Mid-sem past papers, from the first move

Step 1KVL round mesh 1

Clockwise from the bottom-left: up through the 5 V source (a rise of 5), right through 1 Ω (a drop of i₁), down through the I_y source from + to − (a drop of I_y), down through the shared 1 Ω (a drop of i₁ − i₂):

Step 2KVL round mesh 2

Clockwise from the bottom of the middle column: up the shared 1 Ω against i₂ (a drop of i₂ − i₁), up through the I_y source from − to + (a rise of I_y), right through 1 Ω (a drop of i₂), through the I_x source from + to − (a drop of I_x), down the right 1 Ω (a drop of i₂ − i₃):

With I_x = i₁, the i₁ terms cancel: −3i₂ + i₃ + I_y = 0.

Step 3Substitute the two facts you already have

Put i₃ = −1 and I_y = i₂ + 1 into Step 2: −3i₂ − 1 + i₂ + 1 = 0, so i₂ = 0 and I_y = 1 A. Then Step 1: 2i₁ − 0 + 1 = 5, so i₁ = I_x = 2 A.

Check with node voltages (bottom wire = 0): top-left 5 V, after the first 1 Ω 5 − 2 = 3 V, joint below the I_y source 3 − 1 = 2 V, and the shared 1 Ω carries 2 A = i₁ − i₂ ✓. Past the I_x source the top wire sits at 3 − 2 = 1 V, and the right 1 Ω carries 1 A = I_y ✓.

Step 1Right half, backwards

Load: I_L = 3 A, so its voltage is 3 × 2 = 6 V. The 5 Ω carries the same 3 A, so the node before it is at 6 + 5 × 3 = 21 V. The middle 2 Ω then carries 21/2 = 10.5 A. KCL at that node: the current arriving through the 10 Ω is 10.5 + 3 = 13.5 A, and the 10 Ω is in series with the dependent source, so that is the source's current.

Step 2Cross the bridge

2V_x = 13.5, so V_x = 6.75 V. The 5 Ω on the left carries 6.75/5 = 1.35 A. That is the current the source V_o supplies, since everything the source pushes goes through that 5 Ω.

Step 3Left half

The 1.35 A splits between 2 Ω and 4 Ω, which are in parallel (2 × 4/(2 + 4) = 4/3 Ω), so the node after the 5 Ω sits at 1.35 × 4/3 = 1.8 V. KVL round the left loop: V_o = V_x + 1.8 = 6.75 + 1.8 = 8.55 V.

Step 1Open circuit: terminals 1–1′ left open

Let x be the current from N to T through the right 2 Ω. With nothing attached at terminal 1, that current has only one way out: back through the 1 V source and the 1 Ω to A. So the 1 Ω carries x from U (between the 1 Ω and the 1 V) to A, giving V_U = 2 + x. The 1 V source has + on the U side, so V_T = V_U − 1 = 1 + x.

KCL at N (current in from M through 2 Ω = out through the lower 2 Ω + out to T):

So V_oc = V_T = 1 + 1 = 2 V. (Check: V_N = V_T + 2x = 4 V, so V_a = 4 V and V_M = 10 V.)

Step 2Short circuit: wire 1 to 1′, so V_T = 0

Upper path: V_U = V_T + 1 = 1 V, so the 1 Ω carries (2 − 1)/1 = 1 A from A towards T. Middle path, KCL at N with V_T = 0:

Both paths deliver into terminal 1 and down the short: I_sc = 1 + 1 = 2 A.

Step 3Divide

R_N = V_oc/I_sc = 2/2 = 1 Ω. A test source gives the same answer: kill the 2 V and 1 V sources, keep 2V_a, push 1 A into terminal 1, and the terminal voltage comes out as 1 V.

Step 1Before: t < 0, inductor = short

The shorted inductor pins its top node to 0 V, so the 5 Ω has 0 V across it and carries nothing. Seen from the 15 A source, the 24 Ω, the 8 Ω and the 12 Ω (to the 0 V node) are all in parallel: 1/(1/24 + 1/8 + 1/12) = 1/(6/24) = 4 Ω. The top rail sits at 15 × 4 = 60 V, so the 12 Ω carries 60/12 = 5 A, all of it into the inductor: i(0⁻) = 5 A.

Step 2After: t > 0

i(0⁺) = 5 A (inductor current can't jump). No source remains on the right, so i(∞) = 0. The inductor sees the 5 Ω in parallel with 12 Ω + 8 Ω in series: 5 ∥ 20 = 100/25 = 4 Ω, so τ = L/R = 2/4 = 0.5 s.

Step 1Before: capacitor = open, so a plain voltage divider

No current flows into the capacitor, so 3 kΩ and 5 kΩ form a divider: v(0⁻) = 24 × 5/(3 + 5) = 15 V, and so v(0⁺) = 15 V.

Step 2After: only 30 V, 4 kΩ and the capacitor

Once the switch sits at B, the left part plays no role. v(∞) = 30 V (no current in steady state, so no drop across 4 kΩ). τ = RC = 4000 × 0.5 × 10⁻³ = 2 s.

Step 3The two asked-for values

(i) v(1) = 30 − 15e^−0.5 = 30 − 15 × 0.60653 = 20.902 V. (ii) v(4) = 30 − 15e⁻² = 30 − 15 × 0.13534 = 27.970 V.

Step 110 V alone (8 A and 2 A open)

V_N = −10, so V_T = −10 + 3i and V_L = V_T − 4i = −10 − i. With both current sources open, node T has only two branches: the 3 Ω (current i leaving downward) and the 4i/2 Ω column (current V_L/2 leaving downward). KCL at T: i + (−10 − i)/2 = 0, so i = 10 A and V₁′ = 30 V.

Step 28 A alone (10 V shorted, 2 A open)

Now V_N = 0, so V_T = 3i and V_L = −i. The 8 A leaves L towards N, so the current coming down the 4i source into L must feed both the 8 A and the 2 Ω: j = 8 + V_L/2 = 8 − i/2. KCL at T: i + j = 0 ⇒ i/2 + 8 = 0, so i = −16 A and V₁″ = −48 V.

Step 32 A alone (10 V shorted, 8 A open)

Again V_T = 3i, V_L = −i, and the left column carries V_L/2 = −i/2 downward. The 2 A enters T: i − i/2 = 2, so i = 4 A and V₁‴ = 12 V.

Step 4Add, then check with all sources on

Check with all three sources on: V_T = −10 + 3(−2) = −16 V and V_L = −16 − 4(−2) = −8 V. At L, the current arriving down the 4i source must feed the 8 A plus the 2 Ω: j = 8 + (−8)/2 = 4 A. At T, currents leaving are i + j = −2 + 4 = 2 A, which matches the 2 A arriving ✓.

Line 1The ODE

Series loop R–L–C with no source, loop current i entering the capacitor's + plate: L di/dt + Ri + v_C = 0 with i = C dv_C/dt. Dividing by LC:

Line 2Characteristic equation

Try v_C = Ae^st: s² + 2αs + ω₀² = 0 with α = R/2L and ω₀ = 1/√(LC), so s = −α ± √(α² − ω₀²).

Line 3Underdamped means complex roots

If α < ω₀, the square root is imaginary: s = −α ± jω_d with ω_d = √(ω₀² − α²). By Euler's formula, e^{(−α ± jω_d)t} = e^−αt(cos ω_dt ± j sin ω_dt), and the real combinations of the two solutions give

Line 4Constants from the initial values

At t = 0: A₁ = V₀. Differentiating, dv_C/dt(0) = −αA₁ + ω_dA₂, and this must equal I₀/C. So A₂ = (I₀/C + αV₀)/ω_d.

Line 5Inductor current (the same as the loop current)

The most common special case is a capacitor charged to V₀ with no initial current (I₀ = 0). Then v_C = V₀e^−αt(cos ω_dt + (α/ω_d) sin ω_dt) and i = −(V₀/ω_dL) e^−αt sin ω_dt. (The last step uses Cω₀² = 1/L.)

Step 1Classify

α = R/2L = 10/5 = 2 s⁻¹;  ω₀ = 1/√(2.5 × 1/40) = 1/√0.0625 = 4 rad/s. α < ω₀, so the response is underdamped, with ω_d = √(16 − 4) = √12 = 2√3 ≈ 3.464 rad/s.

Step 2Initial slopes at 0⁺

i_L leaves the capacitor's + plate, so i_L = −C dv_C/dt. KVL round the loop: v_C = L di_L/dt + 10i_L. At 0⁺: di_L/dt = (8 − 0)/2.5 = 3.2 A/s, and dv_C/dt = −i_L/C = 0.

Step 3Fit the constants

i_L = e^−2t(B₁ cos ω_dt + B₂ sin ω_dt): i_L(0) = 0 ⇒ B₁ = 0; the slope at 0 is ω_dB₂ = 3.2, so B₂ = 3.2/(2√3) = 0.9238.

v_C = e^−2t(8 cos ω_dt + D sin ω_dt): the slope at 0 is −2 × 8 + ω_dD = 0, so D = 16/(2√3) = 4.619.

Step 1Open circuit: KCL at a and at X

At a (the source pushes 2.5v_x in): (V_a − v_x)/2 + V_a/20 = 2.5v_x, which gives 0.55V_a = 3v_x.
At X: (v_x − 50)/3 + v_x/6 + (v_x − V_a)/2 = 0, and multiplying by 6 gives 6v_x − 3V_a = 100.
Substitute v_x = 0.55V_a/3: 1.1V_a − 3V_a = 100, so V_oc = −100/1.9 = −1000/19 ≈ −52.632 V.

Step 2Short circuit: V_a = 0

At X: (v_x − 50)/3 + v_x/6 + v_x/2 = 0; ×6: 6v_x = 100, so v_x = 50/3 V. The 20 Ω is shorted out. The short collects the 2 Ω current v_x/2 plus the dependent source's 2.5v_x: I_sc = 3v_x = 50 A (from a to b).

Step 3Divide, and don't panic at the sign

R_N = V_oc/I_sc = (−1000/19)/50 = −20/19 ≈ −1.053 Ω. A test source (50 V shorted, 1 A pushed into a) gives the same −20/19 Ω, so the value is right. A negative resistance is genuine here. The dependent source is supplying power, so the network behaves at its terminals like something that pushes current back out when you push current in. This is legal for circuits containing dependent sources; it can never happen with resistors and independent sources alone.

Step 1Interval 1: 0 < t < 4 (only S₁ closed)

One loop: 40 V, 8 Ω, 12 Ω, 10 H. i(0⁺) = 0; i(∞) = 40/(8 + 12) = 2 A; R = 20 Ω, τ₁ = 10/20 = 0.5 s.

i(2) = 2(1 − e⁻⁴) = 2(1 − 0.01832) = 1.9634 A;  i(4) = 2(1 − e⁻⁸) = 1.99933 A.

Step 2Interval 2: t > 4 (both closed): Thevenin at P

Seen from P (looking left and down), there are two sources with series resistors: 40 V/8 Ω and 10 V/4 Ω. Their Thevenin equivalent is V_th = (40/8 + 10/4)/(1/8 + 1/4) = 7.5/0.375 = 20 V, R_th = 8 ∥ 4 = 8/3 Ω. The inductor sees 12 + 8/3 = 44/3 Ω in total.

i(∞) = 20/(44/3) = 15/11 ≈ 1.3636 A;  τ₂ = 10/(44/3) = 15/22 ≈ 0.6818 s.

i(5) = 1.3636 + 0.6357 e^−1.4667 = 1.3636 + 0.6357 × 0.23070 = 1.5103 A.

Step 1(a) The two initial values

v(0⁺) = v(0⁻) = 4 V and i_L(0⁺) = 0 (continuity). KCL at 0⁺ (all three branch currents leaving the top node sum to zero): 1 × dv/dt + 4/0.5 + 0 = 0, so dv/dt(0⁺) = −8 V/s.

Step 2(b) Classify and fit

α = 1/(2 × 0.5 × 1) = 1 s⁻¹; ω₀ = 1/√(0.25 × 1) = 2 rad/s. Since α < ω₀, it is underdamped: ω_d = √(4 − 1) = √3 ≈ 1.732 rad/s.

v = e^−t(A cos √3t + B sin √3t). From v(0) = 4: A = 4. From the slope at 0: −A + √3B = −8, so B = −4/√3 ≈ −2.309.

Step 3(c) The graph: the four numbers that make a sketch earn marks

Start at 4 V, heading down at −8 V/s. It first crosses zero where √3t + 30° = 90°, i.e. t = π/(3√3) = 0.605 s. The first minimum is −1.194 V at 1.209 s; the next zero is 2.418 s. The oscillation lives inside the envelope ±4.619e^−t, and the period is 2π/√3 = 3.63 s.

Step 1Solve the circuit

Node 2: 1.6I_x in = I_x out through the inductor + V₂/8, so V₂ = 4.8I_x. Node 1: −j2 + I_x = V₁/4.8, so V₁ = 4.8I_x − j9.6. Inductor: V₂ − V₁ = j1.92I_x, so j9.6 = j1.92I_x and I_x = 5 A. Then V₂ = 24 V and V₁ = 24 − j9.6 V (|V₁|² = 668.16).

Step 2(a), (b) Powers, and the balance check

(a) 4.8 Ω: 668.16/(2 × 4.8) = 69.6 W; 8 Ω: 24²/16 = 36 W; inductor: 0 W (an ideal L stores and returns energy, it never dissipates).
(b) −j2 A source: ½ Re{(24 − j9.6)(j2)} = ½ × 19.2 = 9.6 W supplied. Dependent source: ½ Re{24 × (1.6 × 5)} = 96 W supplied. Balance: 9.6 + 96 = 105.6 = 69.6 + 36 ✓.

Step 3(c) Thevenin at the 8 Ω's terminals

Remove the 8 Ω. Node 2 KCL becomes 1.6I_x = I_x, so I_x = 0 and V_th = V₁ = −j9.6 V. Test source (independent source open, 1 A into node 2): 1 + 1.6I_x = I_x, so I_x = −5/3, V₁ = 4.8I_x = −8 and V₂ = −8 + j1.92(−5/3) = −8 − j3.2. So Z_th = −8 − j3.2 Ω, and the conjugate-match rule gives Z_L = Z_th* = −8 + j3.2 Ω.

Step 1(a) The four mesh equations

Constraint (CCCS branch): I₃ goes up that branch and I₁ goes down, so I₃ − I₁ = 3i₀ = 3(I₁ − I₂), which rearranges to I₃ − 4I₁ + 3I₂ = 0.
Mesh 2 (top window: 4 kΩ, the VCVS from its + end, 6 kΩ): 4I₂ + 2v₀ + 6(I₂ − I₁) = 0, and with v₀ = 6(I₁ − I₂) this becomes −6I₁ + 2I₂ = 0.
Supermesh 1 + 3 (round the outside of both, skipping the CCCS branch): 12 − 6(I₁ − I₂) + 2v₀ − 3(I₃ − I₄) = 0, which becomes 2I₁ − 2I₂ − I₃ + I₄ = −4.
Mesh 4: 3(I₄ − I₃) + 2I₄ = 0, which becomes 3I₃ − 5I₄ = 0.

Step 2(b) Solve by substitution, one mesh at a time

Mesh 2 gives I₂ = 3I₁. The constraint gives I₃ = 4I₁ − 9I₁ = −5I₁. Mesh 4 gives I₄ = 3I₃/5 = −3I₁. Put all three into the supermesh equation: 2I₁ − 6I₁ + 5I₁ − 3I₁ = −2I₁ = −4, so I₁ = 2 mA.

Step 3(c) The nodal equations

V_X = 12 (the source). Supernode constraint: V_Z − V_Y = 2v₀ = 2(12 − V_Y), which becomes V_Y + V_Z = 24. KCL on the supernode {Y, Z} (currents leaving through resistors = current the CCCS pushes in):

Multiplying by 12 and collecting terms gives 8V_Y + 13V_Z = 132. (The key writes the same equation after moving the 6 kΩ term across, as "= 4i₀".)

Step 4(d) Solve

From the constraint, V_Z = 24 − V_Y. Then 8V_Y + 312 − 13V_Y = 132, so V_Y = 36 V and V_Z = −12 V. Cross-check with the meshes: V_Z = 3(I₃ − I₄) = 3(−4) = −12 ✓.

Step 5(e) Power of each dependent source: say which way it points first

Here i₀ = −4 mA and v₀ = −24 V.
CCCS: it pushes 3i₀ = −12 mA up into Y, which sits at 36 V. Power delivered = (voltage it lifts current through) × (current out of its top) = 36 × (−0.012) = −0.432 W. Negative delivered means 0.432 W absorbed.
VCVS: its value is 2v₀ = −48 V (V_Z − V_Y). The current through it from Y to Z, i.e. out of its + terminal, is I₃ − I₂ = −16 mA. Delivered = (−48)(−0.016) = +0.768 W: 0.768 W delivered.
Balance: 12 V source delivers 12 × 0.002 = 0.024 W; resistors absorb 0.096 + 0.144 + 0.048 + 0.072 = 0.360 W; 0.024 + 0.768 − 0.432 = 0.360 ✓.

Step 1(a) Figure 2a, open circuit

KCL at X: 1 = V_X/8 + (V_X − V_Y)/2, which becomes 5V_X − 4V_Y = 8. KCL at Y: (V_Y − V_X)/2 + (V_Y − 32)/20 = 0, which becomes −10V_X + 11V_Y = 32. Doubling the first and adding: 3V_Y = 48, so V_Th = V_Y = 16 V (V_X = 14.4 V).

Step 2(a) Short circuit, then (b)

With V_Y = 0: V_X = 1/(1/8 + 1/2) = 1.6 V, so the 2 Ω delivers 0.8 A into the short. The source branch delivers 32/20 = 1.6 A. Total I_N = 2.4 A. Then R_Th = 16/2.4 = 6.667 Ω. Cross-check (legal here): (8 + 2) ∥ 20 = 200/30 = 6.667 Ω ✓.

Step 3(d) Maximum power

R = R_Th = 6.667 Ω; I_L = 16/(6.667 + 6.667) = 1.200 A; P_max = V_Th²/(4R_Th) = 256/26.667 = 9.600 W.

Step 4(e) Figure 2b, open circuit

With a–b open, all of I must go down the CCVS branch: I = (V_Y − 2I)/20, so V_Y = 22I, and so V_X = V_Y + 2I = 24I. KCL at X: 1 = 24I/8 + I = 4I, so I = 0.25 A and V_Th = 22 × 0.25 = 5.5 V.

Step 5(e) Figure 2b, short circuit, then R_Th

V_Y = 0: as before V_X = 1.6 V, I = 0.8 A. The CCVS now reads 2 × 0.8 = 1.6 V and pushes 1.6/20 = 0.08 A up into Y. I_sc = 0.8 + 0.08 = 0.88 A, so R_Th = 5.5/0.88 = 6.25 Ω. A test source gives the same 25/4 Ω. Note that (8 + 2) ∥ 20 = 6.667 would be wrong here.

Step 6(f) The new equivalent and its load current

Thevenin: 5.5 V (+ towards a) in series with 6.25 Ω. At maximum power, R = 6.25 Ω and I_L = 5.5/12.5 = 0.440 A, flowing from a down through R to b.

Step 1(a) The three stages

I₁ alone (I₂ open, V_s short): the 8 A splits between the 4 Ω and the 2 + 10 = 12 Ω path. Current divider: I_x1 = 8 × 4/16 = 2 A, so V_x1 = 20 V.
I₂ alone: the 6 A splits at b between 10 Ω and the 2 + 4 = 6 Ω path back through a. The 6 Ω path takes 6 × 10/16 = 3.75 A from b to a, so I_x2 = −3.75 A. The 10 Ω takes 2.25 A, so V_x2 = 22.5 V.
V_s alone (both current sources open): 2/16 = 0.125 A circulates d → b → a → c, so I_x3 = −0.125 A and V_x3 = −1.25 V.

Full nodal check (c = 0): V_a = 39.5, V_b = 43.25, V_d = 2, so V_x = 41.25 ✓ and I_x = (39.5 − 43.25)/2 = −1.875 ✓.

Step 2(b) Stage I₁ (I₂ open), by nodal analysis

With I₂ open, b's only other branch is the 10 Ω, so I_x = V_x/10. Then V_a = V_b + 2I_x = 3V_x + 0.2V_x = 3.2V_x. KCL at a: 8 = V_a/4 + I_x = 0.8V_x + 0.1V_x, so V_x1 = 80/9 = 8.889 V and I_x1 = 8/9 = 0.889 A.

Step 3(b) Stage I₂ (I₁ open), by nodal analysis

KCL at b: the 6 A leaves through the 10 Ω (V_x/10) and back through 2 Ω + 4 Ω to c (V_b/6 = 3V_x/6 = V_x/2). So 6 = 0.6V_x, giving V_x2 = 10 V. Then I_x2 = −V_b/6 = −30/6 = −5 A (it flows b → a).

Full nodal check with the VCVS: V_x = 170/9 = 18.889 ✓.

Step 1(a) Initial voltage and τ

|v_C(0)| = q₀/C = 500 µC/20 µF = 25 V. The source will charge the top plate positive, but the figure shows the top plate currently negative. Measuring v_C top-minus-bottom (the key's convention, the direction the source drives), v_C(0) = −25 V. τ = RC = 1000 × 20 × 10⁻⁶ = 0.020 s.

Step 2(b) The two expressions

v_C(∞) = 50 V, so v_C(t) = 50 + (−25 − 50)e^−t/0.02 = 50 − 75e^−50t V. The current is i(t) = C dv_C/dt = 20 × 10⁻⁶ × 75 × 50 e^−50t = 0.075e^−50t A, clockwise. Check at 0⁺: (50 − (−25))/1000 = 75 mA ✓.

Step 3(c) When does v_C cross zero?

0 = 50 − 75e^−50t₁ gives e^−50t₁ = 2/3, so t₁ = ln(1.5)/50 = 0.405465/50 = 8.109 × 10⁻³ s.

Step 4(e) Classify the RLC

ω₀ = 1/√(0.1 × 50 × 10⁻⁶) = 1/√(5 × 10⁻⁶) = 447.214 rad/s; α = R/2L = 50/0.2 = 250 s⁻¹; ζ = α/ω₀ = 0.559. ζ has no unit, even though the paper asks for one. Since ζ < 1: [B] underdamped, with ω_d = √(ω₀² − α²) = √(200000 − 62500) = √137500 = 370.810 rad/s.

Step 5(f), (g), (h) Natural + forced, then the constants

(f) i_n(t) = e^−250t(A cos 370.810t + B sin 370.810t).
(g) In DC steady state a series capacitor blocks, so i_f = 0.
(h) i(0) = A = 0.5. KVL at 0⁺: v_L(0⁺) = 100 − 50 × 0.5 − 20 = 55 V, so di/dt(0⁺) = 55/0.1 = 550 A/s. The slope of the solution at 0 is −250A + 370.810B = 550, so B = 675/370.810 = 1.820.

Step 1The two node equations

V₁: (V₁ + j40)/2 + V₁/(−j2) + (V₁ − V₂)/(j1) + 10 = 0.
V₂: (V₂ − V₁)/(j1) = 2I + 10 = jV₁ + 10.
The V₂ equation collapses: −jV₂ + jV₁ = jV₁ + 10, so V₂ = j10.

Step 2Back to the time domain

V_Th = V₂ = 10∠90° V, so v_Th(t) = 10 cos(100t + 90°) V. Substituting into the V₁ equation gives V₁ = 20 − j20, so I = V₁/(−j2) = 10 + j10 = 14.142∠45° A and i(t) = 14.142 cos(100t + 45°) A.
Phase differences: Δφ₁ = 90° − (−90°) = 180°; the dependent source is 2I = 28.284∠45° against 10∠0°, so Δφ₂ = 45°.

Step 1Solve the resistive circuit

With the capacitors open, the only path is 9 V → 6 Ω → L₁ → A → 3 Ω → L₂ → 6 Ω → ground, with the 3 A injected at A. KCL at A: I_L2 = I_L1 + 3. KVL round the outer path: 9 = 6I_L1 + 3I_L2 + 6I_L2. Substituting: 9 = 15I_L1 + 27, so I_L1 = −1.2 A and I_L2 = 1.8 A.

Step 2Capacitor voltages, then energies

V_C1 = 9 − 6I_L1 = 16.2 V; V_C2 = 6I_L2 = 10.8 V. Energies: ½(2 mH)(1.2)² = 1.44 mJ; ½(4 mH)(1.8)² = 6.48 mJ; ½(20 µF)(16.2)² = 2.624 mJ; ½(50 µF)(10.8)² = 2.916 mJ.

Step 1Before and after

t < 0: v_c = 120 × 50/60 = 100 V. Separately, the 60 Ω/50 V/200 Ω loop carries i(0⁻) = 50/260 = 0.1923 A. t > 0: C sees 60 ∥ 200 ∥ 50 = 24 Ω, so τ = 24 × 0.05 = 1.2 s. Also v_c(∞) = 50 × (200 ∥ 50)/(60 + 200 ∥ 50) = 50 × 40/100 = 20 V.

Step 2The responses

v_c = 20 + 80e^−t/1.2 V, t ≥ 0. After the switch, the 200 Ω sits directly across C, so i = v_c/200 = 0.1 + 0.4e^−t/1.2 A: it jumps from 0.1923 A to 0.5 A at t = 0.

Step 1Initial values and classification

t < 0: C is open, so no current reaches it and i(0⁻) = 0. The shorted inductor puts C across the lower 2 Ω of a 2 Ω/2 Ω divider: v_C(0⁻) = 4 × 2/4 = 2 V. t > 0: a loop of 2 Ω, 0.5 H and 1 F, so α = 2/(2 × 0.5) = 2 s⁻¹ and ω₀ = 1/√0.5 = 1.414 rad/s. Since α > ω₀: overdamped, s = −2 ± √2 = −0.586, −3.414.

Step 2Slope and constants

KVL at 0⁺: L di/dt = −v_C − Ri = −2 − 0, so di/dt(0⁺) = −4 A/s. With i = A₁e^−3.414t + A₂e^−0.586t: A₁ + A₂ = 0 and −3.414A₁ − 0.586A₂ = −4, so A₁ = √2 = 1.414 and A₂ = −1.414.