Question

Two different coils have self inductance $L_1=8mH,L_2=2mH$. The current in the second coil is also increased at the same constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coil is the same. At that time, the current, the induced voltage and the energy stored in the first coil are $i_1,V_1$ and $W_1$ respectively. Corresponding values for the second coil at the same instant are $i_2,V_2$ and $W_2$ respectively. Then

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

Answer 1

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

The correct options are
A $\frac{i_1}{i_2}=\frac{1}{4}$
C $\frac{W_2}{W_1}=4$
D $\frac{V_2}{V_1}=\frac{1}{4}$

We know

$e=L\frac{di}{dt}$

$e\propto L$

So,

$\frac{e_1}{e_2}=\frac{L_1}{L_2}=\frac{8}{2}=\frac{4}{1}$

Since $P=el=Constant$

Therefore,

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

$P_1=P_2=P$

$e_1{i}_1=e_2{i}_2$

$\therefore \frac{i_1}{i_2}=\frac{e_2}{e_1}=\frac{1}{4}$

Thus, ratio of current is 1:4