Skip to main content

Derivatives at t=0+: 2 H, 100 V, 1 μF

Short answer

At t = 0+, i1 = i2 = 0; di1/dt = 50 A/s, di2/dt = 25 A/s; d²i1/dt² = −125 A/s², d²i2/dt² = −1.25×10⁶ A/s². All results from mesh equations with zero initial inductor current and capacitor voltage.

In the given network of the figure below, both the energy storing elements are initially relaxed i.e. no current is flowing through the inductor and no charge is accumulated across the capacitor before application of voltage. The switch K is closed at t = 0. Find the values of i1, i2, di1/dt, di2/dt, d²i1/dt², d²i2/dt² at t = 0⁺. [Figure: 100 V source; switch K and a 2 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); a 1 μF capacitor on the right]

Assumptions: both elements relaxed. Left mesh (current i₁): 100 V, switch K, L = 2 H, common resistor 10 Ω. Right mesh (current i₂): common 10 Ω, another 10 Ω, C = 1 μF.

Mesh equations

Mesh 1:  100 = 2 di₁/dt + 10(i₁ − i₂)
Mesh 2:  10(i₂ − i₁) + 10 i₂ + (1/C)∫i₂ dt = 0  →  20 i₂ − 10 i₁ + v_C = 0

At t = 0⁺ (i_L = 0, v_C = 0)

i₁(0⁺) = 0
20 i₂ − 0 + 0 = 0  →  i₂(0⁺) = 0

First derivatives

Mesh 1:  100 = 2 i₁′ + 0  →  i₁′(0⁺) = 50 A/s
Mesh 2 differentiated:  20 i₂′ − 10 i₁′ + i₂/C = 0
        20 i₂′ = 10 × 50  →  i₂′(0⁺) = 25 A/s

Second derivatives

Mesh 1 differentiated:  0 = 2 i₁″ + 10(i₁′ − i₂′)
        i₁″ = −10(50 − 25)/2 = −125 A/s²
Mesh 2 differentiated twice:  20 i₂″ − 10 i₁″ + i₂′/C = 0
        i₂″ = [10(−125) − 25/10⁻⁶]/20 = (−1250 − 2.5×10⁷)/20
            = −1.25006×10⁶ A/s²
QuantityValue at 0⁺
i₁0
i₂0
di₁/dt50 A/s
di₂/dt25 A/s
d²i₁/dt²−125 A/s²
d²i₂/dt²−1.25×10⁶ A/s²

Answer: i₁ = i₂ = 0; di₁/dt = 50, di₂/dt = 25; d²i₁/dt² = −125, d²i₂/dt² = −1.25×10⁶ (SI units).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Circuit Theory chapter 2 questions →