JEE Physics 2024 Question with Solution

Given: DC source of $9\,\text{V}$. A $6\,\Omega$ resistor is on the left vertical branch (top), a $3\,\Omega$ resistor on the right vertical branch (bottom). A $3\,\mu\text{F}$ capacitor is at the left bottom to ground, and a $6\,\mu\text{F}$ capacitor is on the right vertical branch. Points $A$ and $B$ are connected by a wire, so $V_A = V_B$.

Find: The charge stored in the $6\,\mu\text{F}$ capacitor.

At steady state for a DC source, capacitors behave as open circuits. Therefore, only the resistors conduct current.

The remaining conducting path has the resistors in series:

So the current is

Voltage drop across the $6\,\Omega$ resistor is

Taking the top rail as $9\,\text{V}$ and ground as $0\,\text{V}$, the potential at point $A$ is

Since points $A$ and $B$ are directly connected,

The $6\,\mu\text{F}$ capacitor is connected between the top rail at $9\,\text{V}$ and point $B$ at $3\,\text{V}$. Hence the potential difference across it is

Now use $Q = C\Delta V$:

Therefore, the charge stored in the capacitor is $36\,\mu\text{C}$.

Given: The capacitors are in a DC circuit and points $A$ and $B$ are shorted.

Find: Charge on the $6\,\mu\text{F}$ capacitor.

In steady state, each capacitor carries no current, so the capacitor branches are treated as open circuits. The only closed path left is through the $6\,\Omega$ and $3\,\Omega$ resistors.

Thus the circuit current is

The drop across the upper $6\,\Omega$ resistor is $6\,\text{V}$, so the common node $A=B$ must be at $3\,\text{V}$ relative to ground.

Since the top plate of the $6\,\mu\text{F}$ capacitor is at $9\,\text{V}$ and the lower plate is at $3\,\text{V}$,

Therefore,

So the numerical answer is 36.