 ohms are in series so we will add these two resistances.

R=10+14 = 24

Step-2 :- By applyying ohm’s law, we get

V=IR

100 = 3 x (24+R)

100= 72+3R

28 = 3R

R= 9.33 Ω

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Ans.

Given :-

To Find :- Current through 4 ohm resistor

Solution :-

Step-i :- Converting the 5A current source into the voltage source and taking the 2Ω in series.

Again transforming 4 V and 2 Ω

Adding 2A and 2A as they are in parallel

By current division rule

I_{4\Omega } =4\times \frac{2}{2+4}=\frac{4}{3}=1.33

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Ans.

Step-1 :- Consider 4 V source only

Short-circuiting the 6 V source

15 || 14.8 = 7.44 Ω

Hence current through 14.8 Ω is,

I_{2}=\frac{15}{15+14.8}\times I I_{2}=\frac{15}{15+14.8}\times 0.537= 0.270 \; A

Same current I₂ will flow through 10 Ω and 4.8 Ω

Hence, current through 8Ω is given as,

I_{3}=\frac{12}{12+8}\times I_{2}= \frac{12}{20}\times 0.270\; A=0.162 \; A

Hemnce, current through 8 Ω is 0.162 A from A to B ,

Step 2:- Consider 6 V source only short circuiting the 4V source.

Hence, current through 8Ω is given as,

I_{1}=\frac{13.75}{13.75+8}\times 0.351= 0.2218 A

Hence, current through 8 Ω is 0.2218 A from A to B

Step-3 :- Hence using superposition theorem , current through 8 Ω is given as,

Current through 8 Ω is = 0.162 + 0.2218 = 0.3838 A (from A to B)

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Ans.

Given :- L₁= 0.04 H , R₁= 25 Ω ,

L₂= 0.2 H , R₂ = 15 Ω , V= 230 V , f= 50 Hz

To Find :- i) power dissipation in each coil =?

ii) Total power factor = ?

Solution :-

According to the given information the circuit can be drawn as below,

Z₁ = 25 +0.04j = 25 ∠ 0.091^o

Z₂ = 15+0.2j = 15.00∠0.76^o

V₁= IZ₁

I_{1}=\frac{V_{1}}{Z_{1}}= \frac{230}{25 +0.04j }= 9.199-0.014j = 9.19\angle -0.091^{o}

I₁= 9.19∠-0.091^o

V₂= IZ₂

230 = I₂ x (15+0.2j)

I₂ = 15.33-0.204j = 15.33∠-0.76^o

i) Calculating the power dissipated in each coil

For this, we need

Z =Z₁ + Z₂ = (25 +0.04j) + (15+0.2j) = 40+0.24j = 40.00∠-0.34

Z=\sqrt{R^{2}+X_{2}}=\sqrt{40^{2}+0.24^{2}}=40

V= 230 V

I=\frac{V}{Z}=\frac{230}{40+0.24j}= 5.74-0.0344j = 5.74\angle -0.34^{o}

Magnitude of current I=\sqrt{(5.74)^{2}+(0.034)^{2}}=5.74 A

i) Power dissipated in coil A = I²R = (5.74)² x R_A = (5.74)² x 25 = 823.69 W

ii) Power dissipated in coil B = I²R_B = (5.74)² x R_B = (5.74)² x 15 = 494.21 W

ii) Calculating the total power factor

P.f = cos\; \Phi =\frac{R}{Z}=\frac{40}{40}=1

Total power factor is unity.

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Ans.

1.There are two types of losses which takes place in a transformer.As in a static transtormer there are no friction or windage losses, Hence, the losses occuring are as follows :-

i) Cu loss

ii) Iron loss

i) Cu loss – This loss is due to the ohmic resistance of the transformer windings.

total Cu loss = I₁²R₁+I₂²R₂= I₁²R_o1 + I₂²R_o2

This shows that the cu loss is proportional to the (current)² or KVA². in other words, cu loss at half the full load is one fourth of that at full load.

These losses are minimum at no load conditions. It is ideally zero and maximum at ‘Full load’ because I₁ and I₂ current is at maximum level.

ii) Iron loss :- In transformer the iron losses are the constant losses and it depends on the voltage. It includes both hysteresis loss and eddy current loss. Because the core flux in a transformer remains practically constant for all loads. The core loss is practically the same at all loads.

2. The ratings of transformer in KVA not in KW because Cu loss of transformer depends on current and iron loss on voltage. Hence, total trasnformer loss depends on volt-ampere(VA) and not on phase angl;e between voltage and current i.e. it is independent of load power factor. that is why rating of trasnformer is in KVA and not in KW.

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Ans.

Ans.

Given figure shows a balanced star-connected load, the load may be assumed to be inductive. Let ????_????????, ????_????????, ????_???????? be the three phase voltages. ????_????,????_????,????_???? be the phase currents. The phase currents will lag behind their respective phase voltages by angle ????. Current through current coil of ????₁=????_????

Voltages across voltage coil of ????₁=????_????????=????_????????+????_????????=????_????????−????_????????

From the phasor diagram, it is clear that the phase angle between ????_???????? and ????_???? is (30°−????)

????₁= ????_????????????_????cos (30°−????)

Current through current coil of ????₂=????_????

Voltage across voltage coil of ????₂=????_????????=????_????????+????_????????=????_????????−????_????????

From phasor diagram, it is clear that phase angle between ????_???????? and ????_???? is (30°+????)

????₂= ????_????????????_????cos (30°+????)

But ????_????,=????_????=????_????

????_????????= ????_????????=????_????

????₁= ????_????????_????cos (30°−????)

????₂= ????_????????_????cos (30°+????)

????₁+????₂= ????_????????_????cos(30°−????)+????_????????_????cos (30°+????)

????₁+????₂= ????_????????_????(2 cos 30° cos ????)

P(active power) = ????₁+????₂= \sqrt{3}V_{L}I_{L}(cos \; \phi )

Thus the sum of two wattmeter reading gives three phase power

The Power fcator can be measured by two wattmeter method.

i) Lagging power factor

PF = cos φ = cos \left ( tan^{-1} \left ( \sqrt{3}\frac{W_{1}+W_{2}}{W_{1}-W_{2}} \right )\right )

ii) Leading Power factor

PF = cos φ = cos \left ( tan^{-1} \left (- \sqrt{3}\frac{W_{1}+W_{2}}{W_{1}-W_{2}} \right )\right )

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Ans.

Given :-

v(t) = 141.4 Sin (377t) V

To Find :-

i) Instantaneous voltage value at t=3ms

ii) The time taken for voltage to reach 70.7 V for first time

Solution :-

As v(t)= 141.4 sin 377 t

Comparing it with

V= V_m sin ωt

We get, ω= 2πf =377

f= 377/2π = 60 Hz

V_m (peak value ) = 141.4 V

V_{rms}=\frac{V_{m}}{\sqrt{2}}= \frac{141.4}{\sqrt{2}}= 99.98\approx 100 \; V

V_avg =0.63 V_m = 0.63 x 141.4 = 89.08 V

i) Instantaneous voltage value at t=3ms

v(t)= 141.4 sin 377 t

V(t) = 141.4 sin (377 x 3 ) = 124.02 V

ii) The time taken for voltage to reach 70.7 V for first time

v(t)= 141.4 sin 377 t

70.7 = 141.4 x sin (377) x t

t=1.71 ms

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Ans.

It states that “The maximum power is delivered from a source to a load when the load resistance is equal to the source resistance”.

The maximum power will be transferred to the load when load resistance is equal to the source resistance .

Proof:

From the fig.(a), I=\frac{V}{R_{s}+R_{L}}

Power delivered to the load RL = P = I² R_L = \left ( \frac{V}{R_{s}+R_{L}} \right )^{2}R_{L}

To determine the value of RL for maximum power to be transferred to the load,

\frac{dP}{dR_{L}}=0 \frac{dP}{dR_{L}}=\frac{d}{dR_{L}}\left ( \frac{V}{R_{s}+R_{L}} \right )^{2}R_{L} \frac{dP}{dR_{L}}=\frac{V^{2}[R_{s}+R_{L}]^{2}-(2R_{L}(R_{s}+R_{L}))}{(R_{s}+R_{L})^{4}}

(Rs+RL)2 – 2RL)Rs+RL)=0

Rs² + R_L² +2R_LR_s – 2RLRs – 2RL² = 0

R_s= R_L

Hence, the maximum power will be transferred to the load when load resistance is equal to the source resistance.
Hence Proved.

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Ans.

Given:- KVA rating = 5 KVA, V₁= 1000 V, V₂ = 200 V , f= 50 Hz

We have

V_o=1000 V , I_o=0.24 A , W_i= 90 W

V_sc= 50 V , I_sc= 5 A , W_sc = 110 W

To Find :- i. Equivalent circuit for transformer with circuit constant.
ii. Regulation at full load at 0.8 lagging
iii. KVA load for maximum efficiency

Solution :-

i) To calculate the equivalent circuit parameters

No load power = W_i= V_oI_o cos φ_o

cos\; \phi {o}=\frac{W{i}}{V_{o}I_{o}}=\frac{90}{1000\times 0.24}=0.375

φ_o= cos^-1(0.375) = 67.97

Sin φ_o= 0.92

I_w= I_o cos φ_o = 0.24 x 0.375 = 0.09 A

I_μ= I_o sin φ_o = 0.24 x 0.92 = 0.2208 A

R_{o}=\frac{V_{1}}{I_{w}}=\frac{1000}{0.09}=11111.11\; \Omega = 11.111 kΩ

X_{o}=\frac{V_{1}}{I_{\mu }}=\frac{1000}{0.22}=4545.45\; \Omega

From S.C test

Z_{02}=Z_{sc}=\frac{V_{sc}}{I_{sc}}=\frac{50}{5}=10 A

R_{02}=R_{sc}=\frac{W_{sc}}{I_{sc}^{2^{2}}}=\frac{50}{5}=2 Ω

X_{02}=\sqrt{(Z_{sc})^{2}-(R_{sc})^{2}}=\sqrt{(10)^{2}-2^{2}}=9.79\; \Omega Z_{01}=\frac{Z_{02}}{K^{2}} R_{01}=\frac{R_{02}}{K^{2}} X_{01}=\frac{X_{02}}{K^{2}}

Where K is transformation ratio

K= V₂/V₁ = 200/1000 = 0.2

Putting value of K to get value of R_o1, X_o1,Z_o1

Z_{01}=\frac{Z_{02}}{K^{2}} = \frac{10}{0.2^{2}} =250 Ω

R_{01}=\frac{R_{02}}{K^{2}} = \frac{2}{0.2^{2}}=50 Ω

X_{01}=\frac{X_{02}}{K^{2}}= \frac{9.79}{0.2^{2}}= 244.75 Ω

ii) Regulation at full load at 0.8 lagging

For this we need, I₁ = KVA/V₁ = 5 x 10³ / 1000 = 5 A

% \; Regulation=\frac{I_{1}(R_{01\; }cos\phi {1}+X{01\; }sin\phi {1})}{V{1}}\times 100 % \; Regulation=\frac{5(50\times 0.375+ 244.75\times 0.92)}{1000}\times 100= 1.21\; %

iii) KVA load for maximum efficiency

maximum efficiency occurs at such a load when P.F =1

Iron loss = Cu loss

\eta {F.L}=\frac{KVA\times 10^{3}\times cos\; \phi }{KVA\times 10^{3}\times cos\; \phi+W{i}+W_{cu}}\times 100

W_i=W_cu

So, W_i=2W_i

\eta _{F.L}=\frac{5\times 10^{3}\times 0.375 }{5\times 10^{3}\times 0.375+2\times90}\times 100= 91.24\; %

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Ans.

Given :- R= 10 Ω , L= 0.04 H , VL= 200 V and f= 50 Hz

For a star connected load ,

V_{ph}=\frac{V_{L}}{\sqrt{3}}=\frac{200}{\sqrt{3}}=115.47

XL = 2πfL = 2π x 50 x 0.4 = 12.6 Ω

Zph = R+jXL = 10 + j 12.6 = 16.08 \angle 51.56^o Ω

Power factor = cos (51.56) =0.273 (lagging)

I_{ph}=\frac{V_{ph}}{Z_{ph}}\frac{115.47}{16.08}=7.2 A

The active, apparent power is given by,

P=\sqrt{3}V_{L}I_{L}cos\phi = 1.73\times 200\times 7.2\times 0.273=680 W Q=\sqrt{3}V_{L}I_{L}sin \phi = 1.73\times 200\times 7.2\times sin (51.56)=2.4 KVAR

S= \sqrt{3}V_{L}I_{L} = 1.73 x 200 x 7.2 = 2.5 KVA

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Q.5.b) In the following circuit find R for maximum power delivered to it. Also find maximum power delivered P_max.

Ans.

According to the maximum power transfer theorem, maximum power will be delivered to R if it is equal to the equivalent resistance seen from R (with R opened) .

Let us find the Req seen from R with R opened and sources replaced by their internal impedances.

Hence, maximum power will be delivered to R, if R= 1 Ω.

To Find the power let us find the thevenin’s voltage of the circuit seen from R.

Now, converting 10 V- 2 Ω and 5 V -2 Ω into equivalent current sources.

Writing KCL equations as C and A

10-\frac{V_{c}}{1}-\frac{V_{c}-V_{A}}{1}-5 =0 2.5-\frac{V_{A}}{2}-\frac{V_{A}-V_{c}}{1}+5 =0

Simplifying we have,

V_A-2V_C = -5

-1.5 V_A + V_C = -2.5

Solving V_A=5 V and Vc = 5 V

Hence, V_th = V_A = 5 V as V_B is ground.

And as we know , Thevenin’s eqivalent ressitance is same as the Req, that we obtain for maximum power

R_th= R_eq = 1 Ω

Hence, thevenin’s equivalent is

P_max delivered to R = I² R

= \left ( \frac{V_{th}}{R_{th}+R} \right )^{2}R=\left ( \frac{5}{1+1} \right )^{2}\times 1=\frac{25}{4}

= 6.25 W

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Ans.

When the transformer secondary is connected to an inductive load, the current flowing in the secondary winding is lagging w.r.t secondary terminal voltage. Let us assume that the current is lagging by an angle of ɵ₂. Let,

R1 = Primary winding Resistance

X1 = Primary winding leakage Reactance

R2 = Secondary winding Resistance

X2 = Secondary winding leakage Reactance
The Phasor Diagram is as follows :

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Ans.

The given diagram is

Given voltage waveform is mathematically presented as ,

V= V_m sin θ

as at θ = π/3 V = 0.866 V_m

and at π=90^o V=V_m

Hence, the given waveform i s

V= V_m sin θ

i) V_{avg}=\frac{1}{\pi }\int_{ 0}^{\pi}V_{m}sin\theta \; d\theta

V_{avg}=\frac{V_{m}}{\pi }\left [ - Cos\;\theta \right ]^{\pi }_{0} V_{avg}=\frac{-V_{m}}{\pi }\left [ cos \pi - Cos\;0 \right ] V_{avg}=\frac{-V_{m}}{\pi }\left [ -1- 1 \right ] V_{avg}=\frac{-V_{m}}{\pi }\left [ -2\right ]

= \frac{2V_m}{\pi} =0.636 V_m

ii) V_RMS

V_{rms}=\sqrt{\frac{1}{\pi }\int_{0}^{\pi }(V_{m}\; sin\theta )^{2}d\theta } V_{rms}=\sqrt{\frac{V_{m}^{2}}{\pi }\int_{0}^{\pi }(\; sin^{2}\theta )d\theta } =\frac{V_{m}}{\sqrt{\pi }}\sqrt{\int_{0}^{\pi }sin^{2}\theta \; d\theta } =\frac{V_{m}}{\sqrt{\pi }}\sqrt{\int_{0}^{\pi }\frac{1-cos 2\theta }{2} \; d\theta } \frac{V_{m}}{\sqrt{\pi }}\sqrt{\frac{1}{2}\int_{0}^{\pi }d\theta -\frac{1}{2}\int_{0}^{\pi }cos \; 2\theta \; d\theta } =\frac{V_{m}}{\sqrt{\pi } } \sqrt{\frac{1}{2}\left [ \theta \right ]^{\pi }{0}-\frac{1}{2}\left [ \frac{Sin 2\theta }{2} \right ]^{\pi }{0}} =\frac{V_{m}}{\sqrt{\pi } } \sqrt{\frac{1}{2}\times \pi -\frac{1}{4}\times 0} =\frac{V_{m}}{\sqrt{\pi } } \times \sqrt{\frac{\pi }{2}} =\frac{V_{m}}{\sqrt{2 } } =0.707 \; V_{m}

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Ans.

Thevenin’s Theorem states that “Any linear circuit containing several voltages and resistances can be replaced by just one single voltage in series with a single resistance or impedance connected across the load“.

Norton’s Theorem states that – A linear active network consisting of independent or dependent voltage source and current sources and the various circuit elements can be substituted by an equivalent circuit consisting of a current source in parallel with a resistance. The current source being the short-circuited current across the load terminal and the resistance being the internal resistance of the source network.

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