Question

A solenoid of resistance \(40\,\Omega\) and inductance \(80\,H\) is connected to a \(200\,V\) battery. How long will it take the current to reach \(50\%\) of its final equilibrium value?

• A. \(0.693\) s
• B. \(3\times0.693\) s
• C. \(\dfrac{1}{2}\times0.693\) s
• D. \(2\times0.693\) s

Hint:

For current growth in an \(RL\) circuit, \[ I=I_0(1-e^{-t/\tau}) \] and \[ \tau=\frac{L}{R}. \] To reach \(50\%\) of the final value, \[ t=\tau\ln2. \]

Answer 1

The Correct Option is D

Concept: For growth of current in an \(RL\) circuit, \[ I=I_0\left(1-e^{-t/\tau}\right) \] where \[ \tau=\frac{L}{R} \] is the time constant.

Step 1: Calculate the time constant. \[ \tau=\frac{L}{R} \] \[ =\frac{80}{40} \] \[ =2\ \text{s} \]

Step 2: Use the condition \(I=\frac{I_0}{2}\). \[ \frac12 = 1-e^{-t/2} \] \[ e^{-t/2} = \frac12 \] Taking natural logarithm, \[ -\frac{t}{2} = \ln\left(\frac12\right) = -\ln2 \] \[ t=2\ln2 \] \[ t=2(0.693) \] \[ t=1.386\ \text{s} \] \[\begin{aligned} \boxed{t=2\times0.693\ \text{s}} \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.