A capacitor is charged by a constant current of 2 mA and results in a voltage incresase of 12 V in a 10 s interval. the value of capacitance is

Answer and solution of A capacitor is charged by a constant current of 2 mA and results in a voltage incresase of 12 V in a 10 s interval. the value of capacitance is ? 

51. In the circuit shown below I_S(t) = 4 sin 4t A. the I_C(t) will be

(a) 4.8 sin 4t A

(b) -4.8 sin 4t A

(c) 9.6 sin 4t A

(d) -9.6 sin 4t A

52. The current in a 50 mH inductor is given in figure. the inductor voltage is
53. The equivalent resistance R_EQ seen by the current source I_S in the circuit is

(a) 3/7

(b) 3/5

(c) 1

(d) 3

54. In the circuit shown in figure, I_in (t) = 300 sin 20t mA, for t > 0 Let C₁ = 40 uf, and C₂ = 30 uf. All capacitors are initially uncharged. the V_IN (t) would be

(a) -0.25 cos 20t V

(b) 0.25 cos 20t V

(c) -36 cos 20t mV

(d) 36 cos 20t mV

55. Both capacitors shown in figure are initally uncharged. At t = 2 ms the current I_C1 is

(a) 0.25 mA

(b) 1 mA

(c) 0.75 mA

(d) 3 mA

56. For the circuit shown in figure, the dependent source

(a) supplies 16 W

(b) absorbs 16 W

(c) supplies 32 W

(d) absorbs 32 W

57. A capacitor is charged by a constant current of 2 mA and results in a voltage incresase of 12 V in a 10 s interval. the value of capacitance is

(a) 0.75 mf

(b) 1.33 mf

(c) 0.6 mf

(d) 1.67 mf

58. The energy required to charge a 10 uf capacitor to 100 v is

(a) 0.10 j

(b) 0.5 j

(c) 5 x 10^-9 j

(d) 10 x 10^-9 j

59. For the circuit in figure the instantaneous current I₁(t) is

(a) 10_/3/2 <90⁰ A

(b) 10_/3/2 < – 90⁰ A

(c) 5 < 60⁰ A

(d) 5 < – 60⁰ A

60.Impedance Z as shown in figure is

(a) j 29

(b) j9

(c) j19

(d) j39

61. For the circuit shown in figure, thevenin’s voltage and thevenin’s equivalent resistance at terminals ab is

(a) 5 V and 2

(b) 7.5 V and 2.5

(c) 4 V and 2

(d) 3 V and 2.5

62. If R₁ = R₂ = R₄ and R₃ = 1. 1R in the bridge circuit shown in figure, then the reading in the ideal voltmeter connected between a and b is

(a) 0.238 V

(b) 0.138 V

(c) -0.238 V

(d) 1 V

63. The h-parameters of the circuit shown in figure are

(a) [0.1  0.1/-0.1  0.3]

(b) 10   1  0.05]

(c) 30  20   20  20]

(d) 10  1  -1  0.50]

64. A square pulse of 3 V amplitude is applied to RC circuit shown in figure. the capacitor is initially uncharged. the output voltage V₀ at time t = 2 s is

(a) 3 V

(b) -3 V

(c) 4 V

(d) -4 V

65. For the lattice circuit shown in figure Z_A = J2 and Z_B = 2 . the values of the open-circuit impedance parameters Z = [Z₁₁ Z₁₂   Z₂₁   Z₂₂] are

(a) [1 – j  1 + j/1 + j  1 + j]

(b) [1 – j + 1 + j/ -1 + j  1 – j]

(c) [1 + j  1 + j/ 1 – j  1 – j]

(d) [1 + j  -1 + j/-1 -j  1 + j]

66. The circuit shown in figure below has initial current I₁ (0^_) = 1 A through the inductor and an initial voltage V_C(0) = – 1 V across the capacitor. for input V(t) = u(t), the laplace transform of the current I(t) for t > 0 is

(a) s/s² + s + 1

(b) s + 2/s² + s + 1

(c) s – 2/s² + s + 1

(d) s – 2/s² + s – 1

67. In the circuit shown below, the switch is opened at t = 0 after long time. the current 1 I(t) for t > 0 is

(a) e^{-2.306 t}  + e^-0.869t A

(b) -e^-2.306t + 2e^-0.869t A

(c)e-^4.431t + e^-0.903t A

(d) 2e^-4.431t -e^-0.903t A

68. In the circuit shown below, switch is moved from position a to b at t = 0.

The I_L (t) for t > 0 is

(a) (4 – 6t) e^4t A

(b) (3 – 6t)e^4t A

(c) (3 – 9t) e^-5t A

(d) (3 – 8t) e^-5t A

69. In the circuit shown below a steady state has been established before switch closed.

Then I(t) for t > 0 is

(a) 0.73 e^-2t sin 4.58t A

(b) 0.89 e^-2t sin 6.38t A

(c) 0.73 e^-4t sin 4.58t A

(d) 0.89 e^-4t sin 6.38t A

70. The switch is closed after long time in the circuit shown below. the V(t) for t > 0 is

(a) -8 + 6e^-3t sin 4t V

(b) -12 + 4e^-3t cos 4t V

(c) -12 + (4 cos 4t + 3 sin 4t) e^-3t V

(d) -12 + (4 cos 4t + 6 sin 4t) e^-3t V

71. In the following circuit, I_L(0⁺) is

(a) zero

(b) -2 A

(c) 1 A

(d) -1 A

72. Parameters for a RLC circuit are R = 2, L = 1 H, C = 1F. if these are connected in series first and then in parallel. the system reponse for both the circuits will be

(a) under damped, undamped

(b) critically damped, over damped

(c) critically damped, under damped

(d) under-damped, critically-damped

73. In the given circuit below, values of I_L (0⁺) and dV_R(0⁺) /dt are, respectively

(a) 0.8 A, zerp

(b) zero, 320 V/s

(c) 0, zero

(d) 0.8 , 320 V/s

74. In the following circuit below, switch S is closed at t = 0. if response V_C(t) is critically damped then value of K is

(a) 3

(b) 2

(c) 1

(d) for any value of k, system is critically damped

75. In the given circuit below, values of dV_C(0⁺)/dt and V_L(0⁺) are

(a) 2 V/s, 20 V

(b) 2 V/s, 40 V

(c) zero, 40 V

(d) 2 V/s, zero

answers and solution

51. (d)

V_L = L dI                _L/dt, I_C = C dV_C/dt

V_C = 3V_L  I_C = 3LC d²i_l/dt = – 9.6 sin 4t A

52. (d)

V_L = L dI_L/dt

for 2 < t < 4,

V_L = (0.05) (-100 – 0/2) = – 2.5 V

For 4 < t < 8,

V_L = (0.05) (100 + 100/4) = 2.5 V

For 8 < t < 10.

V_L = (0.05) (0 – 100/2) = – 2.5 V

53. (d)

in the circuit R_EQ is given by

R_EQ = V_S/I_S

V_S = (I_S + V_X) (2 + 1) = 3I_S + 3V_X …………(1)

Current in 2 resistor

V_X/2 = I_S + V_X

-V_X/2 = I_S, V_X = – 2I ……………….(2)

From eq. (1) and (2), we get

V_S = 3I_S + 3(-2I_S)

V_S = 3I_S – 6I_S, V_S = – 3I_S

R_EQ = V_S/I_S = 3

54. (c)

the circuit is shown in figure below,

C_A = 30 x 60/30 + 60 = 20 uF

C_B = 30(20 + 40)/30 + 20 + 0 = 20 uF

We can say C_D = 20 uF and

C_EQ = 20 + 40 = 60 mF

V_C = 1/C

= 1/60 m (-300/20 cos 20t) x 10^-3

= – 0.25 cos 20t V

55. (c)

I_C1 = I_INC₁/C₁ + C₂ = 0.8 sin 600 t mA

at t = 2 ms, I_C1 = 0.75 mA

56. (d)

power P = VI = 2I_X x I_X = 2I_X²

I_X = 4 A, P = 32 W

57.(d)

V_T2 – V_T1 = 1/C |Idt

12 = 1/C 2m (t₂ – t₁)

12C = 2 m x 10

C = 1.67 mF

58. (b)

E = 1/2 CV²

5 x 10^-6 x 100² = 0.05 j

59. (a)

applying KCL at node A,

5 <0⁰ + (t) = 10 <60⁰

(t) = 10 <60⁰ – 5 <0⁰

= 5 + j5 /3 – 5

5/3 = 5/3 <90⁰ A

Alternate Method

in steady stete, capacitor is open-circuited while at t = 0 or initial capacitor is short-circuited.

I = 4/5 x 10 = 8 = 4

I = R_EQ C_EQ

= (|K||4 K) (4 uF||uF)

= 4/5 K x 5 uF

= 4 ms

B = initial value = 0 (given)

[capacitor is initially uncharged]

volt = final value – (final-initial) e^-t/t

= A – (A – B) e^-t/t = 8 – (8 – 0) e^-t/0.004

r₀(t) = 8 (1 – e^-t/0.004) V

60. (b)

in L₁ current enters in undotted terminal terminal whereas in L₃ current enters in dotted terminal.

hence, mutual inductance due to L₁ and L₃ is – 2 x j10. in L₂ current enters in dotted terminal and also in L₃ current enters in dotted terminal.

Hence, mutual inductance due to L₂ and L₃ is + 2 x j10.

impedance Z = j5 + j2 + j2 – j20 + j20 = j9

61. (b)

let V_OC be the open-circuit voltage and I_SC short-circuit current

V_TH = V_OC

R_TH = V_OC/I_SC

To calculate V_OC

Let node as reference node.

applying KCL at node

V_OC/5 + V_OC – 10/5 = 1

2V_OC /5 = 3

V_OC = 15/2 V

To calcuate I_SC

5 can be removed.

hence,  1A = I_SC = – 10/5

I_SC = 1 + 10/5 = 3A

R_TH = V_OC/I_SC = 15/2 x 3 = 5/2 = 2.5

62. (c)

voltage across R₂, i.e., V_A = 10 x R₂/R₁ + R₂

R₁ = R₂ = R

V_A = 5V

Voltage across R₃ , i.e., V_B  = 10 x R₃/R₄ + R₃

R₃ = 1.1 R₄

V_B = 10 x 1.1 x R₄(1.1 + 1)R₄ = 10 x 1.1/2.1 = 5.2380 V

Reading of ideal voltmeter = V_A – V_B = – 0.238 V

63. (d)

the voltage and current express h-parameters

[h₁₁ h₁₂  h₂₁  h₂₂] as

V₁ = h₁₁ I₁ + h₁₂V₂

I₂ = h₂₁ I₁ + h₂₂V₂

h₁₁ = V₁/I₁|_{V2 = 0}   = 10 = 10 [20 shorted]

h₁₂ = V₁/V₂|_{I1 = 0}  = 1[V₁ = V₂]

h₂₁ = I₂/I₁|_{V2 = 0}    =- 1[I₁ = – I₂]

h₂₂ = I₂/V₂|_{I1 = 0}   = 1/20 = 0.05

[h] = [10  1/ -1  0.05]

64. (b)

RC = 0.1 x 10^-6 x 1 x 10³

= 1 x 10^-4 = 0.1 ms

as (=2 s) >> RC

hence, steady state reached in 2 s.

the capacitor gets opne-circuited with voltage 3 V.

65. (d)

V₀ = 3 V

The voltage and current with Z-parameters as

V₁ = Z₁₁I₁ + Z₁₂I₂

V₂ = Z₂₁I₁ + Z₂₂I₂

Z₁₁ = V₁/I₁|_{I2 = 0}

As two similar circuits divide I, hence I = I₁/2

V₁ = I₁/2 Z_A + Z_B

Z₁₁ = Z_A + Z_N/2

Z₁₂ = V₁/I₂|_{I1 = 0}

I = I₂/2

V₁ = Z_A I₂/2 – Z_B I₂/2

Z₁₂ = 1/2 [Z_A – Z_B]

Simiarly, Z₂₁ = 1/2 [Z_A – Z_B]

Z₂₂ = 1/2 [Z_A + Z_B]

Putting the values Z_A and Z_B,

[Z] = [1 + J  J-1/J – 1 1 + J]

66. (b)

applying KVL in the loop,

V(t) = 1 x I(t) + 1 x LdI(t)/dt + 1/1

u(t) = I(t) + L dI/dt + |I dt

u(t) = I(t) + L dI/dt + I dt

taking laplace transform,

1/s = I(s) + sI (s) – I(0⁺) + I(s)/s + V(0⁺)/s

1/s = I(s) + sI (s) – 1 + I(s)/s – 1/s

I(s) = s + 2/s² + s + 1

67. (c)

I(0⁺) = 2 A, V_L (0⁺) = 2 x 10 x 5/10 + 15 = 4 V

R = 5||(10 + 10) = 4

3/4 dI(0⁺)/dt = 4 – 4 x 2 = – 4

0_o = 1/1_/3 3_/4 = 2

a = R/2L = 4/2^3/4 = 8/3

s = – 8/3 + 64/9 – 4 = – 4.431, – 0.903

I(t) = Ae^-4.431t + Be^-0.903t

I(0⁺) = 2 = A + B

dI(0⁺)/dt = -16/3 = – 4.431 A – 0.903 B

A = 1, B = 1

68. (c)

V_C(0) = 0, L (0) = 4 x 6/6 + 2 = 3

0.02 dV_C(0)/dt = I_L(0) = 3

dV_C(0)/dt = 150

a = 6 + 14/2 x 2 = 5, 0₀ = 1/2 x 0.02 = 5

a = 0_o critically damped

V(t) = 12 + (A + Bt) e^-5t

0 = 12 + A, 150 = – 5A + B

A = – 12, B = 90

V(t) = 12 + (90t – 12)e^-5t

I_L(t) = 0.02 (-5) e^-5t (90t – 12) + 0.02(90)e^-5t

= (3 – 9t)e^-5t

69. (a)

V(0⁺) = 100 x 5/5 + 5 + 20 = 50/3, I_L(0⁺) = 0

I_F = 0

dI_L(0⁺)/dt = 20 – 50/3 = 10/3

a = 4/2 x 1 = 2, 0_o = 1/1 x 1_/25 = 5

s = 2 + 4 – 25 = – 2 + j4.58

I(t) = (A cos 4.58t + B sin 4.58t) E^-2t

70. (c)

I_L(0⁺) = 0, V_L(0⁺) = 4 – 12 = – 8

1/25 dV_L(0⁺)/dt = I_L (0⁺) = 0

a = 6/2 = 3, 0_o = 1/1 x 1_/25 = 5

= – 3 + 9 – 25 = – 3 + j4

V₁(t) = – 12 + (A cos 4t + B sin 4t) e^-3t

V_L(0) = – 8 = 12 + A

A = 4

dI_L(0)/dt = 0 = – 3A + 4 B

B = 3

71. (d)

before switch is closed, equivalent circuit is (t < 0)

I_A = 10/2 = 5 A

Current I_L(0) is

I_L(0) = 10 – 3I_A/5

= 10 – 3 x 5/5 = – 5/5 = – 1A

I_L(0⁺) = I_L(0^–) = – 1 A

72. (c)

for series resonant circuit, damping factor

= R/2 C/L

2/2 1/1 = 1

For parallel resonant circuit

= 1/2R L/C = 1/2 x 2 1/1 = 1/4 < 1

73. (d)

for t < 0 equivalent circuit is

current I_L(0) = 1/40 + 10 x 40 = 0.8 A

V_C (0) = 0

At t = 0⁺, I_L and V_C does not change abruptly and equivalent circuit is

I_L(0⁺) = I_L(0) = 0.8 A

V_C(0⁺) = V_C(0) = 0

In the circuit,

-1 = V_R(0⁺)/40 = V_R(0⁺)/10 = 5/40 V_R(0⁺)

V_R (0⁺) = – 8 V

dV_R(t)/dt = d/dt [V_R(t) = d/dt [10 x I_R(t)]

= d/dt [10 x (-40 – V_C(t)/50]

= -0.2 dV_C (t)/dt = 0.2I_C(t)/C

I_C(0⁺) = -8/10 – I_L (0⁺)

= – 0.8 – 0.8 = – 1.6 A

So, V_R(0⁺)/dt = 0.2I_C(0⁺)/C

dV_R(0⁺)/dt = -0.2 x -1.6/1 x 10^-3 = 320 V/s

74. (a)

we can obtain thevenin equivalent across L and C as following (for t > 0)

writing node equation,

I_S = V_S – 12/1 x 10³ + V_S KV₁/1 x 10³

V₁ = 12 – V_S

So, I_S 1 x 10³ = V_S – 12 + V_S – K (12 – V_S)

1000I_S = (2 + K)V_S – (1 + K) 12

V_S = (1000/2 + K)I_S + (1 + K)/(2 + K) 12

For thevenin equivalent,

V_S = R_THI_S + V_TH

So, R_TH = 1000/2 + K

The equivalent circuit is

for critically damped response,

R²C = 4L

(1000/2 + K)² x 0.1 x 10^-6 = 4 x 10^-3

25 = (2 + k)², 2 + k = 5  k = 3

75. (a)

for t > 0 the circuit is sourcefree.

u(t) = 0, t < 0

I_L(0) = 0, V_C(0) = 0

At t = 0, the circuit is

u(t) = 1, t < 0

I_L and V_C cannot change instantaneously.

I_L = (0⁺) = I_L (0^–) = 0

V_C(0⁺) = V_C(0^–) = 0

V_A = (0⁺) = 4 x 10 = 40 v

I_C(0⁺) = V_A(0⁺) = 4 mA

I_C(0⁺) = V_A(0⁺) – 5000I_C

= 40 – 5000 x 4 x 10^-3 = 20 V

I_C(0⁺) = C dV_C(0⁺)/dt

4 x 10^-3 = 2 x 10^-3 dV_C(0⁺)

dV_C(0⁺)/dt = 2 V/s