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

Step 1: Current growth law.
In an $RL$ circuit the current rises as \[ I=I_0\left(1-e^{-t/\tau}\right),\quad \tau=\frac{L}{R}. \]

Step 2: Find the time constant.
With $L=80$ H and $R=40\,\Omega$, \[ \tau=\frac{80}{40}=2\ \text{s}. \]

Step 3: Set the half-value condition.
We want $I=\tfrac12 I_0$, so \[ \frac12=1-e^{-t/\tau}. \]

Step 4: Isolate the exponential.
\[ e^{-t/\tau}=\frac12. \]

Step 5: Take logarithm.
\[ \frac{t}{\tau}=\ln 2=0.693, \] so $t=0.693\,\tau$.

Step 6: Put the value of $\tau$.
\[ t=0.693\times2=2\times0.693\ \text{s}. \]
\[ \boxed{t=2\times0.693\ \text{s}} \]