(a) (i) For the circuit shown in Figure 2(a)(i), initial current through the coil is zero. The switch is closed at time t = 0. Find the time domain expression of current flowing through the coil for t ≥ 0. How long will it take for the coil current t

Question

(a) (i) For the circuit shown in Figure 2(a)(i), initial current through the coil is zero. The switch is closed at time t = 0. Find the time domain expression of current flowing through the coil for t ≥ 0. How long will it take for the coil current to reach 95% of its final value and what is the final value of this current? (10 marks)
(ii) Obtain the Thevenin's equivalent of the network shown in Figure 2(a)(ii), across the terminals XY. (10 marks)
(b) (i) Determine the Inverse Laplace Transform of the following function: F(s) = (s³ + 7s² + 14s + 11)/(s³ + 6s² + 11s + 6) (6 marks)
(ii) Find the initial and final value of the current whose Laplace Transform is given below: I(s) = 0.32/[s(s² + 2.42s + 0.672)] (4 marks)
(iii) Solve the following differential equation: d²i/dt² + di/dt = t² + 2t. Given that i(0-) = 4 and (di/dt)₀₋ = -2. (10 marks)
(c) A single phase AC bridge rectifier as shown in Figure 2(c) is operating at firing delay angle α = 45°. The thyristor T₃ gets damaged and behaves as an open circuit. Calculate the value of load resistance R if load current is 3·1556 A. (10 marks)

UPSC Answer Check · Accepted Answer

(a)(i) RL Transient: Coil Current

Circuit for t ≥ 0. Once S closes, the 100 V source with its 50 Ω series resistor drives the 100 Ω resistor and the 4 H coil, which are in parallel. i_L(0⁻) = 0 and the inductor current cannot jump, so i_L(0⁺) = 0.

Thevenin equivalent seen by the coil:

V_Th = 100 × 100/(50 + 100) = 66.67 V

R_Th = 50 ∥ 100 = (50 × 100)/150 = 33.33 Ω

Time constant: τ = L/R_Th = 4/33.33 = 0.12 s

Final value. In steady state the coil is a short circuit, which also shorts out the 100 Ω:

i_L(∞) = V_Th/R_Th = 66.67/33.33 = 2 A

Time-domain expression. For a first-order circuit, i_L(t) = i_L(∞) + [i_L(0⁺) − i_L(∞)]e^(−t/τ):

i_L(t) = 2(1 − e^(−t/0.12)) = 2(1 − e^(−8.33t)) A, t ≥ 0

Time to reach 95 %:

0.95 = 1 − e^(−t/τ) ⇒ t = τ ln 20 = 0.12 × 2.996 = 0.36 s (≈ 3τ)

The coil current rises exponentially to 2 A and is within 5 % of that value after about 0.36 s.

(a)(ii) Thevenin Equivalent across XY

Circuit (Figure 2(a)(ii)). A 4 V source in series with 2 kΩ feeds node P. A dependent current source Vₓ/4 kΩ is connected from the bottom rail up into P. From P, 3 kΩ runs to terminal X. Y is on the bottom rail, and Vₓ = V_XY.

Open-circuit voltage. With X–Y open, no current flows in the 3 kΩ, so Vₓ = V_P. KCL at P, with the source current Vₓ/4k entering P:

(4 − V_P)/2k + V_P/4k = 0 ⇒ 2(4 − V_P) + V_P = 0 ⇒ V_P = 8 V

V_Th = V_XY(oc) = 8 V

This is larger than the 4 V source because the controlled source feeds current back into P.

Short-circuit current. With X–Y shorted, Vₓ = 0, so the dependent source is dead:

I_sc = 4/(2k + 3k) = 0.8 mA

R_Th = V_Th/I_sc = 8/0.8 mA = 10 kΩ

Check by test source (4 V source replaced by a short, V applied at X, current I into X):
KCL at P: I + V/4k = V_P/2k ⇒ V_P = 2k·I + V/2
KVL: V = 3k·I + V_P = 5k·I + V/2 ⇒ V/I = 10 kΩ ✓

The controlled source roughly doubles the passive resistance (3k + 2k = 5 kΩ).

(b)(i) Inverse Laplace Transform

Long division. The degrees of numerator and denominator are equal, so divide first:

F(s) = 1 + (s² + 3s + 5)/[(s + 1)(s + 2)(s + 3)]

since s³ + 6s² + 11s + 6 = (s + 1)(s + 2)(s + 3).

Residues:
s = −1: (1 − 3 + 5)/[(1)(2)] = 3/2
s = −2: (4 − 6 + 5)/[(−1)(1)] = −3
s = −3: (9 − 9 + 5)/[(−2)(−1)] = 5/2

F(s) = 1 + (3/2)/(s + 1) − 3/(s + 2) + (5/2)/(s + 3)

f(t) = δ(t) + [1.5e^(−t) − 3e^(−2t) + 2.5e^(−3t)] u(t)

(b)(ii) Initial and Final Values

Initial value theorem:

i(0⁺) = lim(s→∞) sI(s) = lim 0.32/(s² + 2.42s + 0.672) = 0 A

Final value theorem. First check that it applies: s² + 2.42s + 0.672 = (s + 0.32)(s + 2.10). Both poles of sI(s) lie in the left half-plane, so the theorem is valid.

i(∞) = lim(s→0) sI(s) = 0.32/0.672 = 0.476 A

(b)(iii) Solution of the Differential Equation

Equation: i″ + i′ = t² + 2t, with i(0⁻) = 4 and i′(0⁻) = −2.

Complementary function. The characteristic equation is m² + m = 0, so m = 0 and m = −1:

i_c = A + Be^(−t)

Particular integral. A constant is already part of i_c (m = 0 is a root), so try a polynomial one degree higher, i_p = at³ + bt² + ct:

i_p′ = 3at² + 2bt + c, i_p″ = 6at + 2b

i_p″ + i_p′ = 3at² + (6a + 2b)t + (2b + c) = t² + 2t

Equating coefficients:
3a = 1 ⇒ a = 1/3
6a + 2b = 2 ⇒ b = 0
2b + c = 0 ⇒ c = 0

So i_p = t³/3.

Applying the initial conditions to i(t) = A + Be^(−t) + t³/3 (no impulses act at t = 0, so the 0⁻ values carry over):
i(0) = A + B = 4
i′(0) = −B + 0 = −2 ⇒ B = 2, A = 2

i(t) = 2 + 2e^(−t) + t³/3, t ≥ 0

Check by Laplace: (s² + s)I − 4s − 2 = 2/s³ + 2/s², which gives I(s) = 2/s + 2/(s + 1) + 2/s⁴. This is the same result.

(c) Bridge Rectifier with T₃ Open

Healthy bridge (Figure 2(c)). The upper terminal of the 200 V supply goes to the T₁–T₄ junction and the lower terminal to the T₃–T₂ junction. In the positive half-cycle T₁ and T₂ conduct. In the negative half-cycle T₃ and T₄ conduct.

With T₃ open-circuited, the negative-half path is broken. T₄ alone cannot complete a circuit, so the load receives only the positive half-cycles, from α to π, through T₁–T₂. The bridge behaves as a single-phase half-wave controlled rectifier.

Average output voltage (resistive load, ideal thyristors, V_m = √2 × 200 = 282.8 V):

V₀ = (1/2π) ∫ from α to π of V_m sin ωt d(ωt) = (V_m/2π)(1 + cos α)

= (282.8/2π)(1 + 0.7071) = 45.02 × 1.7071 = 76.85 V

Load resistance. The load current of a rectifier is taken as its average (dc) value:

R = V₀/I₀ = 76.85/3.1556 = 24.35 Ω

Before the fault the same R would have carried 2 × 3.1556 = 6.31 A. The loss of one thyristor halves the dc output and adds a large 50 Hz ripple component. If the 3.1556 A were read as an rms value, the half-wave rms voltage of 134.8 V would give R = 42.7 Ω.

(a) (i) For the circuit shown in Figure 2(a)(i), initial current through the coil is zero. The switch is closed at time t = 0…

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(a)(i) RL Transient: Coil Current

(b)(i) Inverse Laplace Transform

(c) Bridge Rectifier with T₃ Open

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