Question

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

• A. $100\ln \left(2\right)\mathrm{\Omega }$
• B. $100(1-\ln \left(2\right))\mathrm{\Omega }$
• C. $100\ln \left(4\right)\mathrm{\Omega }$
• D. $100\left(1-\ln \left(4\right)\right)\mathrm{\Omega }$

Answer 1

The correct option is C

$100\ln \left(4\right)\mathrm{\Omega }$

Step 1:Given

At time($t$)=$0$

voltage across inductor=$20V$

inductance of inductor=$2H$

At time($t$)=$20milliseconds$

Voltage after decreasing=$5V$

Step 2:Formulae applied

Voltage becomes ${\frac{1}{4}}^{\mathrm{th}}$ in $20\mathrm{ms}$

which is equal to 2 half-lives

Thus, two half-lives equal to $20\mathrm{ms}$.

So, one half-life is $10\mathrm{ms}$.

Step 3:Calculation

${\mathrm{t}}_{\frac{1}{2}}=\ln \left(2\right)\frac{\mathrm{L}}{\mathrm{R}}$

$\Rightarrow$$\mathrm{R}=\ln \left(2\right)\frac{\mathrm{L}}{{\mathrm{t}}_{{\displaystyle \frac{1}{2}}}}$

$\Rightarrow$$\mathrm{R}=\ln \left(2\right)\frac{2}{10\times {10}^{-3}}$

$\Rightarrow$$\mathrm{R}=100\ln \left(4\right)\mathrm{\Omega }$

Hence option $\left(C\right)$ is the correct option.