The charge stored by the capacitor C in the given circuit in the steady state is ________ mutextC$\mu\text{C}$.

### Related Formula Q = C V_c$Q = C V_c$ where V_c$V_c$ is the steady-state voltage across the capacitor. ### Core Logic In steady state, the capacitor acts as an open circuit (blocks DC current). We must analyze the active branches.

The branch with the reversed-biased diode will carry no current. Current flows through the outer loop via the forward-biased diode. Total active resistance R_texteq = 1\,Omega + 4\,Omega = 5\,Omega$R_{\text{eq}} = 1\,\Omega + 4\,\Omega = 5\,\Omega$. Current in the steady state: i = fracVR_texteq = frac2.55 = 0.5 text A$$i = \frac{V}{R_{\text{eq}}} = \frac{2.5}{5} = 0.5 \text{ A}$$ ### Step 1: Voltage Calculation The voltage across the capacitor V_c$V_c$ is equal to the voltage drop across the 4\,Omega$4\,\Omega$ resistor, because the branch is connected in parallel. V_c = i times 4\,Omega$$V_c = i \times 4\,\Omega$$ V_c = 0.5 times 4 = 2 text V$$V_c = 0.5 \times 4 = 2 \text{ V}$$ ### Step 2: Final Conclusion The charge stored is: Q = C V_c$Q = C V_c$ Q = 5\,mutextF times 2text V = 10\,mutextC$$Q = 5\,\mu\text{F} \times 2\text{ V} = 10\,\mu\text{C}$$ ### Pattern Recognition Capacitor in steady-state DC = open wire. Analyze only the paths where current can physically flow, check diode polarity, find nodal voltage across the capacitor terminals, apply Q=CV. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics Class 12 Physics: Current Electricity Class 12 Physics: Semiconductor Electronics