Question

Two different coils have self-inductance \(3L\) and \(L\). The current in both the coils is increased at the same constant rate. At certain instant of time, the power given to the two coils is same. At that time there was current and voltage induced in the two coils. At the same instant, the ratio of energy stored in the first coil to that in the second coil is

• A. \(1:9\)
• B. \(1:3\)
• C. \(3:1\)
• D. \(9:1\)

Hint:

Power is L i (di/dt); equal power and equal di/dt give the ratio of currents.

Answer 1

The Correct Option is B

Step 1: Power Delivered to an Inductor:
The induced emf is \(\varepsilon=L\dfrac{di}{dt}\), and the power given to the coil is \(P=\varepsilon i=L\dfrac{di}{dt}i\).

Step 2: Equal Power, Equal Rate:
Both currents rise at the same rate, so \(\dfrac{di}{dt}\) is the same. Equal power gives
\[ (3L)i_1=(L)i_2\Rightarrow i_2=3i_1 \]

Step 3: Energy Stored:
\(U=\dfrac12Li^2\).
\[ \frac{U_1}{U_2}=\frac{\frac12(3L)i_1^2}{\frac12L(3i_1)^2}=\frac{3}{9}=\frac13 \]

Step 4: Check the Options:
\(1:9\) would be the ratio if we compared \(i^2\) alone, and \(3:1\) and \(9:1\) are inverses or wrong. So (B) is correct.

Final Answer:
The energy ratio is \(1:3\), option (B). \[ \boxed{\text{(B) } 1:3} \]