Question

Two different coils have self-inductance $L_1$ = 8mH and $L_2$ = 2mH respectively. The current in both the coils is increased at a same rate. At a certain instant of time, the power being supplied to both the coils is equal. At that instant, the current, the induced voltage, and the energy stored in $L_1$ and $L_2$ are $I_1$ and $I_2$, $V_1$ and $V_2$, and $W_1$ and $W_2$ respectively. Then

• A. $\frac{i_1}{i_2}=\frac{1}{4}$
• B. $\frac{i_1}{i_2}=48$
• C. $\frac{W_2}{W_1}=4$
• D. $\frac{V_2}{V_1}=\frac{1}{4}$

Answer 1

Answer: (A), (C), (D)

$\frac{V_2}{V_1}=\frac{1}{4}$

Power supplied (P) to an inductor of an inductance (L) is ${\frac{d}{dt}}^{\left(\frac{1}{2}L{i}^2\right)}=\frac{Lidi}{dt}$
Say $i_1$ is the current in $L_1$, $P_1$ is the power supplied to $L_1$, $i_2$ is the current in $L_2$ and $P_2$ is power supplied to $L_2$.
Given that $P_1=P_2$
$\Rightarrow L_1{i}_1\frac{d{i}_1}{dt}=L_2{i}_2\frac{d{i}_2}{dt}$
Also given that $\frac{d{i}_1}{dt}=\frac{d{i}_2}{dt}$
$\Rightarrow L_1{i}_1=L_2{i}_2$
$\Rightarrow \frac{i_1}{i_2}=\frac{L_2}{L_1}=\frac{1}{4}$
and $\frac{V_1}{V_2}=\frac{L_1\frac{d{i}_1}{dt}}{L_2\frac{d{i}_2}{dt}}=\frac{L_1}{L_2}=4$
$\frac{W_1}{W_2}=\frac{\frac{1}{2}{L}_1{i}_1^2}{L_2{i}_2^2}=\frac{L_1{i}_1}{L_2{i}_2}\times \frac{i_1}{i_2}$
$=\frac{i_1}{i_2}$
$=\frac{1}{4}$