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Alternating Current

Question

An inductor (L = 20 H), a resistor (R = 100 Ω) and a battery (E = 10 V) are connected in series. After a long time, the circuit is short-circuited and then the battery is disconnected. Find the current in the circuit at 1 ms after short circuiting.

4.5 x 10⁵ A

3.2x 10^-⁵ A

9.8 × 10^-5 A

6.7 × 10^-4 A

Solution

6.7 × 10^-4 A

The initial current i = i₀ = $\frac{E}{R}$

= $\frac{10}{100}$

i = 0.10 A

the time constant

$τ = \frac{L}{R}$

= $\frac{20 mH}{100 Ω}$

$τ = 0.20 ms$

The current at t = 1 ms is

i = i₀ e ^{-t /Τ}

⇒ i = ( 0.10 ) $e^{(\frac{- 1 ms}{0.20 ms})}$

⇒ i = ( 0.10 ) e^-⁵

⇒ i = (0.10) × 0.0067

⇒ i = 6.7 × 10^-4 A