In the circuit shown switch is closed at time t = 0 . Let I1... | PHYSICS - EFFECT OF DIELECTRIC ON CAPACITANCE | SolveeIt | Solveeit

Today, August 19

Question

In the circuit shown switch is closed at time t = 0 . Let I₁ and I₂ be the currents in both branches at t = 0, then ratio of $\frac{I_{1}}{I_{2}}$ .

A

$\frac{I_{1}}{I_{2}}$ = 2 : 1

B

= 1 : 2

C

$\frac{I_{1}}{I_{2}}$ = 1 : 1

D

$\frac{I_{1}}{I_{2}}$ = $\frac{3RC}{RC}$

Answer

$\frac{I_{1}}{I_{2}}$ = 1 : 1

Explanation

Solution

At t = 0 capacitors acts as short circuit and hence current in both branch I = $\frac { \mathrm { V } } { \mathrm { R } }$ .