Question

An inductor-coil of inductance $20\text{mH}$ having resistance of $10\Omega$ is joined to an ideal battery of emf $5\text{V}.$ Find the emf induced across the inductor at $t=0\text{s.}$

• A. $5\text{V}$
• B. $2.5\text{V}$
• C. $1\text{V}$
• D. $0.5\text{V}$

Answer 1

The correct option is A $5\text{V}$

Given, $L=20\text{mH};R=10\Omega ;E=5\text{V}$

The induced emf across an inductor is,

$E_i=-L\left(\frac{di}{dt}\right)$

$\left|\begin{array}{c}{E}_i\end{array}\right|=L\left(\frac{di}{dt}\right)....\left(1\right)$

The time constant of the given $L-R$ circuit is,

$\tau =\frac{L}{R}=\frac{20\times {10}^{-3}}{10}=2\times {10}^{-3}\text{s}$

At any instant time $t$ current in $L-R$ growth circuit is,

$i=i_0[1-e^{(-\frac{t}{\tau })}]$

$i=\frac{E}{R}[1-e^{(-\frac{t}{\tau })}]$

$i=\frac{5}{10}[1-e^{\left(\frac{-t}{2\times {10}^{-3}}\right)}]$

$i=0.5[1-e^{\left(\frac{-t}{2\times {10}^{-3}}\right)}]$

Differentiating the above relation with time $t$ we get,

$\frac{di}{dt}=\frac{d}{dt}\left(0.5[1-e^{\left(\frac{-t}{2\times {10}^{-3}}\right)}]\right)$

$=[0.5\times -e^{\left(\frac{-t}{2\times {10}^{-3}}\right)}\times \frac{-1}{2\times {10}^{-3}}]$

Putting this value in (1) we get,

$\left|\begin{array}{c}{E}_i\end{array}\right|=20\times {10}^{-3}\times [0.5\times -e^{\left(\frac{-t}{2\times {10}^{-3}}\right)}\times \frac{-1}{2\times {10}^{-3}}]$

$=5\times -e^{\left(\frac{-t}{2\times {10}^{-3}}\right)}$

At time $t=0\text{s}$

$\left|\begin{array}{c}{E}_i\end{array}\right|=5\text{V}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.

Alternate solution: At $t=0$, the inductor will behave like an infinite resistance. Hence, $E$ will be $5\text{V}$ as no current passes through the circuit.