Question

A coil of inductance 8.4 mH and resistance $6\Omega$ is connected to 12 V battery. The current in the coil is 1 A at approximately the time :

• A. 500 s
• B. 20 s
• C. 35 ms
• D. 1 ms

Answer 1

Answer: (D)

1 ms

For a DC L-R circuit, current in the inductor grows as per below relationship,

$I\left(t\right)=I_0(1-e^{-t/\tau })$

Here, $I_0=V/R=12/6=2A$, is the steady-state current through the circuit.

$\tau =L/R=8.4m/6=1.4ms$, is the time constant.

So, $I\left(t\right)=1A⟹2(1-e^{-t/1.4m})=1$

$⟹e^{-t/1.4m}=0.5$

Taking logarithm of both sides,

$-t/1.4m=-0.693⟹t=0.97ms\approx 1ms$

Option D is correct.