A two-port network is represented by ABCD-parameters given by If port 2 is terminated by RL, the input impedance seen at port 1 is given by

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solution and answer of question A two-port network is represented by ABCD-parameters given by If port 2 is terminated by RL, the input impedance seen at port 1 is given by ? 

  1. In the circuit shown below voltage source 6 V has been applied for a long time. At t = 0 switch is opened. the VO (t) for t > 0 is

(a) 3e-4t V

(b) -6e – 4t V

(c) 6e – 4t V

(d) -3e -4t V

  1. In the following circuit, the switch S is closed at t = 0. the rate of change of current dI/dt (0+) is given by

(a) zero

(b) RSIS/L

(c) (R + RS)IS/L

(d) infinite

  1. In the following circuit, the switch is closed at t = 0 with the capacitor uncharged. the value of dil /dt and d2 ll/dt2 at t = 0+ is

(a) -100 a/s, 105 a/s2

(b) 105 a/s, – 107 a/s2

(c) -100 a/s, -105 a/s2

(d) 105 a/s, 107 a/s2

  1. The circuit shown below reaches a steady state with the switch closed. At t = 0, switch is opened. for t > 0, VO (t) is

(a) 9.6 e-9.6t V

(b) -9.6 e9.6t V

(c) 2.4 e-2.4t V

(d) -2.4 e2.4t V

Statements for linked answer questions 30 and 31

In the following circuit shown below, the 80 v source has been applied for a long time. the switch is opened at t = 90 ms.

  1. At t = 0+ , the current I1 (0+) is

(a) 0.25 A

(b) 0.17 A

(c) 0.05 A

(d) 0.2 A

  1. At t = 80 ms the current I1 (80 ms) is

(a) 20.3 mA

(b) 8.25 mA

(c) 1.84 mA

(d) 6.98 mA

  1. In the circuit shown below, if VIN = 25u(t) – 25u(-t) v, the IL(t) for t > 0 is

(a) 0.4 (1 – e-250t) A

(b) 0.2 (1 – e6250t) A

(c) 0.2 (1 – e-250t) A

(d) 0.4 (1 – e-6250t) A

  1. In the circuit shown below the switch is opened at t = 0 after a long time. the voltage V0(t) for t > 0 is

(a) (6 + 2e-2.5t) V

(b) 10 – 6e-2.5t) V

(c) (6 + 4e-2.5t) V

(d) (10 – 2e2.5t) V

  1. In the network shown below, the switch is opened at t = 0. at t = 0+, the value of d2VR Idt2 is

(a) -1010 v/s2

(b) 1010 v/s2

(c) 10+13 v/s2

(d) -10+13 v/s2

  1. In the following circuit, the current source 6 ma has been applied for a long tirme before the switch open at t = 0. the current IL(0+) is

(a) 4 A

(b) 4.8 A

(c) 2 A

(d) R and L are required

  1. In the following circuit, the 12 V source has been applied for a long time before the switch opens at t = 0. at t = 10 us the current IL (at 10 us) is

(a) 4.36 mA

(b) 4.96 mA

(c) 8.38 mA

(d) 7.49 mA

  1. In the figure the switch was closed for a long time before opening at t = 0. the voltage VX at t = 0+ is

(a) 25 V

(b) 50 V

(c) -50 V

(d) zero

  1. In the circuit shown below, the current excitation is IIN(t) = 4u (-t) A. the IL (t) for t > 0 is

(a) 4e-5000t A

(b) 2e-5000t A

(c) 6 – 4e-5000t A

(d) 6e-5000t A

  1. In the circuit shown below, the voltage source 8 v has been applied for a long time before the switch opens at t = 0. the current IR (0+) is

(a) 1 mA

(b) 2 mA

(c) 3 mA

(d) C is required

  1. In the circuit shown below, the current excitation is given by 6u(-t) + 2u(t) A. for t>0 inductor current IL (t) is

(a) 4 – e-2rt A

(b) 1 + 2e-2rt A

(c) 1 – 2e-rt/2 A

(d) R is required

  1. A two-port network is represented by ABCD-parameters given by

If port 2 is terminated by RL, the input impedance seen at port 1 is given by

(a) A + BRL/C + DRL

(b) ARL + C/BRL + D

(c) DRL + A/BRL + C

(d) B + ARL/D + CRL

  1. In the two-port network shown in the figure below Z12 and Z21 are respectively,

(a) rc and br0

(b) 0 and -br0

(c) 0 and br0

(d) r0 and -br0

  1. The first and the last critical frequencies (singularities) of a driving point impedance function of a passive network having two kinds of elements are a pole and a zero, respectively. the above property will be satisfied by

(a) RL network only

(b) RC network only

(c) LC network only

(d) RC as well as RL networks

  1. A 2 mH inductor with some initial current can be represented as shown below, where s is the laplace transform variable. the value of initial current is

(a) 0.5 A

(b) 2.0 A

(c) 1.0 A

(d) zero

  1. In the figure shown below, assume that all the capacitors are initially uncharged. if V1(t) = 10 u(t) V, V0(t) is given by

(a) 8e-0.004t V

(b) 8 (1 – e-t/0.004) V

(c) 8 u(t) V

(d) 8 V

  1. A negative resistance RNEG is connected to a passive network N having driving point impedence ZS(s) to be positive real

(a) |RNEG| < Re Z1(J0),

(b) |RNEG|<Z1(J0)

(c) |RNEG|<ImZ1(Jo),

(d) |RNEG|<Z1(J0),

  1. The current in a 100 uf capacitor is shown in figure. if capacitor is initially uncharged, then the waveform for the voltage across it is
  2. The voltage across a 100 uf capacitor is shown in figure. the waveform for the current in the capacitor is
  3. The waveform for the current in a 200 uf capacitor is shown in figure. the waveform for the capacitor voltage is
  4. The induced voltage across a 10 mH inductor is V(t) = 120 cos 377t V. the expression for the power is

(a) 1910 sin 754t W

(b) 5.07 sin2 377t W

(c) 10.14 sin2 377t W

(d) 3820 sin 754t W

Answers and solution

  1. (b)

RTH = VOC/ISC = 12/1.5 = 8

  1. (b)

imitially I (0+) = 0. Due to inductor I(0+) = 0. Thus, all current IS will flow in resistor RS and voltage across inductor will be

VL(0+) = ISR3 but VL(0+) = L dI(0+)/dt

thus, dI(0+)/dt = ISRS/L

  1. (a)

IC(0+) = 100/1K = 0.1 A

103IC + 1/10-6 ICdt = 50

103dIC/dt + IC/10-6 = 0 ………………..(1)

At t = 0+

dIC(0+)/dt = 0.1/10-6 x 103 = – 100 a/s

  1. (b)

IL(0) = 24/4 = 6 A

REQ = 6||6||12 = 2.4

IL(t) = 6e-2.4 x 4t A

VL(t) = 1/4 x (-2.4 x 4) 6e-9.6t

= – 14.4 e-9.6t V

VO(t) = 8VL/4 + 8 = – 9.6e-9.6t V

  1. (b)

VC(0) = 80 x 3/400 + VC(0)/200

3/400 + 1/200 + 1/100

VC(0) = 34.39 V = VC (0+)

I1(0+) = 34.29/200 = 0.17 A

  1. (d)

the circuit is as shown below.

VTEST/13.33||200 + VTEST – 0.25 VTEST/100 = 1

VTEST = 50 V

RTH = 50

50 x 0.5 m

VC(t) = 34.29et/50×0.5 m

I1(80 m) = 34.29e-80m/25m/200 = 6.98 mA

  1. (b)

in steady state,

VIN = 100IL + 50 (IL – VX/200)

VX = 100IL

VIN = 100IL + 50 (I1 – 0.5IL) = 125IL

t > 0, -25 = 125IL

IL = – 0.2 A

t > 0,25 = 125IL

IL = 0.2 A

resistance seen by inductor for t > 0

VOC = 25 V, ISC = 0.2 A

RTH = 25/0.2 = 125

IL(t) = 0.2 + (-0.2 – 0.2) e125 t/20 m

= 0.2 (1 – 2e-6250 t) A

  1. (a)

V0(0) = 4 K/4 K + 2 K (12) = 8 V

V0 (00) = 12/2 = 6 V

REQ = (4 K ||4 K + 2 K) = 4 K

VO(t) = 6 + (8 – 6) et/100+4 = 6 + 2e-2.5t V

  1. (a)

at t = 0+ VR = 0

VR/1 K + 10-6 dVR/dt = 10 ……………(i)

t = 0+ dVR/dt = 107 v/s

differentiating eq. (1).

10-6 d2VR/dt2 + 1/1 dVR/dt = 0

at t = 0

d2VR/dt2 = -106/103 x 107 = – 1010 v/s2

  1. (a)

IL(0) = 6(2R)/2R + R = 4A = IL (0+)

  1. (c)

IL(0) = (1/800) = 12/100 /1/100 + 1/1333 + 1/800 = 12.5 MA

IL(t) = 12.5e-800t/20 m = 12.5e-4 x 104t mA

IL(10 us) = 12.5 e-0.4 = 8.38 mA

  1. (c)

when switch was closed, in steady state

IL (0) = 2.5 A

At t = 0+ all current of inductor will pass through 2 resistor

VX = – 2.5 x 20 = – 50 V

  1. (b)

IL(0) = 4/2 = 2 A

REQ = (400 + 400) ||200 = 160

IL = 0 + (2 – 0) e160/32 m = 2e-5000t A

  1. (a)

VC(t) = 8 x 6/6 + 2 = 6 V = VC(0)

IR(0) = 6/6 K = 1 mA

  1. (a)

IL (0+) = 6/2 = 3 A = IL(0+)

IL (00) = 2/2 = 1 A

REQ = (R + R)||2R = R

IL (t) = 1 + (3 – 1)e-rt/0.5

= 1 + 2e-2rt A

  1. (d)

ABCD-parameter

[V1 I1] = [A B C D][V2 -I2]

V1 = AV2 – BI2……………(1)

I1 = CV2 – DI2 ………………(2)

The input impedance seen at port (1) obtained by dividing eq. (1) from eq. (2)

ZIN = V1/I1 = AV2 – BI2/CV2 – DI2

From circuit

V2 = – I2RL

ZIN = -I2RLA – BI2/-I2RLC – DI2 = ARL + B/CRL + D

  1. (b)

the impedance parameters are given by

V1 = Z11I1 + Z12I2

V2 = Z21I1 + Z22I2

V1 = I1rc

V2 = (I2 – BI1) or = – Br0c1 + roI2

  1. (b)

pole and zeros are critical frequency. at poles the network function becomes infinite and at zero network function be zero.

given criteria that pole and zero both exist in driving point impedance function is satisfied if the network having R and C. in LC and RL networks either pole exists or zero.

Z(s) = 1/sC + R = 1 + sRC/sC

pole exits at origin whereas zero at (-1/RC).

  1. (b)

REQ = 1K ||4 K = 4/b k

CEQ = 1UF ||4UF = 5UF

Time constant = 4/5 x 103 x 5 x 10-6 = 4 x 10-3 s

as all capacitors are initiall uncharged. the charging current flows in the circuit.

charging current

I(t) = VI/1 K + 4 K (1 – e-t/t)

= 10/5 k (1 – e-t/t)

output voltage VO(t) = 10/5 k x 4 k (1 – e-t/t)

= 8 (1 – e-t/0.004) V

  1. (a)

we can write

Z2(s) = RNEG + Z1(s)

RNEG = Re Z1(s) + jImZ1(s)

ReZ2(s) + jImZ2 (s)

Equating real and imaginary parts,

Re Z2(s) = RNEG + ReZ1(s)

given, Re Z2(s) > 0

RNEG + ReZ1(s) > 0

ReZ1(s) >|RNEG|

ReZ1(j) >|RNEG|

  1. (d)

VC = 1/C|Idt

10 x 10-3/100 x 10-6 (2 x 10-3) = 0.2 V

this 0.2 V increases linearly from 0 to 0.2 v. then current is zero. so, capaccitor holds this voltage.

  1. (d)

I = C dV/dt

for 0 > t < 1,

C dV/dt = 100 x 10 x 10-6 x 6 – 0/103 – 0

= 600 mA

  1. (b)

for 0 < t < 4

VC = 1/C

= 1/200 x 10-6 |5/4 tdt = 3125t2

at t = 4 ms, VC = 0.05 V

It will be parabolic path, at t = 0 t-axis will be tangent.

  1. (a)

I = 1/L |Vdt = 0.1 |120 cos 3t dt

= 12000/377 sin 377t

P = VI 12000 x 120/377 (sin 377t) (cos 377t)

= 1910 sin 754t W