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

\[= \int\limits_0^t i_0 = \int\limits_0^t i_0 (1 - e^{- t/\tau} )dt\]

\[ = i_0 \left[ \int\limits_0^t dt - \int\limits_0^t e^{- t/\tau} dt \right]\]

\[ = i_0 \left[ t - \left\{ \left( - \tau \right) \left| e^{- t/\tau} \right|^t_0 \right\} \right]\]

\[ = i_0 \left[ t + \tau\left\{ e^{- t/\tau} - 1 \right\} \right]\]

The steady-state current and the time constant for the given circuit are as follows:-

\[i_0  = \frac{E}{R_{total}} = \frac{6}{10 + 2} = \frac{6}{12} = 0 . 5  A\]

\[\tau = \frac{L}{R} = \frac{120}{12} = 0 . 01  s\]

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