1. **Stating the problem:** We have a circuit with a voltage source $V$, resistors of $3\Omega$, $6\Omega$, and $2\Omega$, and a capacitor $C$. We need to find: - The total current in the circuit at $t=0$ (just after connection). - The total current in the circuit as $t \to \infty$ (steady state). - Identify the circuit behavior after a long time. - Determine the voltage across the capacitor after a long time. - Calculate the voltage $V$ after a long time and the current in the conductor. - Find the impulse when the capacitor starts charging. 2. **Key concepts and formulas:** - At $t=0$, the capacitor behaves like a short circuit (since it is uncharged). - At $t \to \infty$, the capacitor behaves like an open circuit (fully charged, no current through it). - Use Ohm's law: $I = \frac{V}{R}$. - For series and parallel resistors, calculate equivalent resistance. 3. **Step 1: Total current at $t=0$** - Capacitor acts as a short circuit. - The $2\Omega$ resistor and capacitor branch become just $2\Omega$ resistor in parallel with a short (0 $\Omega$), so effectively the branch is a short. - So the circuit reduces to $3\Omega$ and $6\Omega$ resistors in series with the shorted branch. - Total resistance $R_{total} = 3 + 6 + 0 = 9\Omega$. - Total current $I_0 = \frac{V}{9}$. 4. **Step 2: Total current at $t \to \infty$** - Capacitor acts as open circuit. - The $2\Omega$ resistor is now disconnected (open circuit), so only $3\Omega$ and $6\Omega$ resistors remain in series. - Total resistance $R_{\infty} = 3 + 6 = 9\Omega$. - Total current $I_{\infty} = \frac{V}{9}$. 5. **Step 3: Identify circuit after long time** - Capacitor is fully charged, no current through capacitor branch. - Circuit is a simple series of $3\Omega$ and $6\Omega$ resistors. 6. **Step 4: Voltage across capacitor after long time** - Voltage drop across $2\Omega$ resistor branch is zero (open circuit). - Voltage across capacitor equals voltage across the branch it is in. - Since no current flows through $2\Omega$ resistor, voltage across capacitor equals voltage across that branch. - Voltage across capacitor $V_C = V - I_{\infty} \times (3 + 6) = V - V = 0$ (This suggests capacitor voltage equals source voltage, but since no current flows, capacitor voltage equals voltage across open branch, which is $V$). 7. **Step 5: Voltage $V$ after long time and current in conductor** - Voltage $V$ is constant source voltage. - Current $I_{\infty} = \frac{V}{9}$. 8. **Step 6: Impulse when capacitor starts charging** - At $t=0$, capacitor voltage is zero. - Current changes from $I_0$ to $I_{\infty}$ as capacitor charges. - The impulse corresponds to the charging transient, which depends on $RC$ time constant. **Final answers:** - Total current at $t=0$: $I_0 = \frac{V}{9}$. - Total current at $t \to \infty$: $I_{\infty} = \frac{V}{9}$. - Capacitor voltage at $t \to \infty$: $V_C = V$. - Circuit behaves as series $3\Omega$ and $6\Omega$ resistors after long time.