Two Multiple Choice Questions on Electromagnetic Induction

(1) The initial rate of increase of current when a battery of emf 6V is connected in series with an an inductance of 2H and a resistance of 12 Ω is

(a) 0.5 A s^–1 (b) 1 A s^–1 (c) 3 A s^–1 (d) 3.5 A s^–1 (e) 4.5 A s^–1

Initially the current in the circuit is zero but the rate of increase of current is maximum. There will be no voltage drop (which is equal to R×I) across the resistance and the entire voltage will appear across the inductance. The magnitude of the voltage across the inductance will be L(dI/dt) so that initially we have

L(dI/dt) = V, from which dI/dt = V/L = 6/2 = 3 A s^–1.

[You can calculate the rate of increase of current at instant using the expression for the current in an LR circuit during the rise, given by I = I₀(1– e^–Rt/L) where ‘e’ is the base of the natural logarithm, ‘I’ is the current at the instant ‘t’ and I₀ is the final (maximum) current equal to V/R.

Differentiating the above equation, dI/dt = (I₀R/L)e^–Rt/L.

Initially, since t = 0, dI/dt = I₀R/L = (V/R)×R/L = (6/12)×12/2 = 3 As^–1

The rate of decrease of current at any instant ‘t’ when the current in an LR circuit is switched off is – (I₀R/L)e^–Rt/L since the current during the decay is given by I = I₀ e^–Rt/L]

(2) The emf induced in a coil at time t = 0.5 s when the magnetic flux (Φ) linked with the coil changes as Φ = t² – 4t + 2 weber is

(a) 3V (b) 4V (c) 5V (d) 6.5V (e) 8V

The emf (V) induced is given V = – dΦ/dt = –d[t² – 4t + 2]/dt = –2t + 4.

On substituting for t (=0.5 s), V = 3 volt