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

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 ? 

26. 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

27. 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_SI_S/L

(c) (R + R_S)I_S/L

(d) infinite

28. In the following circuit, the switch is closed at t = 0 with the capacitor uncharged. the value of di_l /dt and d² l_l/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

29. 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.

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

(a) 0.25 A

(b) 0.17 A

(c) 0.05 A

(d) 0.2 A

31. 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

32. 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

33. 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 – 2e^2.5t) V

34. In the network shown below, the switch is opened at t = 0. at t = 0⁺, the value of d²V_R Idt² is

(a) -10¹⁰ v/s²

(b) 10¹⁰ v/s²

(c) 10⁺¹³ v/s²

(d) -10⁺¹³ v/s²

35. 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

36. 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

37. 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

38. 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

39. 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

40. 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

41. 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

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

(a) r_c and br₀

(b) 0 and -br₀

(c) 0 and br₀

(d) r₀ and -br₀

43. 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

44. 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

45. 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

46. 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 Z₁(J₀),

(b) |R_NEG|<Z₁(J₀)

(c) |R_NEG|<ImZ₁(J_o),

(d) |R_NEG|<Z₁(J₀),

47. 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
48. The voltage across a 100 uf capacitor is shown in figure. the waveform for the current in the capacitor is
49. The waveform for the current in a 200 uf capacitor is shown in figure. the waveform for the capacitor voltage is
50. 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

26. (b)

R_TH = V_OC/I_SC = 12/1.5 = 8

27. (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_SR₃ but V_L(0⁺) = L dI(0⁺)/dt

thus, dI(0⁺)/dt = I_SR_S/L

28. (a)

I_C(0⁺) = 100/1K = 0.1 A

10³I_C + 1/10^-6 I_Cdt = 50

10³dI_C/dt + I_C/10^-6 = 0 ………………..(1)

At t = 0⁺

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

29. (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

30. (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⁺)

I₁(0⁺) = 34.29/200 = 0.17 A

31. (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}

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

32. (b)

in steady state,

V_IN = 100I_L + 50 (I_L – V_X/200)

V_X = 100I_L

V_IN = 100I_L + 50 (I₁ – 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

33. (a)

V₀(0) = 4 K/4 K + 2 K (12) = 8 V

V₀ (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

34. (a)

at t = 0⁺ V_R = 0

V_R/1 K + 10^-6 dV_R/dt = 10 ……………(i)

t = 0⁺ dV_R/dt = 10⁷ v/s

differentiating eq. (1).

10^-6 d²V_R/dt² + 1/1 dV_R/dt = 0

at t = 0

d²V_R/dt² = -10⁶/10³ x 10⁷ = – 10¹⁰ v/s²

35. (a)

I_L(0) = 6(2R)/2R + R = 4A = I_L (0⁺)

36. (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

37. (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

38. (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

39. (a)

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

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

40. (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

41. (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)

Z_IN = V₁/I₁ = AV₂ – BI₂/CV₂ – DI₂

From circuit

V₂ = – I₂R_L

Z_IN = -I₂R_LA – BI₂/-I₂R_LC – DI₂ = AR_L + B/CR_L + D

42. (b)

the impedance parameters are given by

V₁ = Z₁₁I₁ + Z₁₂I₂

V₂ = Z₂₁I₁ + Z₂₂I₂

V₁ = I₁r_c

V₂ = (I₂ – BI₁) or = – Br₀c₁ + r_oI₂

43. (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).

46. (b)

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

C_EQ = 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) = 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

46. (a)

we can write

Z₂(s) = R_NEG + Z₁(s)

R_NEG = Re Z₁(s) + jImZ₁(s)

ReZ₂(s) + jImZ₂ (s)

Equating real and imaginary parts,

Re Z₂(s) = R_NEG + ReZ₁(s)

given, Re Z₂(s) > 0

R_NEG + ReZ₁(s) > 0

ReZ₁(s) >|R_NEG|

ReZ₁(j) >|R_NEG|

47. (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.

48. (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

49. (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.

50. (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