Question:
The current in an inductor is given by I = (3t + 8) where t is in second. The magnitude of induced emf produced in the inductor is 12 mV. The self inductance of the inductor _______ mH.
The current in an inductor is given by I = (3t + 8) where t is in second. The magnitude of induced emf produced in the inductor is 12 mV. The self inductance of the inductor _______ mH.
Updated On: Jan 13, 2026
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Correct Answer: 4
Approach Solution - 1
The emf (\( \varepsilon \)) induced in an inductor is related to the rate of change of current and self-inductance (\( L \)) by the formula:
\[ |\varepsilon| = L \frac{dI}{dt}. \]
Step 1: Determine the rate of change of current
The given current is:
\[ I = 3t + 8. \]
Differentiate \( I \) with respect to \( t \):
\[ \frac{dI}{dt} = 3 \, \text{A/s}. \]
Step 2: Substitute the given values
The magnitude of induced emf is \( |\varepsilon| = 12 \, \text{mV} = 12 \times 10^{-3} \, \text{V} \). Substituting into the formula:
\[ 12 \times 10^{-3} = L \cdot 3. \]
Step 3: Solve for \( L \)
Rearrange to find \( L \):
\[ L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \, \text{H}. \]
Convert to millihenries:
\[ L = 4 \, \text{mH}. \]
Thus, the self-inductance of the inductor is \( L = 4 \, \text{mH} \).
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Approach Solution -2
Step 1: Given data.
Current through the inductor, \( I = 3t + 8 \)
Induced emf, \( e = 12 \, \text{mV} = 12 \times 10^{-3} \, \text{V} \)
Step 2: Formula for induced emf in an inductor.
\[ e = L \frac{dI}{dt} \] where \( L \) is the self-inductance.
Step 3: Find \( \frac{dI}{dt} \).
Given \( I = 3t + 8 \), differentiate with respect to \( t \):
\[ \frac{dI}{dt} = 3 \, \text{A/s} \]
Step 4: Substitute the values in the emf equation.
\[ 12 \times 10^{-3} = L \times 3 \] \[ L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \, \text{H} \] \[ L = 4 \, \text{mH} \]
Step 5: Final Answer.
The self inductance of the inductor is:
\[ \boxed{4 \, \text{mH}} \]
Final Answer: 4 mH
Current through the inductor, \( I = 3t + 8 \)
Induced emf, \( e = 12 \, \text{mV} = 12 \times 10^{-3} \, \text{V} \)
Step 2: Formula for induced emf in an inductor.
\[ e = L \frac{dI}{dt} \] where \( L \) is the self-inductance.
Step 3: Find \( \frac{dI}{dt} \).
Given \( I = 3t + 8 \), differentiate with respect to \( t \):
\[ \frac{dI}{dt} = 3 \, \text{A/s} \]
Step 4: Substitute the values in the emf equation.
\[ 12 \times 10^{-3} = L \times 3 \] \[ L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \, \text{H} \] \[ L = 4 \, \text{mH} \]
Step 5: Final Answer.
The self inductance of the inductor is:
\[ \boxed{4 \, \text{mH}} \]
Final Answer: 4 mH
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