Two capacitors C_1 and C_2 = 2,C_1 are connec | vedclass

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

Question

In steady state, both the capacitors are at the same potential,

i.e., $ \quad \frac{Q_{1}}{C_{1}}=\frac{Q_{2}}{C_{2}} $ or $ \frac{Q_{1}}{C}=\frac{

Vedclass · Accepted Answer

$\frac{Q}{2},\frac{{2Q}}{3}$

Vedclass · Answer

In steady state, both the capacitors are at the same potential,

i.e., $ \quad \frac{Q_{1}}{C_{1}}=\frac{Q_{2}}{C_{2}} $ or $ \frac{Q_{1}}{C}=\frac{Q_{2}}{2 C} $ or $ Q_{2}=2 Q_{1}$

Also, $\mathrm{Q}_{1} +\mathrm{Q}_{2}=\mathrm{Q}$

$	herefore $ $\mathrm{Q}_{1}=\frac{\mathrm{Q}}{3}, \mathrm{Q}_{2}=\frac{2 \mathrm{Q}}{3}$

Two capacitors $C_1$ and $C_2 = 2\,C_1$ are connected in a circuit with a switch between them as shown in the figure. Initially the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state, the charge on capacitors will be

Similar Questions

When a proton is accelerated through $1\,V,$ then its kinetic energy will be........$eV$

Five balls marked a to $e$ are suspended using separate threads. Pairs $(b, c)$ and $(d, e)$ show electrostatic repulsion while pairs $(a, b),(c, e)$ and $(a, e)$ show electrostatic attraction. The ball marked a must be

If potential at centre of uniformaly charged ring is $V_0$ then electric field at its centre will be (assume radius $=R$ )

Two parallel metal plates having charges $+ Q$ and $-Q$ face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will

Consider the situation shown. The switch $S$ is opened for a long time and then closed. The charge flown through the battery when $S$ is closed