The charge stored by the capacitor C in the given circuit in the steady state is 1cm.

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Given: The capacitor is in a resistive circuit connected to a source of $2.5 \, \text{V}$ . The capacitance is $C = 5 \, \mu \text{F}$ .

Find: The charge stored on the capacitor in steady state.

In steady state, no current flows through the capacitor branch, so the capacitor behaves as an open circuit. The remaining network is treated as a potential divider bridge.

Potential at the left node is

V_L = 2.5 \times \frac{4}{1+4} = 2.5 \times 0.8 = 2.0 \, \text{V}

Potential at the right node is

V_R = 2.5 \times \frac{2}{3+2} = 2.5 \times 0.4 = 1.0 \, \text{V}

Therefore, the potential difference across the capacitor is

\Delta V = |V_L - V_R| = 2.0 - 1.0 = 1.0 \, \text{V}

Now use

Q = C\Delta V

So,

Q = 5 \, \mu \text{F} \times 1.0 \, \text{V} = 5 \, \mu \text{C}

Therefore, the charge stored by the capacitor is $5 \, \mu \text{C}$ . The correct option is C.

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