Question

An inductor-coil of resistance 10 Ω and inductance 120 mH is connected across a battery of emf 6 V and internal resistance 2 Ω. Find the charge which flows through the inductor in (a) 10 ms, (b) 20 ms and (c) 100 ms after the connections are made.

Answer 1

Given:
Inductance, L = 120 mH = 0.120 H
Resistance, R = 10 Ω
Emf of the battery, E = 6 V
Internal resistance of the battery, r = 2 Ω

The current at any instant in the LR circuit is given by
i = i₀(1 − e^−t/τ)
Charge dQ flown in time dt is given by
dQ = idt = i₀(1 − e^−t/τ)dt
Q = ∫ dQ
$$\begin{gathered} =\underset{0}{\overset{t}{\int }}{i}_0=\underset{0}{\overset{t}{\int }}{i}_0(1-e^{-t/\tau })dt \\ =i_0\left[\underset{0}{\overset{t}{\int }}dt-\underset{0}{\overset{t}{\int }}{e}^{-t/\tau }dt\right] \\ =i_0\left[t-\left\{\left(-\tau \right){{\left|e^{-t/\tau }\right|}^t}_0\right\}\right] \\ =i_0\left[t+\tau \left\{e^{-t/\tau }-1\right\}\right] \end{gathered}$$
The steady-state current and the time constant for the given circuit are as follows:
$$\begin{gathered} i_0=\frac{E}{R_{\mathrm{total}}}=\frac{6}{10+2}=\frac{6}{12}=0.5\mathrm{A} \\ \tau =\frac{L}{R}=\frac{120}{12}=0.01\mathrm{s} \end{gathered}$$

Now,
(a) At time t = 0.01 s,
Q = 0.5 [0.01 + 0.01(e^−0.1/0.01 − 0.01)]
= 0.00108 = 1.8 × 10⁻³ = 1.8 mΩ

(b) At t = 20 ms = 2 × 10⁻² s = 0.02 s,
Q = 0.5 [0.02 + 0.01(e^−0.02/0.01 − 0.01)]
= 0.005676 = 5.7 × 10⁻³ C
= 5.7 mC

(c) At t = 100 ms = 0.1 s,
Q = 0.5 [0.1 + 0.1 (e^−0.1/0.01 − 0.01)]
= 0.045 C = 45 mc