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

By  

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 ? 

  1. In the circuit shown below IS(t) = 4 sin 4t A. the IC(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

  1. The current in a 50 mH inductor is given in figure. the inductor voltage is
  2. The equivalent resistance REQ seen by the current source IS in the circuit is

(a) 3/7

(b) 3/5

(c) 1

(d) 3

  1. In the circuit shown in figure, Iin (t) = 300 sin 20t mA, for t > 0 Let C1 = 40 uf, and C2 = 30 uf. All capacitors are initially uncharged. the VIN (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

  1. Both capacitors shown in figure are initally uncharged. At t = 2 ms the current IC1 is

(a) 0.25 mA

(b) 1 mA

(c) 0.75 mA

(d) 3 mA

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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]

  1. 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

  1. For the lattice circuit shown in figure ZA = J2 and ZB = 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]

  1. The circuit shown in figure below has initial current I1 (0_) = 1 A through the inductor and an initial voltage VC(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

  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

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

The IL (t) for t > 0 is

(a) (4 – 6t) e4t A

(b) (3 – 6t)e4t A

(c) (3 – 9t) e-5t A

(d) (3 – 8t) e-5t A

  1. 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

  1. 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

  1. In the following circuit, IL(0+) is

(a) zero

(b) -2 A

(c) 1 A

(d) -1 A

  1. 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

  1. In the given circuit below, values of IL (0+) and dVR(0+) /dt are, respectively

(a) 0.8 A, zerp

(b) zero, 320 V/s

(c) 0, zero

(d) 0.8 , 320 V/s

  1. In the following circuit below, switch S is closed at t = 0. if response VC(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

  1. In the given circuit below, values of dVC(0+)/dt and VL(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

  1. (d)

VL = L dI                L/dt, IC = C dVC/dt

VC = 3VL  IC = 3LC d2il/dt = – 9.6 sin 4t A

  1. (d)

VL = L dIL/dt

for 2 < t < 4,

VL = (0.05) (-100 – 0/2) = – 2.5 V

For 4 < t < 8,

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

For 8 < t < 10.

VL = (0.05) (0 – 100/2) = – 2.5 V

  1. (d)

in the circuit REQ is given by

REQ = VS/IS

VS = (IS + VX) (2 + 1) = 3IS + 3VX …………(1)

Current in 2 resistor

VX/2 = IS + VX

-VX/2 = IS, VX = – 2I ……………….(2)

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

VS = 3IS + 3(-2IS)

VS = 3IS – 6IS, VS = – 3IS

REQ = VS/IS = 3

  1. (c)

the circuit is shown in figure below,

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

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

We can say CD = 20 uF and

CEQ = 20 + 40 = 60 mF

VC = 1/C

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

= – 0.25 cos 20t V

  1. (c)

IC1 = IINC1/C1 + C2 = 0.8 sin 600 t mA

at t = 2 ms, IC1 = 0.75 mA

  1. (d)

power P = VI = 2IX x IX = 2IX2

IX = 4 A, P = 32 W

57.(d)

VT2 – VT1 = 1/C |Idt

12 = 1/C 2m (t2 – t1)

12C = 2 m x 10

C = 1.67 mF

  1. (b)

E = 1/2 CV2

5 x 10-6 x 1002 = 0.05 j

  1. (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 = REQ CEQ

= (|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

  1. (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

  1. (b)

let VOC be the open-circuit voltage and ISC short-circuit current

VTH = VOC

RTH = VOC/ISC

To calculate VOC

Let node as reference node.

applying KCL at node

VOC/5 + VOC – 10/5 = 1

2VOC /5 = 3

VOC = 15/2 V

To calcuate ISC

5 can be removed.

hence,  1A = ISC = – 10/5

ISC = 1 + 10/5 = 3A

RTH = VOC/ISC = 15/2 x 3 = 5/2 = 2.5

  1. (c)

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

R1 = R2 = R

VA = 5V

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

R3 = 1.1 R4

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

Reading of ideal voltmeter = VA – VB = – 0.238 V

  1. (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]

  1. (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.

  1. (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 ZA + ZB

Z11 = ZA + ZN/2

Z12 = V1/I2|I1 = 0

I = I2/2

V1 = ZA I2/2 – ZB I2/2

Z12 = 1/2 [ZA – ZB]

Simiarly, Z21 = 1/2 [ZA – ZB]

Z22 = 1/2 [ZA + ZB]

Putting the values ZA and ZB,

[Z] = [1 + J  J-1/J – 1 1 + J]

  1. (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

  1. (c)

I(0+) = 2 A, VL (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

0o = 1/1/3 3/4 = 2

a = R/2L = 4/23/4 = 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

  1. (c)

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

0.02 dVC(0)/dt = IL(0) = 3

dVC(0)/dt = 150

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

a = 0o 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

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

= (3 – 9t)e-5t

  1. (a)

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

IF = 0

dIL(0+)/dt = 20 – 50/3 = 10/3

a = 4/2 x 1 = 2, 0o = 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

  1. (c)

IL(0+) = 0, VL(0+) = 4 – 12 = – 8

1/25 dVL(0+)/dt = IL (0+) = 0

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

= – 3 + 9 – 25 = – 3 + j4

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

VL(0) = – 8 = 12 + A

A = 4

dIL(0)/dt = 0 = – 3A + 4 B

B = 3

  1. (d)

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

IA = 10/2 = 5 A

Current IL(0) is

IL(0) = 10 – 3IA/5

= 10 – 3 x 5/5 = – 5/5 = – 1A

IL(0+) = IL(0) = – 1 A

  1. (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

  1. (d)

for t < 0 equivalent circuit is

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

VC (0) = 0

At t = 0+, IL and VC does not change abruptly and equivalent circuit is

IL(0+) = IL(0) = 0.8 A

VC(0+) = VC(0) = 0

In the circuit,

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

VR (0+) = – 8 V

dVR(t)/dt = d/dt [VR(t) = d/dt [10 x IR(t)]

= d/dt [10 x (-40 – VC(t)/50]

= -0.2 dVC (t)/dt = 0.2IC(t)/C

IC(0+) = -8/10 – IL (0+)

= – 0.8 – 0.8 = – 1.6 A

So, VR(0+)/dt = 0.2IC(0+)/C

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

  1. (a)

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

writing node equation,

IS = VS – 12/1 x 103 + VS KV1/1 x 103

V1 = 12 – VS

So, IS 1 x 103 = VS – 12 + VS – K (12 – VS)

1000IS = (2 + K)VS – (1 + K) 12

VS = (1000/2 + K)IS + (1 + K)/(2 + K) 12

For thevenin equivalent,

VS = RTHIS + VTH

So, RTH = 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

  1. (a)

for t > 0 the circuit is sourcefree.

u(t) = 0, t < 0

IL(0) = 0, VC(0) = 0

At t = 0, the circuit is

u(t) = 1, t < 0

IL and VC cannot change instantaneously.

IL = (0+) = IL (0) = 0

VC(0+) = VC(0) = 0

VA = (0+) = 4 x 10 = 40 v

IC(0+) = VA(0+) = 4 mA

IC(0+) = VA(0+) – 5000IC

= 40 – 5000 x 4 x 10-3 = 20 V

IC(0+) = C dVC(0+)/dt

4 x 10-3 = 2 x 10-3 dVC(0+)

dVC(0+)/dt = 2 V/s