solution and answer of question **A two-port network is represented by ABCD-parameters given by ****If port 2 is terminated by R _{L}, the input impedance seen at port 1 is given by ? **

- In the circuit shown below voltage source 6 V has been applied for a long time. At t = 0 switch is opened. the V
_{O}(t) for t > 0 is

(a) 3e^{-4t} V

(b) -6e^{ – 4t} V

(c) 6e – ^{4t} V

(d) -3e ^{-4t} V

- 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) R_{S}I_{S}/L

(c) (R + R_{S})I_{S}/L

(d) infinite

- In the following circuit, the switch is closed at t = 0 with the capacitor uncharged. the value of di
_{l}/dt and d^{2}l_{l}/dt^{2}at t = 0^{+}is

(a) -100 a/s, 10^{5} a/s^{2}

(b) 10^{5} a/s, – 10^{7 }a/s^{2}

(c) -100 a/s, -10^{5} a/s^{2}

(d) 10^{5} a/s, 10^{7} a/s^{2}

- The circuit shown below reaches a steady state with the switch closed. At t = 0, switch is opened. for t > 0, V
_{O}(t) is

(a) 9.6 e^{-9.6t }V

(b) -9.6 e^{9.6t} V

(c) 2.4 e^{-2.4t} V

(d) -2.4 e^{2.4t} 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.

- At t = 0
^{+}, the current I_{1}(0^{+}) is

(a) 0.25 A

(b) 0.17 A

(c) 0.05 A

(d) 0.2 A

- At t = 80 ms the current I
_{1}(80 ms) is

(a) 20.3 mA

(b) 8.25 mA

(c) 1.84 mA

(d) 6.98 mA

- In the circuit shown below, if V
_{IN}= 25u(t) – 25u(-t) v, the I_{L}(t) for t > 0 is

(a) 0.4 (1 – e^{-250t}) A

(b) 0.2 (1 – e^{6250t}) A

(c) 0.2 (1 – e^{-250t}) A

(d) 0.4 (1 – e^{-6250t}) A

- In the circuit shown below the switch is opened at t = 0 after a long time. the voltage V
_{0}(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 – 2e^{2.5t}) V

- In the network shown below, the switch is opened at t = 0. at t = 0
^{+}, the value of d^{2}V_{R}Idt^{2}is

(a) -10^{10} v/s^{2}

(b) 10^{10} v/s^{2}

(c) 10^{+13} v/s^{2}

(d) -10^{+13} v/s^{2}

- 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 I
_{L}(0^{+}) is

(a) 4 A

(b) 4.8 A

(c) 2 A

(d) R and L are required

- 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 I
_{L}(at 10 us) is

(a) 4.36 mA

(b) 4.96 mA

(c) 8.38 mA

(d) 7.49 mA

- In the figure the switch was closed for a long time before opening at t = 0. the voltage V
_{X}at t = 0^{+}is

(a) 25 V

(b) 50 V

(c) -50 V

(d) zero

- In the circuit shown below, the current excitation is I
_{IN}(t) = 4u (-t) A. the I_{L}(t) for t > 0 is

(a) 4e^{-5000t} A

(b) 2e^{-5000t} A

(c) 6 – 4e^{-5000t} A

(d) 6e^{-5000t} A

- 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 I
_{R }(0^{+}) is

(a) 1 mA

(b) 2 mA

(c) 3 mA

(d) C is required

- In the circuit shown below, the current excitation is given by 6u(-t) + 2u(t) A. for t>0 inductor current I
_{L}(t) is

(a) 4 – e^{-2rt} A

(b) 1 + 2e^{-2rt} A

(c) 1 – 2e^{-rt/2} A

(d) R is required

**A two-port network is represented by ABCD-parameters given by**

**If port 2 is terminated by R _{L}, the input impedance seen at port 1 is given by**

(a) A + BR_{L}/C + DR_{L}

(b) AR_{L} + C/BR_{L} + D

(c) DR_{L} + A/BR_{L} + C

(d) B + AR_{L}/D + CR_{L}

- In the two-port network shown in the figure below Z
_{12}and Z_{21}are respectively,

(a) r_{c} and br_{0}

(b) 0 and -br_{0}

(c) 0 and br_{0}

(d) r_{0} and -br_{0}

- 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

- 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

- In the figure shown below, assume that all the capacitors are initially uncharged. if V
_{1}(t) = 10 u(t) V, V_{0}(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

- A negative resistance R
_{NEG }is connected to a passive network N having driving point impedence Z_{S}(s) to be positive real

(a) |R_{NEG}| < Re Z_{1}(J_{0}),

(b) |R_{NEG}|<Z_{1}(J_{0})

(c) |R_{NEG}|<ImZ_{1}(J_{o}),

(d) |R_{NEG}|<Z_{1}(J_{0}),

- 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
- The voltage across a 100 uf capacitor is shown in figure. the waveform for the current in the capacitor is
- The waveform for the current in a 200 uf capacitor is shown in figure. the waveform for the capacitor voltage is
- 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 sin^{2} 377t W

(c) 10.14 sin^{2} 377t W

(d) 3820 sin 754t W

### Answers and solution

- (b)

R_{TH} = V_{OC}/I_{SC} = 12/1.5 = 8

- (b)

imitially I (0^{+}) = 0. Due to inductor I(0^{+}) = 0. Thus, all current I_{S} will flow in resistor R_{S} and voltage across inductor will be

V_{L}(0^{+}) = I_{S}R_{3} but V_{L}(0^{+}) = L dI(0^{+})/dt

thus, dI(0^{+})/dt = I_{S}R_{S}/L

- (a)

I_{C}(0^{+}) = 100/1K = 0.1 A

10^{3}I_{C} + 1/10^{-6} I_{C}dt = 50

10^{3}dI_{C}/dt + I_{C}/10^{-6} = 0 ………………..(1)

At t = 0^{+}

dI_{C}(0^{+})/dt = 0.1/10^{-6} x 10^{3} = – 100 a/s

- (b)

I_{L}(0) = 24/4 = 6 A

R_{EQ} = 6||6||12 = 2.4

I_{L}(t) = 6e^{-2.4 x 4t} A

V_{L}(t) = 1/4 x (-2.4 x 4) 6e^{-9.6t}

= – 14.4 e^{-9.6t} V

V_{O}(t) = 8V_{L}/4 + 8 = – 9.6e^{-9.6t} V

- (b)

V_{C}(0) = 80 x 3/400 + V_{C}(0)/200

3/400 + 1/200 + 1/100

V_{C}(0) = 34.39 V = V_{C} (0^{+})

I_{1}(0^{+}) = 34.29/200 = 0.17 A

- (d)

the circuit is as shown below.

V_{TEST}/13.33||200 + V_{TEST} – 0.25 V_{TEST}/100 = 1

V_{TEST} = 50 V

R_{TH} = 50

50 x 0.5 m

V_{C}(t) = 34.29e^{t/50×0.5 m}

I_{1}(80 m) = 34.29e^{-80m/25m}/200 = 6.98 mA

- (b)

in steady state,

V_{IN} = 100I_{L} + 50 (I_{L} – V_{X}/200)

V_{X} = 100I_{L}

V_{IN} = 100I_{L} + 50 (I_{1} – 0.5I_{L}) = 125I_{L}

t > 0, -25 = 125I_{L}

I_{L} = – 0.2 A

t > 0,25 = 125I_{L}

I_{L} = 0.2 A

resistance seen by inductor for t > 0

V_{OC} = 25 V, I_{SC} = 0.2 A

R_{TH} = 25/0.2 = 125

I_{L}(t) = 0.2 + (-0.2 – 0.2) e^{125 t/20 m}

= 0.2 (1 – 2e^{-6250 t}) A

- (a)

V_{0}(0) = 4 K/4 K + 2 K (12) = 8 V

V_{0} (_{00}) = 12/2 = 6 V

R_{EQ} = (4 K ||4 K + 2 K) = 4 K

V_{O}(t) = 6 + (8 – 6) e^{t/100+4} = 6 + 2e^{-2.5t} V

- (a)

at t = 0^{+} V_{R} = 0

V_{R}/1 K + 10^{-6} dV_{R}/dt = 10 ……………(i)

t = 0^{+} dV_{R}/dt = 10^{7} v/s

differentiating eq. (1).

10^{-6} d^{2}V_{R}/dt^{2} + 1/1 dV_{R}/dt = 0

at t = 0

d^{2}V_{R}/dt^{2} = -10^{6}/10^{3} x 10^{7} = – 10^{10} v/s^{2}

- (a)

I_{L}(0) = 6(2R)/2R + R = 4A = I_{L} (0^{+})

- (c)

I_{L}(0) = (1/800) = 12/100 /1/100 + 1/1333 + 1/800 = 12.5 MA

I_{L}(t) = 12.5e^{-800t/20 m} = 12.5e^{-4 x 104t} mA

I_{L}(10 us) = 12.5 e^{-0.4} = 8.38 mA

- (c)

when switch was closed, in steady state

I_{L} (0) = 2.5 A

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

V_{X} = – 2.5 x 20 = – 50 V

- (b)

I_{L}(0) = 4/2 = 2 A

R_{EQ} = (400 + 400) ||200 = 160

I_{L} = 0 + (2 – 0) e^{160/32 m} = 2e^{-5000t} A

- (a)

V_{C}(t) = 8 x 6/6 + 2 = 6 V = V_{C}(0)

I_{R}(0) = 6/6 K = 1 mA

- (a)

I_{L} (0^{+}) = 6/2 = 3 A = I_{L}(0^{+})

I_{L} (_{00}) = 2/2 = 1 A

R_{EQ} = (R + R)||2R = R

I_{L} (t) = 1 + (3 – 1)e^{-rt/0.5}

= 1 + 2e^{-2rt} A

- (d)

ABCD-parameter

[V_{1} I_{1}] = [A B C D][V_{2} -I_{2}]

V_{1} = AV_{2} – BI_{2}……………(1)

I_{1} = CV_{2} – DI_{2} ………………(2)

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

Z_{IN} = V_{1}/I_{1} = AV_{2} – BI_{2}/CV_{2} – DI_{2}

From circuit

V_{2} = – I_{2}R_{L}

Z_{IN} = -I_{2}R_{L}A – BI_{2}/-I_{2}R_{L}C – DI_{2} = AR_{L} + B/CR_{L} + D

- (b)

the impedance parameters are given by

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

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

V_{1} = I_{1}r_{c}

V_{2} = (I_{2} – BI_{1}) or = – Br_{0}c_{1} + r_{o}I_{2}

- (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).

- (b)

R_{EQ} = 1K ||4 K = 4/b k

C_{EQ} = 1UF ||4UF = 5UF

Time constant = 4/5 x 10^{3} 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) = V_{I}/1 K + 4 K (1 – e^{-t/t})

= 10/5 k (1 – e^{-t/t})

output voltage V_{O}(t) = 10/5 k x 4 k (1 – e^{-t/t})

= 8 (1 – e^{-t/0.004}) V

- (a)

we can write

Z_{2}(s) = R_{NEG} + Z_{1}(s)

R_{NEG} = Re Z_{1}(s) + jImZ_{1}(s)

ReZ_{2}(s) + jImZ_{2} (s)

Equating real and imaginary parts,

Re Z_{2}(s) = R_{NEG} + ReZ_{1}(s)

given, Re Z_{2}(s) > 0

R_{NEG} + ReZ_{1}(s) > 0

ReZ_{1}(s) >|R_{NEG}|

ReZ_{1}(j) >|R_{NEG}|

- (d)

V_{C} = 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.

- (d)

I = C dV/dt

for 0 > t < 1,

C dV/dt = 100 x 10 x 10^{-6} x 6 – 0/10^{3} – 0

= 600 mA

- (b)

for 0 < t < 4

V_{C} = 1/C

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

at t = 4 ms, V_{C} = 0.05 V

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

- (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