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- 2074 Chaitra · 8 marks
Derivatives at t=0+: 1 H, 1 F, 10 V
Short answer
At t = 0+, i1 = i2 = 0, di1/dt = 10 A/s, di2/dt = 5 A/s, d²i1/dt² = −50 A/s², d²i2/dt² = −25.25 A/s². Inductor current and capacitor voltage start at zero; derivatives from mesh equations.
Obtain the value of i1, i2, di1/dt, di2/dt, d²i1/dt² and d²i2/dt² at t = 0⁺, if the switch is closed at t = 0 in the circuit shown in the figure below. [Figure: 10 V source; switch (closes at t = 0) and a 1 H inductor in the top wire of the left mesh (i1); a 10 Ω resistor in the middle vertical branch; a 10 Ω resistor in the top wire of the right mesh (i2) and a 1 F capacitor on the right]
Assumption: the inductor and capacitor are initially relaxed. i₁ flows in the left mesh (10 V, switch, L = 1 H, common R = 10 Ω); i₂ flows in the right mesh (common 10 Ω, 10 Ω, C = 1 F).
Mesh equations (t > 0)
Mesh 1: 10 = L di₁/dt + 10(i₁ − i₂)
Mesh 2: 10(i₂ − i₁) + 10 i₂ + (1/C)∫i₂ dt = 0
At t = 0⁺
- Inductor current cannot change: i₁(0⁺) = 0.
- Capacitor voltage cannot change: v_C(0⁺) = 0.
Mesh 2: 20 i₂ − 10 i₁ + 0 = 0 → i₂(0⁺) = 0
First derivatives
Mesh 1: 10 = 1·i₁′ + 10(0 − 0) → di₁/dt(0⁺) = 10 A/s
Mesh 2 differentiated: 20 i₂′ − 10 i₁′ + i₂/C = 0
20 i₂′ = 10(10) − 0 → di₂/dt(0⁺) = 5 A/s
Second derivatives
Mesh 1 differentiated: 0 = i₁″ + 10(i₁′ − i₂′)
i₁″ = −10(10 − 5) = −50 A/s²
Mesh 2 differentiated twice: 20 i₂″ − 10 i₁″ + i₂′/C = 0
20 i₂″ = 10(−50) − 5 = −505 → i₂″ = −25.25 A/s²
| Quantity | Value at 0⁺ |
|---|---|
| i₁ | 0 |
| i₂ | 0 |
| di₁/dt | 10 A/s |
| di₂/dt | 5 A/s |
| d²i₁/dt² | −50 A/s² |
| d²i₂/dt² | −25.25 A/s² |
Answer: as tabulated above.
✓ Answer checked by Claude Opus 5.5 (high effort)
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