Question

Two different coils have self inductance $L_1=8$$mH$ and $L_2=2$$mH$. The currents in both are increasing at the same constant rate. At a certain instant of time, the power given to the two coils is the same. At this moment the current, the induced voltage and energy stored in the first coil are $i_1$ , $V_1$ and $U_1$ respectively. The corresponding values in the second coil are $i_2$, $V_2$ and $U_2$ respectively. Then the values of $\frac{i_1}{i_2},\frac{V_1}{V_2}$ and $\frac{U_1}{U_2}$ are respectively :

• A. $\frac{1}{4},4,\frac{1}{4}$
• B. $4,\frac{1}{4},\frac{1}{4}$
• C. $\frac{1}{4},\frac{1}{4},4$
• D. $4,4,\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{1}{4},4,\frac{1}{4}$

$L_1=8mH,L_2=2mH$

Rate of increasing current is same in both and is constant.
At a instant when power is same

$\therefore i_1{L}_1\frac{d{i}_1}{dt}=L_2\frac{d{i}_2}{dt}{i}_2$

As $\frac{d{i}_1}{dt}=\frac{d{i}_2}{dt}$

Hence $4i_1=i_2$
Energy $=\frac{L}{2}{I}^2$

$\frac{U_1}{U_2}=\frac{L_1{I}_1^2}{L_2{I}_2^2}=\frac{8}{2}\times \frac{1}{16}=\frac{1}{4}$
Voltage induced$=L\frac{di}{dt}$

$\frac{V_1}{V_2}=\frac{L_1\frac{d{i}_1}{dt}}{L_2\frac{d{i}_2}{dt}}=\frac{4}{1}$