An inductor of inductance L = 400 mH and resistors of resistance $R_{1} = 2 Ω$ and $R_{2} = 2 Ω$ are connected to a battery of emf 12 V 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

6 e^{- 5 t} V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{12}{t} e^{- 3 t} V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6 (1 - e^{- t / 0.2}) V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12 e^{- 5 t} V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $12 e^{- 5 t} V$
$I_{1} = \frac{E}{R_{1}} = \frac{12}{2} = 6 A E = L \frac{d l_{2}}{d t} + R_{2} \times I_{2} I_{2} - I_{0} (1 - e^{- t / t_{0}}) \Rightarrow I_{0} = \frac{E}{R_{2}} = \frac{12}{2} = 6 A t_{0} = \frac{L}{R} = \frac{400 \times 10^{- 3}}{2} 0.2 I_{2} = 6 (1 - e^{- t / 0.2})$
potential drop across $L = E - R_{2} l_{2} = 12 - 2 \times 6 (1 - e^{- b t}) = 12 e^{- 5 t}$

Suggest Corrections

Similar questions

An inductor of inductance L = 400 mH and resistors of resistance $R_{1} = 2 Ω$ and $R_{2} = 2 Ω$ areconnected to a battery of emf 12 V as shown in the figure. Theinternal 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