A source of constant voltage V is connected t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $i$ be the current through the resistance $R$,and $i_1$ and $i_2$ be the currents through inductors $L_1$ and $L_2$ respectively.At $t=0$,the

Vedclass · Accepted Answer

$A, B, C$

Vedclass · Answer

(A) Let $i$ be the current through the resistance $R$,and $i_1$ and $i_2$ be the currents through inductors $L_1$ and $L_2$ respectively.At $t=0$,the inductors act as open circuits because the current cannot change instantaneously. Thus,the current through the resistance $R$ is $0$.For $t > 0$,the voltage across the parallel combination of $L_1$ and $L_2$ is the same. Therefore,$V_{L1} = V_{L2} \implies L_1 \frac{di_1}{dt} = L_2 \frac{di_2}{dt}$.Integrating both sides with respect to time,we get $L_1 i_1 = L_2 i_2$ (assuming initial currents are zero). This implies $\frac{i_1}{i_2} = \frac{L_2}{L_1}$,so the ratio of currents is fixed at all times.After a long time,the inductors act as short circuits (ideal inductors). The total current $i = \frac{V}{R}$.Since $i_1 + i_2 = i = \frac{V}{R}$ and $L_1 i_1 = L_2 i_2$,we have $i_2 = \frac{L_1}{L_2} i_1$.Substituting this into the current equation: $i_1 + \frac{L_1}{L_2} i_1 = \frac{V}{R} \implies i_1 \left( \frac{L_1+L_2}{L_2} \right) = \frac{V}{R} \implies i_1 = \frac{V}{R} \frac{L_2}{L_1+L_2}$.Similarly,$i_2 = \frac{V}{R} \frac{L_1}{L_1+L_2}$.Thus,options $A, B,$ and $C$ are correct.

Similar Questions

$A$ small bar magnet of dipole moment $M$ is moving with speed $v$ along the $x$-direction towards a small closed circular conducting loop of radius $a$ with its centre $O$ at $x=0$ (see figure). Assume $x >> a$ and the coil has a resistance $R$. Which of the following statement$(s)$ is/are true?

The dimension of the ratio of magnetic flux and the resistance is equal to that of:

The magnetic field varies as $B = B_0 e^{-t}$. The coil has a radius $r$ and resistance $R$. What is the power dissipated when the key $K$ is closed at $t = 0$?

The switch $S$ shown in the circuit is closed at $t=0$. The ratio of the current drawn from the battery by the circuit at $t=0$ and $t=\infty$ is:

In the circuit shown in the figure,a conducting wire $HE$ is moved with a constant speed $V$ towards the left. The complete circuit is placed in a uniform magnetic field $\vec B$ perpendicular to the plane of the circuit and directed inward. The current in $HKDE$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module