Question

A constant voltage is applied to a series $R-L$ circuit by closing the switch. The voltage across inductor $(L=2H)$ is $20V$ at $t=0$ and drops to $5V$ at $20ms$. The value of $R$ in $\Omega$ is :

• A. $100\ln 2\Omega$
• B. $100(1-\ln 2)\Omega$
• C. $100\ln 4\Omega$
• D. $100(1-\ln 4)$

Answer 1

The correct option is C $100\ln 4\Omega$

Value remains 1/4th in $20ms$ times. hence two half lives are equal to $20ms$. So, one half-life is $10ms$.
$t_{1/2}=\left(\ln 2\right){\tau }_C=\left(\ln 2\right)\frac{L}{R}$
$\therefore R=\frac{\left(\ln 2\right)L}{t_{1/2}}=\frac{\left(\ln 2\right)\left(2\right)}{10\times {10}^{-3}}=\left(100\ln 4\right)\Omega$