Question

Two different coils have self-inductance $ L_{1} = 9 \,mH $ and $ L_{2} = 3\, mH $ . At a certain instant, the current in the two coils is increasing at the same rate and the power supplied to the coils is also the same. The ratio of the energy stored in the two coils $ (U_{1} / U_{2}) $ at that instant is

• A. $ 1/3 $
• B. $ 1 $
• C. $ 3 $
• D. $ 27 $

Answer 1

Answer: (A)

Given that $L_{1}=9\,mH$
$L_{2}=3\,mH$
$\frac{dI_{1}}{dt}=\frac{dI_{2}}{dt} \ldots\left(i\right)$
Power $P_{1}=P_{2}$
$e_{1}I_{1}=e_{2}I_{2}$
$\frac{e_{1}}{e_{2}}=\frac{I_{2}}{I_{1}}$
$\frac{L_{1}\frac{dI_{1}}{dt}}{L_{2}\frac{dI_{2}}{dt}}=\frac{I_{2}}{I_{1}}$
Hence, $\frac{L_{1}}{L_{2}}=\frac{I_{2}}{I_{1}}$
or $\frac{L_{2}}{L_{1}}=\frac{I_{1}}{I_{2}} \ldots\left(ii\right)$
$\therefore \frac{U_{1}}{U_{2}}=\frac{\frac{1}{2}L_{1} I_{1}^{2}}{\frac{1}{2}L_{2}I_{2}^{2}}$
$=\frac{L_{1}}{L_{2}}\left(\frac{I_{1}}{I_{2}}\right)^{2}$ [from Eq. $\left(ii\right)$]
$=\frac{L_{2}}{L_{1}}=\frac{3}{9}=\frac{1}{3}$