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 ? **

- 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

- The current in a 50 mH inductor is given in figure. the inductor voltage is
- 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

- In the circuit shown in figure, I
_{in}(t) = 300 sin 20t mA, for t > 0 Let C_{1}= 40 uf, and C_{2}= 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

- 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

- 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

**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

- 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

- For the circuit in figure the instantaneous current I
_{1}(t) is

(a) 10_{/}3/2 <90^{0} A

(b) 10_{/}3/2 < – 90^{0} A

(c) 5 < 60^{0} A

(d) 5 < – 60^{0} A

60.Impedance Z as shown in figure is

(a) j 29

(b) j9

(c) j19

(d) j39

- 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

- If R
_{1}= R_{2}= R_{4}and R_{3}= 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

- 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]

- A square pulse of 3 V amplitude is applied to RC circuit shown in figure. the capacitor is initially uncharged. the output voltage V
_{0}at time t = 2 s is

(a) 3 V

(b) -3 V

(c) 4 V

(d) -4 V

- For the lattice circuit shown in figure Z
_{A}= J2 and Z_{B}= 2 . the values of the open-circuit impedance parameters Z = [Z_{11}Z_{12}Z_{21}Z_{22}] 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]

- The circuit shown in figure below has initial current I
_{1}(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/s^{2} + s + 1

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

(c) s – 2/s^{2} + s + 1

(d) s – 2/s^{2} + s – 1

- 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

- 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

- 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

- 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

- In the following circuit, I
_{L}(0^{+}) is

(a) zero

(b) -2 A

(c) 1 A

(d) -1 A

- 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

- 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

- 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

- 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

- (d)

V_{L} = L dI _{L}/dt, I_{C} = C dV_{C}/dt

V_{C} = 3V_{L} I_{C} = 3LC d^{2}i_{l}/dt = – 9.6 sin 4t A

- (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

- (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

- (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

- (c)

I_{C1} = I_{IN}C_{1}/C_{1} + C_{2} = 0.8 sin 600 t mA

at t = 2 ms, I_{C1} = 0.75 mA

- (d)

power P = VI = 2I_{X} x I_{X} = 2I_{X}^{2}

I_{X} = 4 A, P = 32 W

57.(d)

V_{T2} – V_{T1} = 1/C |Idt

12 = 1/C 2m (t_{2} – t_{1})

12C = 2 m x 10

C = 1.67 mF

- (b)

E = 1/2 CV^{2}

5 x 10^{-6} x 100^{2} = 0.05 j

- (a)

applying KCL at node A,

5 <0^{0} + (t) = 10 <60^{0}

(t) = 10 <60^{0} – 5 <0^{0}

= 5 + j5 /3 – 5

5/3 = 5/3 <90^{0} 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}

r_{0}(t) = 8 (1 – e^{-t/0.004}) V

- (b)

in L_{1} current enters in undotted terminal terminal whereas in L_{3} current enters in dotted terminal.

hence, mutual inductance due to L_{1} and L_{3} is – 2 x j10. in L_{2} current enters in dotted terminal and also in L_{3} current enters in dotted terminal.

Hence, mutual inductance due to L_{2} and L_{3} is + 2 x j10.

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

- (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

- (c)

voltage across R_{2}, i.e., V_{A} = 10 x R_{2}/R_{1} + R_{2}

R_{1} = R_{2} = R

V_{A} = 5V

Voltage across R_{3} , i.e., V_{B } = 10 x R_{3}/R_{4} + R_{3}

R_{3} = 1.1 R_{4}

V_{B} = 10 x 1.1 x R_{4}(1.1 + 1)R_{4} = 10 x 1.1/2.1 = 5.2380 V

Reading of ideal voltmeter = V_{A} – V_{B} = – 0.238 V

- (d)

the voltage and current express h-parameters

[h_{11} h_{12} h_{21} h_{22}] as

V_{1} = h_{11} I_{1} + h_{12}V_{2}

I_{2} = h_{21} I_{1} + h_{22}V_{2}

h_{11} = V_{1}/I_{1}|_{V2 = 0} = 10 = 10 [20 shorted]

h_{12} = V_{1}/V_{2}|_{I1 = 0} = 1[V_{1} = V_{2}]

h_{21} = I_{2}/I_{1}|_{V2 = 0} =- 1[I_{1} = – I_{2}]

h_{22} = I_{2}/V_{2}|_{I1 = 0} = 1/20 = 0.05

[h] = [10 1/ -1 0.05]

- (b)

RC = 0.1 x 10^{-6} x 1 x 10^{3}

= 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.

- (d)

V_{0} = 3 V

The voltage and current with Z-parameters as

V_{1} = Z_{11}I_{1} + Z_{12}I_{2}

V_{2} = Z_{21}I_{1} + Z_{22}I_{2}

Z_{11} = V_{1}/I_{1}|_{I2 = 0}

As two similar circuits divide I, hence I = I_{1}/2

V_{1} = I_{1}/2 Z_{A} + Z_{B}

Z_{11} = Z_{A} + Z_{N}/2

Z_{12} = V_{1}/I_{2}|_{I1 = 0}

I = I_{2}/2

V_{1} = Z_{A} I_{2}/2 – Z_{B} I_{2}/2

Z_{12} = 1/2 [Z_{A} – Z_{B}]

Simiarly, Z_{21} = 1/2 [Z_{A} – Z_{B}]

Z_{22} = 1/2 [Z_{A} + Z_{B}]

Putting the values Z_{A} and Z_{B},

[Z] = [1 + J J-1/J – 1 1 + J]

- (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/s^{2} + s + 1

- (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

- (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, 0_{0} = 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}

- (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}

- (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

V_{1}(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

- (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

- (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

- (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

- (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 10^{3} + V_{S} KV_{1}/1 x 10^{3}

V_{1} = 12 – V_{S}

So, I_{S} 1 x 10^{3} = 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_{TH}I_{S} + V_{TH}

So, R_{TH} = 1000/2 + K

The equivalent circuit is

for critically damped response,

R^{2}C = 4L

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

25 = (2 + k)^{2}, 2 + k = 5 k = 3

- (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