Question:
In a coil the current varies from \(-3\,A\) to \(+3\,A\) in \(4\,s\), and induces an emf of \(0.2\,V\). The self inductance of the coil is
In a coil the current varies from \(-3\,A\) to \(+3\,A\) in \(4\,s\), and induces an emf of \(0.2\,V\). The self inductance of the coil is
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Always calculate total current change carefully:
\[
\Delta I = I_f-I_i
\]
Negative current values are extremely important in inductance problems.
Updated On: Jun 17, 2026
- \(0.133\,H\)
- \(0.266\,H\)
- \(0.65\,H\)
- \(0.532\,H\)
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The Correct Option is A
Solution and Explanation
Concept:
The induced emf in an inductor is:
\[
e=L\frac{dI}{dt}
\]
where:
• \(L\) = self inductance
• \(\dfrac{dI}{dt}\) = rate of change of current
Step 1: Calculate change in current. Initial current: \[ I_1=-3A \] Final current: \[ I_2=+3A \] Therefore: \[ \Delta I = 3-(-3)=6A \]
Step 2: Calculate rate of change of current. Time interval: \[ \Delta t=4s \] Thus: \[ \frac{dI}{dt}=\frac{6}{4}=1.5\,A/s \]
Step 3: Use emf formula. Given: \[ e=0.2V \] Using: \[ e=L\frac{dI}{dt} \] \[ 0.2=L(1.5) \] \[ L=\frac{0.2}{1.5} \] \[ L=0.133H \] Hence: \[ \boxed{0.133H} \]
• \(L\) = self inductance
• \(\dfrac{dI}{dt}\) = rate of change of current
Step 1: Calculate change in current. Initial current: \[ I_1=-3A \] Final current: \[ I_2=+3A \] Therefore: \[ \Delta I = 3-(-3)=6A \]
Step 2: Calculate rate of change of current. Time interval: \[ \Delta t=4s \] Thus: \[ \frac{dI}{dt}=\frac{6}{4}=1.5\,A/s \]
Step 3: Use emf formula. Given: \[ e=0.2V \] Using: \[ e=L\frac{dI}{dt} \] \[ 0.2=L(1.5) \] \[ L=\frac{0.2}{1.5} \] \[ L=0.133H \] Hence: \[ \boxed{0.133H} \]
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