Question

In $0.1s$, the current in a coil increases from $1A$ to $1.5A$. If the inductance of the coil is $60mH$, then induced current in external resistance of $3\Omega$ will be

• A. $1A$
• B. $0.5A$
• C. $0.2A$
• D. $0.1A$

Answer 1

The correct option is D

$0.1A$

Step 1: Given,

The inductance of the coil, $L=60mH$

$$\begin{gathered} i_1=1A \\ {i}_2=1.5A \end{gathered}$$

Change in current:

$$\begin{gathered} ∆i=i_2-i_1 \\ ∆i=(1.5-1)=0.5A \end{gathered}$$

Time interval $∆t=0.1s$

External resistance $R=3\Omega$

Step 2: Formula used

$emf=L\frac{di}{dt}$

$I=\frac{emf}{R}$

Where $L$ is the inductance, $\frac{di}{dt}$ is the rate of change of current, $I$ is the current and $R$ is the resistance.

Step 3: Finding $emf$

The coefficient of self inductance $L$ of a coil is numerically equal to $emf$ induced in the coil when the rate of change of current $\frac{∆i}{∆t}$is unity.

$$\begin{gathered} \Rightarrow emf=L\frac{∆i}{∆t} \\ \Rightarrow emf=(60\times {10}^{-3})\frac{0.5}{0.1} \end{gathered}$$

⇒ $emf=0.3V$

Step 4: Finding $I$ in the external resistor

⇒ $emf=IR$

⇒ $0.3=I\left(3\right)$

⇒ $I=0.1A$

Hence, the correct option is D.