1. **Stating the problem:** We analyze the given electrical circuit at different times: at the moment the voltage source is connected ($t=0$), after a long time ($t \to \infty$), and during the capacitor discharge. 2. **At $t=0$ (just connected):** The capacitor behaves like a short circuit (wire) because it is initially uncharged. 3. **Circuit simplification at $t=0$:** - Capacitor replaced by a wire. - The branch with capacitor is a short. - The resistors $2R$ and $R$ in series remain. - The $3R$ resistor is in parallel with the short circuit (capacitor branch), so it is bypassed. 4. **Calculate total resistance at $t=0$:** - The $3R$ resistor is bypassed by the short, so it does not affect the circuit. - The total resistance is $R$ (top branch) in parallel with $2R + R = 3R$ (bottom branch). 5. **Formula for parallel resistors:** $$ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} $$ 6. **Calculate $R_{eq}$:** $$ R_{eq} = \frac{R \times 3R}{R + 3R} = \frac{3R^2}{4R} = \frac{3R}{4} $$ 7. **Total current at $t=0$:** Using Ohm's law $I = \frac{V}{R_{eq}}$, if the source voltage is $V$: $$ I_0 = \frac{V}{\frac{3R}{4}} = \frac{4V}{3R} $$ 8. **At $t \to \infty$ (steady state):** The capacitor behaves like an open circuit (no current through it). 9. **Circuit simplification at $t \to \infty$:** - Capacitor branch is open. - The $3R$ resistor is now in series with the parallel combination of $R$ and $(2R + R)$. 10. **Calculate parallel resistance of $R$ and $3R$:** $$ R_p = \frac{R \times 3R}{R + 3R} = \frac{3R^2}{4R} = \frac{3R}{4} $$ 11. **Total resistance at $t \to \infty$:** $$ R_{total} = R_p + 3R = \frac{3R}{4} + 3R = \frac{3R}{4} + \frac{12R}{4} = \frac{15R}{4} $$ 12. **Total current at $t \to \infty$:** $$ I_\infty = \frac{V}{R_{total}} = \frac{V}{\frac{15R}{4}} = \frac{4V}{15R} $$ 13. **Charge stored in capacitor at $t \to \infty$:** Voltage across capacitor equals voltage across $2R + R = 3R$ resistor branch. 14. **Current through $3R$ resistor branch:** Current divides between $R$ and $3R$ branches, but capacitor branch is open, so current through capacitor is zero. 15. **Voltage division:** Voltage across $3R$ resistor branch is: $$ V_C = I_\infty \times 3R = \frac{4V}{15R} \times 3R = \frac{12V}{15} = \frac{4V}{5} $$ 16. **Charge on capacitor:** $$ Q = C \times V_C = C \times \frac{4V}{5} $$ 17. **Discharge of capacitor after source removal:** The capacitor discharges through the resistors. 18. **Time constant $\tau$ of discharge:** Equivalent resistance seen by capacitor is the series of $2R + R = 3R$ and $3R$ resistors in parallel with $R$. 19. **Calculate equivalent resistance for discharge:** The capacitor discharges through $3R$ resistor and the parallel combination of $R$ and $3R$. 20. **Parallel of $R$ and $3R$:** $$ R_p = \frac{R \times 3R}{R + 3R} = \frac{3R^2}{4R} = \frac{3R}{4} $$ 21. **Total resistance for discharge:** $$ R_{discharge} = 3R + R_p = 3R + \frac{3R}{4} = \frac{12R}{4} + \frac{3R}{4} = \frac{15R}{4} $$ 22. **Time constant:** $$ \tau = R_{discharge} \times C = \frac{15R}{4} C $$ 23. **Half-life time $t_{1/2}$:** For exponential decay: $$ Q(t) = Q_0 e^{-t/\tau} $$ Half charge when: $$ \frac{Q_0}{2} = Q_0 e^{-t_{1/2}/\tau} \Rightarrow e^{-t_{1/2}/\tau} = \frac{1}{2} $$ Taking natural log: $$ -\frac{t_{1/2}}{\tau} = \ln \frac{1}{2} = -\ln 2 \Rightarrow t_{1/2} = \tau \ln 2 $$ 24. **Final expression for half-life:** $$ t_{1/2} = \frac{15R}{4} C \ln 2 $$ **Summary:** - $I_0 = \frac{4V}{3R}$ - $I_\infty = \frac{4V}{15R}$ - $Q = C \times \frac{4V}{5}$ - $t_{1/2} = \frac{15R}{4} C \ln 2$