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

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

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

The current I in the circuit shown is

(a) -j1A

(b) j1A

(d) 20 A

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

(a) zero

(b) 5 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

(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^-125tu(t) mA

(b) 20 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

(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

(d) s² + s + 1/s² + 2s + 1

The driving point impedance of the following network is given by Z(s) = 0.2s/s² + 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

(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

(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

(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

(d) 12 V

For the circuit shown in figure below the value of R_TH is

(a) 100

(b) 136.4

(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

(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 S₁ and S₂ are mechanically coupled and connected as follows For 2n T < t <(2n + 1) T, (n = 0,1,2,……..) S₁ to P₁ and S₂ to P₂

For (2n + 1) T < t < (2n + 2) T, (n = 0, 1, 2,……) S₁ to Q₁ and S₂ to Q₂

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)

(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

(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)]

(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 B₁ and that of filter 2 be B₂ . the value of B₁/B₂ is

(a) 4

(b) 1

(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

(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

(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 < 30⁰

(d) 17 <30⁰

In the following circuit capacitor is initially uncharged. At t = 0⁺ the value of d²v_c/dt² and d³v_c/dt³

(a) 2 V/s², -8 V/S³

(b) -8 V/S² , 26 V/S³

(d) 8 V/S^2,-26 V/S³

A square pulse of 3 V amplitude is applied to CR circuit shown below. the capacitor is initially uncharged. the output voltage V₂ at time t = 2 s is

(a) 3 V

(b) -3 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

(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

(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/1.2V

(d) 38 – 8e^-t/1.2V

Answers with solution

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 10³/15 = 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

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 10³ x 20 x 10^-3 (1 + j x 10³ x 50 x 10^-6+ 1)

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

= V/20j – 1 + 1 = 20/20j = – j1A

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

current I = V/Z_L + Z_S = 20 <0⁰/8 + 6J

= 20/8² + 6² = <0⁰/<tan^-13/4

= 20/10 < – tan^-13/4

2 < – tan^-13/4

power consumed by load = |I|²Z_L

= 4 (7 + 4J)

= 28 + J16

The reactive power = 16 VAR

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

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 = I₁ + I₂

Applying KVL in lower loop.

100 – 8I₁ = 0

I₁ = 100/8 = 25/2

V_X = – 4I₁ = – 4 x 25/2 = – 50 V

100 + V_X – 4I₂ = 0

I₂ = 100 – 50/4 = 25/2

I_SE = I₁ + I₂

= 25/2 + 25/2 = 25

R_TH = V_OC/I_SC = 100/25 4

R_L = R_TH = 4

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

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

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

(R/1 + sRC) = sRL/S²RLC + sL + R

R/1 + sRC + sL

Z(s) = 0.2s/s² + 0.1s + 2

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(I₁ – I₂) + 300 (I₁ – I₂) + 900(I₁)

V_S = (600 + 300 + 900)I₁ – 600 I₂ – 300 I₃

V_S = 1800I₁– 600I₂ – 300I₃

I₁ = I_S,I₂ = 0.3 V_S

I₃ = 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 = 90I₂

For thevenin equivalent

V_S = R_TH I_S + V_OC

So, thevenin voltage V_OC = 0 V

Resistance R_TH = 90

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

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

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

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.

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.

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/s₂ + 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

in frequency domain

V_R(s) = 1. V_I(s)/1_/s + 1 + s

s/s² + 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]

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

B₁ = R/L₁

B₂ = R/L₂ = R/L₁/4 = 4R/L₁

B₁/B₂ = 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

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

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 <30⁰ x 3.4

= 17 <30⁰

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), d²V_C/dt² + 3 dV_C/dt = – 2e^-t …………..(3)

at t = 0⁺, d²V_C(0⁺)/dt² = -2 – 6 = – 8 v/s²

differentiating eq. (ii), d²V_C/dt³ + 3 d²V_C/dt = 2e^-t

at t = 0⁺, d³v_c(0⁺)/dt² = 2 + 24 = 26 v/s³

RC = 0.1 x 10^-6 x 10³ = 10^-4 s

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

thus, V_C = 3 V and V₂ = -V_C = – 3 V

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_THI_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

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_SR_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

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