Q140 / 150 · MHT CET 2026 (17 April Shift 1) · Physics
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
- A$\dfrac{L_1^2}{L_2^2}$
- B$\dfrac{L_2^2}{L_1^2}$
- C$\dfrac{L_1}{L_2}$
- Correct answerD$\dfrac{L_2}{L_1}$
Solution
Power supplied to an inductor is given by $P = |\varepsilon| i = L \dfrac{di}{dt} i$
Given that the current is increasing at a constant rate, $\dfrac{di_1}{dt} = \dfrac{di_2}{dt} = k$
Since the power dissipated in both coils is the same, $P_1 = P_2$
$L_1 k i_1 = L_2 k i_2 \Rightarrow L_1 i_1 = L_2 i_2 \Rightarrow \dfrac{i_1}{i_2} = \dfrac{L_2}{L_1}$
The energy stored in an inductor is $U = \dfrac{1}{2} L i^2$
The ratio of energy stored is $\dfrac{U_1}{U_2} = \dfrac{\dfrac{1}{2} L_1 i_1^2}{\dfrac{1}{2} L_2 i_2^2} = \dfrac{L_1}{L_2} \left( \dfrac{i_1}{i_2} \right)^2$
Substituting the ratio of currents, we get:
$\dfrac{U_1}{U_2} = \dfrac{L_1}{L_2} \left( \dfrac{L_2}{L_1} \right)^2 = \dfrac{L_2}{L_1}$
Answer: $\dfrac{L_2}{L_1}$
Given that the current is increasing at a constant rate, $\dfrac{di_1}{dt} = \dfrac{di_2}{dt} = k$
Since the power dissipated in both coils is the same, $P_1 = P_2$
$L_1 k i_1 = L_2 k i_2 \Rightarrow L_1 i_1 = L_2 i_2 \Rightarrow \dfrac{i_1}{i_2} = \dfrac{L_2}{L_1}$
The energy stored in an inductor is $U = \dfrac{1}{2} L i^2$
The ratio of energy stored is $\dfrac{U_1}{U_2} = \dfrac{\dfrac{1}{2} L_1 i_1^2}{\dfrac{1}{2} L_2 i_2^2} = \dfrac{L_1}{L_2} \left( \dfrac{i_1}{i_2} \right)^2$
Substituting the ratio of currents, we get:
$\dfrac{U_1}{U_2} = \dfrac{L_1}{L_2} \left( \dfrac{L_2}{L_1} \right)^2 = \dfrac{L_2}{L_1}$
Answer: $\dfrac{L_2}{L_1}$
Hence, the correct answer is option: 'D'.