Question:
Two coils have self inductance \(L_1\) and \(L_2\). The current through them is increasing at constant rate. If the power dissipated in both the coils is same, then the ratio of energy stored in the coil having inductance \(L_1\) to that in \(L_2\) is
Two coils have self inductance \(L_1\) and \(L_2\). The current through them is increasing at constant rate. If the power dissipated in both the coils is same, then the ratio of energy stored in the coil having inductance \(L_1\) to that in \(L_2\) is
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Power in an inductor is \(P=LI\frac{dI}{dt}\), and energy is \(U=\frac12LI^2\).
Updated On: Oct 1, 2026
- \(\frac{L_1^2}{L_2^2}\)
- \(\frac{L_2^2}{L_1^2}\)
- \(\frac{L_1}{L_2}\)
- \(\frac{L_2}{L_1}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
With a steady rate of change of current, the power in an inductor is \(P=LI\dfrac{dI}{dt}\). The stored energy is \(U=\tfrac12LI^2\).
Step 2: Key Formula or Approach
Equal power with the same \(dI/dt\) gives \(L_1I_1=L_2I_2\), so \(I\propto\dfrac1L\).
Step 3: Detailed Explanation
\[ \frac{U_1}{U_2}=\frac{L_1I_1^2}{L_2I_2^2}=\frac{L_1}{L_2}\cdot\frac{L_2^2}{L_1^2}=\frac{L_2}{L_1} \]
Final Answer:
The energy ratio is \(\frac{L_2}{L_1}\), option (D). \[ \boxed{\dfrac{L_2}{L_1}\ \text{(D)}} \]
With a steady rate of change of current, the power in an inductor is \(P=LI\dfrac{dI}{dt}\). The stored energy is \(U=\tfrac12LI^2\).
Step 2: Key Formula or Approach
Equal power with the same \(dI/dt\) gives \(L_1I_1=L_2I_2\), so \(I\propto\dfrac1L\).
Step 3: Detailed Explanation
\[ \frac{U_1}{U_2}=\frac{L_1I_1^2}{L_2I_2^2}=\frac{L_1}{L_2}\cdot\frac{L_2^2}{L_1^2}=\frac{L_2}{L_1} \]
Final Answer:
The energy ratio is \(\frac{L_2}{L_1}\), option (D). \[ \boxed{\dfrac{L_2}{L_1}\ \text{(D)}} \]
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