Question

An inductor of inductance L = 400 mH and resistors of resistances R₁ = 2W and R₂ = 2W are connected to a battery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is –

• A. 6 e^–5t V
• B. $\frac{12}{t}e^{- 3t}V$
• C. $6\left( 1 - e^{- t/0.2} \right)V$
• D. 12 e^–5t V

Answer 1

Answer: (D)

12 e^–5t V

Answer 2

I = I₀ (1 – e^–t/t ) (t = L/R²)

e =$\frac{LdI}{dt}$ = $\frac{+ LI_{0}e^{–t/\tau}}{\tau}$

e = $\frac{E_{0} \times L}{R \times L/R}$ e^–t/t

e = E ₀ e^–t/t

e = 12e^–5t

So option (4) is correct

Note : Smart work : At t = 0 inductor acts as open CRt and P.D across inductor will be 12V which is possible only in option (4).