solution or answer :

An AC source of rms voltage 20 V with internal impedance Z_{S} = (1 + 2j) feeds a load of impedance Z_{L} = (7 + 4J) in the figure below. the reactive power consumed by the load is ?

**Question and Answers given below (with solution)** :

- In the circuit shown, the switch S is open for a long time and is closed at t = 0. the current I(t) for t > 0
^{+}is

(a) I(t) = 0.5 – 0.125 e^{-1000t} A

(b) I(t) = 1.5 – 0.125 e^{-1000t} A

(c) I(t) = 0.5 – 0.5 e^{-1000t} A

(d) I(t) = 0.375 e^{-1000t} A

- The current I in the circuit shown is

(a) -j1A

(b) j1A

(c) zero

(d) 20 A

- In the circuit shown, the power supplied by the voltage source is

(a) zero

(b) 5 W

(c) 10 W

(d) 100 W

**An AC source of rms voltage 20 V with internal impedance Z**_{S}= (1 + 2j) feeds a load of impedance Z_{L}= (7 + 4J) in the figure below. the reactive power consumed by the load is

(a) 8 VAR

(b) 16 VAR

(c) 28 VAR

(d) 32 VAR

- The switch in the circuit shown was on position a for a long time and is moved to position b at time t = 0. the current I(t) for t > 0 is given by

(a) 0.2 e^{-125t}u(t) mA

(b) 20 e^{-1250t} u(t) mA

(c) 0.2 e^{-1250t}u(t) mA

(d) 20 e^{-1000t} u(t) mA

- In the circuit shown, what value of R
_{L}maximizes the power delivered to R_{L}?

(a) 2.4

(b) 8/3

(c) 4

(d) 6

- The thevenin equivalent impedance Z
_{th}between the nodes P and Q in the following circuit is

(a) 1

(b) 1 + s + 1/s

(c) 2 + s + 1/s

(d) s^{2} + s + 1/s^{2} + 2s + 1

- The driving point impedance of the following network is given by Z(s) = 0.2s/s
^{2}+ 0.1s + 2 the component values are

(a) L = 5 H, R = 0.5, C = 0 1F

(b) L = 0.1 H, R = 0.5, C = 5 F

(c) L = 5 H, R = 2, C = 0.1 F

(d) L = 0.1 H, R = 2, C = 5 F

- In the following circuit the value of open circuit volage and thevenin resistance at terminal ab are

(a) V_{OC} = 100 V, R_{TH} = 1800

(b) V_{OC} = 0 V, R_{TH} = 270

(c) V_{OC} = 100 V, R_{TH} = 90

(d) V_{OC} = 0 V, R_{TH} = 90

- In the following linear circuit it is given that V
_{AB}= 4 V for R_{L}10 K and V_{AB}= 1 V for R_{L}= 2 K.

The values of thevenin resistance and voltage for the network N are

(a) 16 k, 30 V

(b) 30 k, 16 V

(c) 3 k, 6 V

(d) 50 k, 30 V

- Consider the following linear circuit. three sources have fixed value only I
_{S1}can be varied.

It is given that if I_{S1} = 3 mA then V_{OUT} = 16 V and if I_{S1} = 1 mA then V_{OUT} = 8 V. if I_{S1} = 0.5 mA, then the value of V_{OUT} is

(a) 6 V

(b) 8 V

(c) 4 V

(d) 12 V

- For the circuit shown in figure below the value of R
_{TH}is

(a) 100

(b) 136.4

(c) 200

(d) 272.8

- Consider the following circuits shown below.

The relation betwwwn I_{A} and I_{B} is

(a) I_{B} = I_{A} + 6

(b) I_{B} = I_{A} + 2

(c) I_{B} = 1.5I_{A}

(d) I_{B} = I_{A}

- The circuit shown in the figure is used to charge the capacitor C alternately from two current sources as indicated. the switches S
_{1}and S_{2}are mechanically coupled and connected as follows For 2n T < t <(2n + 1) T, (n = 0,1,2,……..) S_{1}to P_{1}and S_{2}to P_{2}

For (2n + 1) T < t < (2n + 2) T, (n = 0, 1, 2,……) S_{1} to Q_{1} and S_{2} to Q_{2}

Assume that the capacitor has zero initial charge. given that u(t) is a unit step function, the voltage V_{C} (t) across the capacitor is given by

(a) _{n – 0} (-1)^{n} tu (t – nT)

(b) u(t) + 2 _{n = 0} (-1)^{n} u (t – nT)

(c) tu (t) + 2 _{n = 0}b(-1)^{n} (t – nT) U (t – nt)

(d) _{n = 0} [0.5 – e^{-(t – 2nt}) + 0.5 e^{-(t – 2nT – T)}]

The following series RLC circuit with zero initial condition is excited by a unit impulse function (t).

- For t > 0, the output voltage V
_{C}(t) is

(a) 1/3 (e^{-2/3t} – e^{3/2t})

(b) 2/3 te^{-1/2t}

(c) 2/3 e^{-1/2 t} cos (3/2t)

(d) 2/3 e^{-1/2t} sin (3/2t)

- For t > 0. the voltage across the resistor is

(a) 1/3 (e^{-3/2t} – e^{-1/2t})

(b) e^{-1/2t} [cos (3t/2) – 1/3 sin (3/2)]

(c) 2/3 e^{-1/2t} sin (3t/2)

(d) 2/3 e^{-1/2t} cos (3t/2)

- Two series resonant filters are as shown in the figure. let the 3 dB bandwidth of filter 1 be B
_{1}and that of filter 2 be B_{2}. the value of B_{1}/B_{2}is

(a) 4

(b) 1

(c) 1/2

(d) 1/4

- For the circuit shown in the figure, the thevenin voltage and resistance looking into X-Y are

(a) 4/3 V, 2

(b) 4 V, 2/3

(c) 4/3 V, 2/3

(d) 4V, 2

- In the circuit shown V
_{C}is 0 volt at = 0 s. for t < 0, the capacitor current I_{C}(t) where t is in second, is given by

(a) 0.50 exp (-25t) mA

(b) 0.25 exp (-25t) mA

(c) 0.50 exp (-12.5t) mA

(d) 0.25 exp (-6.25t) mA

- In the AC network shown in the figure, the phasor voltage V
_{AB}(in volt) is

(a) zero

(b) 5 < 30^{0}

(c) 12.5 <30^{0}

(d) 17 <30^{0}

- In the following circuit capacitor is initially uncharged. At t = 0
^{+}the value of d^{2}v_{c}/dt^{2}and d^{3}v_{c}/dt^{3}

(a) 2 V/s^{2}, -8 V/S^{3}

(b) -8 V/S^{2} , 26 V/S^{3}

(c) -2 V/S^{2}, 8 V/S^{3}

(d) 8 V/S^{2, }-26 V/S^{3}

- A square pulse of 3 V amplitude is applied to CR circuit shown below. the capacitor is initially uncharged. the output voltage V
_{2}at time t = 2 s is

(a) 3 V

(b) -3 V

(c) 4 V

(d) -4 V

- In the given circuit below initially capacitor is uncharged. the V
_{C}(t) for t >0 is

(a) (8 – 8e^{-t}) V

(b) (8 + 8e^{-t}) V

(c) 8 V

(d) (-8 + 8e^{t}) V

- In the circuit given below V
_{IN}(t) = 10 u(t) , the current I_{L}(t) is

(a) (-0.01 + 0.01e^{-5000t})A

(b) 0.1 A

(c) (-0.1 – 0.1e^{-5000t}) A

(d) 0.2 A

- In the following circuit the initial values are V
_{C1}(0^{–}) = 6 V and V_{C2}(0^{–}) = 24 V. the voltage V_{O}(t) for t > 0 is

(a) 38 – 8e^{ -t/1.2} V

(b) 8 + 22e^{ -t/1.2 }V

(c) 8 + 22e^{ – 3.75t} V

(d) 38 – 8e^{ -t/1.2 }V

### Answers with solution

- (a)

at steady state,

inductor becomes short-circuit.

I(t) at steady state

I(t) = 1.5/3 = 0.5

L_{EQ} = 15 mH

R_{EQ} = 5 + 10 = 15

L = L_{EQ}/R_{EQ} = 15 x 10^{3}/15 = 1/1000

I(t) A – (A – B)e^{-t/10}

= 0.5 – (0.5 – B) e^{-1000t}

= 0.5 (0.5 – 0.375) e^{-1000t}

I(t) = 0.5 – 0.125 e^{-1000t}

when switch is closed for a long time then inductor is short and current through inductor 1.5/2 = 0.75 A

current through inductor remains same after closing the switch at t = 0.

so, current through 10

= 0.75/2 = 0.375

- (a)

X_{EQ} = sL + R x 1/sC/ R + 1/sC

= sL + R / 1 + sRC

I_{O} = V/X_{EQ}

I = X_{C}/X_{C} + R I_{O}

= 1/sC/1_{/}sC + R x V/sL (1 + sRC) + R/1 + sRC

= 1/1 + sRC x V/sL (1 + sRC) + R/(1 + sRC)

= V/ sL (1 + sRC) + R

= V/j x 10^{3} x 20 x 10^{-3} (1 + j x 10^{3} x 50 x 10^{-6 }+ 1)

V/20j (1 + j50 x 10^{-3}) + 1

= V/20j – 1 + 1 = 20/20j = – j1A

- (a)

let current supplied by voltage source. the current in different branch is indicated in circuit using Kirchhoff’s current law.

applying KVL in outer loop,

10 – (I + 3) x (1 + 1) – (I + 2) x 2 = 0

10 – 2(I + 3) – 2 (I – 2) = 0

I = 0

VI = 0

- (b)

current I = V/Z_{L} + Z_{S} = 20 <0^{0}/8 + 6J

= 20/8^{2} + 6^{2} = <0^{0}/<tan^{-1}3/4

= 20/10 < – tan^{-1}3/4

2 < – tan^{-1}3/4

power consumed by load = |I|^{2}Z_{L}

= 4 (7 + 4J)

= 28 + J16

The reactive power = 16 VAR

- (c)

C_{EQ} = 0.8 x 0.2/0.8 + 0.2 = 0.16

when the switch is at position a for a long time then voltage across capacitor

V_{C}(t = 0) = 100 V

as capacitor acts as open-circuit

at t > 0

the discharging current

I(t) = V_{O}/R e^{-t/rc} for t > 0

= 100/5000 e^{-t/5 x 103 x 0.16 x 10-6}

= 0.020 e^{-125t} A

on comparing RL = 0.2

RLC = 1

L = 0.1 H

R = 2

C = 5 F

- (c)

Maximum power delivered to R_{L} when load resistance equals to thevenin resistance of circuit.

R_{L} = R_{TH} = V_{OC}/I_{SC}

due to open-circuit V_{OC} = 100 V

I_{SC} = I_{1} + I_{2}

Applying KVL in lower loop.

100 – 8I_{1} = 0

I_{1} = 100/8 = 25/2

V_{X} = – 4I_{1} = – 4 x 25/2 = – 50 V

100 + V_{X} – 4I_{2} = 0

I_{2} = 100 – 50/4 = 25/2

I_{SE} = I_{1} + I_{2}

= 25/2 + 25/2 = 25

R_{TH} = V_{OC}/I_{SC} = 100/25 4

R_{L} = R_{TH} = 4

- (a)

to calculate thevenin resistance, all the current sources get open-circuited and voltage sources short-circuited. writing in impedance of inductor and capacitor.

R_{TH} = (1/S + 1) ||(1 + S)

(1 + S) (1/S + 1)/1 + S + 1/S + 1 = (1/S + 1 + 1 + S)/(1/S + 1 + 1 + S)

R_{TH} = 1

- (d)

dividing point impedance = R ||1/sC||sL

= {(R) (1/sC)/R + 1/sC}||sL

(R/1 + sRC) = sRL/S^{2}RLC + sL + R

R/1 + sRC + sL

Z(s) = 0.2s/s^{2} + 0.1s + 2

- (d)

by writing loop equation for the circuit,

V_{S} = V_{X},I_{S} = I_{X}

I_{S} = Test source to find out thevenin equivalent

V_{S} = 600(I_{1} – I_{2}) + 300 (I_{1} – I_{2}) + 900(I_{1})

V_{S} = (600 + 300 + 900)I_{1} – 600 I_{2} – 300 I_{3}

V_{S} = 1800I_{1}– 600I_{2} – 300I_{3}

I_{1} = I_{S},I_{2} = 0.3 V_{S}

I_{3} = 3I_{S} + 0.2V_{S}

V_{S} = 1800I_{S} – 600(0.01V_{S}) -300 (3I_{S} + 0.01V_{S})

V_{S} = 1800I_{S} – 6V_{S} – 900I_{S} – 3V_{S}

10V_{S} = 900I_{S}

V_{S} = 90I_{2}

For thevenin equivalent

V_{S} = R_{TH} I_{S} + V_{OC}

So, thevenin voltage V_{OC} = 0 V

Resistance R_{TH} = 90

- (b)

the circuit is shown in figure below.

when R_{L} = 10 K and V_{AB} = 4 V,

Current in circuit

I – V_{AB}/R_{L} = 4/10 = 0.4 mA

thevenin voltage is given by

V_{TH} = I(R_{TH} + R_{L})

V_{TH} = 0.4 (R_{TH} + 10)

V_{TH} = 0.4 R_{TH} + 4…………..(1)

Simiarly,

R_{L} = 2 K, V_{AB} = 1 V

I = 1/2 = 0.5 mA

V_{TH} = 0.5 (R_{TH} + 2)

V_{TH} = 0.5R_{TH} + 1……………(2)

From eq. (1) and (2)

0.1R_{TH} = 3

R_{TH} = 30 K

V_{TH} = (12 + 4) = 16 V

- (a)

for a linear circuit, we can write

V_{OUT} = AI_{S1} + BV_{S1} + CI_{S2} + DI_{S3}

= AI_{S1} + (BV_{S1} + CI_{S3} + DI_{S3})

V_{OUT} = AI_{S1} + K

(V_{S1}, I_{S2} and I_{S3} are fixed)

16 = A x 3 + K

8 = A x 1 + K

2A = 8

A = 4

4 + k = 8, K = 4 V

so, the equation for V_{OUT} is

V_{OUT} = 4I_{S1} + 4

I_{S1} = 0.5 mA

V_{OUT} = 4 x 0.5 + 4 = 6 V

- (a)

The circuit is shown below,

I_{X} = 1 A, V_{X} = V_{TEST}

V_{TEST} = 100(1 – 2I_{X}) + 300 (1 – 2I_{X} – 0.01V_{X}) + 800

V_{TEST} = 1200 – 800I_{X} – 3V_{TEST}

4V_{TEST} = 1200 – 800 = 400

V_{TEST} = 100 V

R_{TH} = V_{TEST}/1 = 100

- (c)

in circuit (b) transforming the 3 A source is to 18 V source all sources are 1.5 times of that in circuit (a). hence, I_{B} = 1.5 I_{A}.

- (c)

for 2nt < t < (2n + 1) t, n = 0, 1, 2, 3…………….

the capacitor gets charge to its peak value. as the constant current is flowing, the voltage across capacitor is ramp function.

for (2n + 1) T < t (2n + 2) T

the capacitor get discharge from peak value with same slope as for charging to zero value.

- (d)

in frequency domain,

V_{C}(S) = t/s/1_{/}s + 1 + s

V_{I} (t) = s(t)

V_{I}(s) = 1

V_{C}(s) = 1/s_{2} + s + 1

= 2/3 |B_{/}2/(s + 1_{/}2)^{2} + (3_{/}2)^{2}|

taking inverse laplace transform,

V_{C}(t) = 2/3 sin (3/2 t) .e^{-1/2t}

- (b)

in frequency domain

V_{R}(s) = 1. V_{I}(s)/1_{/}s + 1 + s

s/s^{2} + s + 1

V_{R}(s) = s + 1_{/}2 – 1_{/}2/(s + 1_{/}2)^{2} + 3_{/}4

= (s + 1_{/}2)/(s + 1_{/}2)^{2} + (3_{/}2)^{2} – 1_{/}2 /(s + 1_{/}2)^{2} + (3_{/}2)^{2}

taking inverse laplace transform,

V_{R}(t) = e^{-t/2} cos 3/2 t -1/2 x 2/3 e^{-t/2} sin 3/2 t

= e^{-t/2} [cos 3/2 + 1/3 sin 3/2 t]

- (d)

for series resonant circuit, 3 dB bandwidth is R/L

B_{1} = R/L_{1}

B_{2} = R/L_{2} = R/L_{1}/4 = 4R/L_{1}

B_{1}/B_{2} = 1/4

Note bandwidth of series resonant circuit is independent from value of capacitor.

R_{TH} = V_{OC}/I_{SC}

V_{TH} = V_{OC}

Applying KCL at node A,

2I – V_{TH}/1 + 2 = I + V_{TH}/2 ……………(1)

I = V_{TH}/1

Putting 2V_{TH} – V_{TH} + 2 = V_{TH} + V_{TH}/2

V_{TH} = 4 V

When XY get shorted, 2A current flows through short-circuited path.

R_{TH} = 4/2 = 2

- (a)

the capacitor voltage

V_{C}(T) = V_{C}(_{00}) – (V_{C}(_{00}) – V_{C}(0)) e^{-t/req c}

R_{EQ} = 20 K || 20 K

= 10 K

V_{C} (_{00}) = 10 x 20/20 + 20 = 5 V

given V_{C}(0) = 0

V_{C}(t) = 5 – (5 – 0) e^{-t/10 x 4 x 10-6 x 103}

V_{C}(t) = 5 (1 – e^{-25t})

I_{C}(t) = C dV_{C}(t)/dt = 4 x 10^{-6} d/dt 5 (1 – e^{-25t})

= 4 x 10^{-6} x 5 x 25 e^{-25t}

I_{L}(T) = 0.50 e^{-2.5t} mA

- (d)

equivalent impedance

= (5 + j3) ||(5 – 3)

= 5 + j3) (5 – j3)/5 + j3 + 5 – j3

= 25 + 9/1 = 3.4

V_{AB} = Current x impedance

= 5 <30^{0} x 3.4

= 17 <30^{0}

- (b)

at t = 0^{+}, V_{C}(0^{+}) = 0

V_{C}/20 + V_{C} – e^{-1}/10 + 1/20 . dV_{C}/dt = 0 ……………(1)

dV_{C}/dt + 3V_{C} = 2e^{-t} ……………….(2)

at t = 0^{+} dV_{C}(0^{+})/dt = 2 v/s

differentiating eq. (1), d^{2}V_{C}/dt^{2} + 3 dV_{C}/dt = – 2e^{-t} …………..(3)

at t = 0^{+}, d^{2}V_{C}(0^{+})/dt^{2} = -2 – 6 = – 8 v/s^{2}

differentiating eq. (ii), d^{2}V_{C}/dt^{3} + 3 d^{2}V_{C}/dt = 2e^{-t}

at t = 0^{+}, d^{3}v_{c}(0^{+})/dt^{2} = 2 + 24 = 26 v/s^{3}

- (b)

RC = 0.1 x 10^{-6} x 10^{3} = 10^{-4} s

as RC is very small, so steady state will be reached in 2 s.

thus, V_{C} = 3 V and V_{2} = -V_{C} = – 3 V

- (d)

by source transformation, we have equivalent circuit shown below.

thevenin equivalent for above circuit can be obtained as (by putting a test source)

by writing KVL,

V_{S} – I_{S} 0.5 – 9V_{X} + 8 – 500I_{S} = 0

V_{S} = 5I_{S} + 9V_{X} – 8 + 500I_{S} …………….(1)

V_{X} = V_{S} + 8 – 500I_{S}

V_{S} = 5I_{S} + 9[V_{S} + 8 – 500I_{S}] – 8 + 500I_{S}

V_{S} = 1000I_{S} + 64 + 9V_{S}

V_{S} = 125 I_{S} – 8

For thevenin’s equivalent circuit

V_{S} = R_{TH}I_{S} + V_{TH}

V_{TH} = – 8 V, R_{TH} = – 125

The circuit is

V_{C}(t) = V_{C}(_{00}) + [V (0) – V(_{00})]e^{-1/rc}

V_{C}(t) = – 8 + (0 + 8) e^{-t/-125 x 8 x 10-3}

here, time constant is negative.

V_{C}(t) = – 8 + 8e^{t}

- (a)

thevenin equivalent across the inductor for t > 0 can be obtained as

by applying node equation,

I_{S} = 0.01V_{S} + (V_{S} – V_{A} + V_{IN})/100

100I_{S} = 1. V_{A} + V_{S} – V_{A }+ V_{IN}

V_{S} = – V_{IN} + 100I_{S}

For thevenin’s equivalent circuit,

V_{S} = V_{TH }+ I_{S}R_{TH}

V_{TH} = – V_{IN} = – 10 V.

R_{TH} = 100

t < 0, I_{L}(0^{–}) = 0

I_{L}(_{00}) = -10/100 = – 0.1A

I_{L}(t) = 0.1 [1 – e^{-t}] where = time constant

= L/R_{TH} = 20 x 10^{-3}/100

I_{L}(t) = 0.1 [1 – e^{-5000t}]

= (-0.1 + 0.1 e^{-5000t}) A

- (a)

let 1/sensitivity = 1/20 = 50 uA

for 0 – 10 V scale, R_{M} = 10 x 20 = 200 k

for 0 – 50 V scale, R_{M} = 50 x 20 = 1 m

for 4 V reading I = 4/10 x 50 = 20 uA

V_{TH} = 20 R_{TH} + 20 x 200 = 4 + 20uR_{TH}……………..(1)

For 5 V reading I = 5/50 x 50 = 5uA

V_{TH} = 5 x R_{TH} + 5 x 1m = 5 + 5R_{TH} ……………….(2)

Solving eqs. (1) and (2)

V_{TH} = 16/3 V. R_{TH} = 200/3 K