Let's break down the problem step by step and solve each part of the question.

D. Voltage across the capacitors at $t = 9.5$ seconds:

The switch is closed at $t = 9$ seconds, and the system likely follows an RC charging or discharging behavior.

The time constant $\tau$ for the RC circuit is given by: $\tau = R_{\text{eq}} C_{\text{eq}}$ Where:

• $R_{\text{eq}}$ is the equivalent resistance of the circuit.
• $C_{\text{eq}}$ is the equivalent capacitance.

For this circuit:

• Two $1k\Omega$ resistors are in parallel, so the equivalent resistance of this combination is: $R_{\text{parallel}} = \frac{1k \Omega}{2} = 500 \Omega$ The total resistance is the sum of the $2k\Omega$ resistor in series: $R_{\text{eq}} = 500 \Omega + 2k\Omega = 2.5k\Omega = 2500 \Omega$

• The two $10\mu F$ capacitors are in parallel, so the equivalent capacitance is: $C_{\text{eq}} = 10\mu F + 10\mu F = 20\mu F$

The time constant $\tau$ is then: $\tau = 2500 \Omega \times 20 \times 10^{-6} F = 0.05 \text{ seconds}$

Now, we know the capacitor voltage $V_C(t)$ during charging (if the capacitors are charging) is: $V_C(t) = V_{\text{final}} \left( 1 - e^{-t/\tau} \right)$ where $V_{\text{final}}$ is the final voltage after charging.

E. Expression for current across the $2k\Omega$ resistor:

The current through the $2k\Omega$ resistor at any time $t$ can be found using Ohm’s law: $I(t) = \frac{V(t)}{R}$ where $V(t)$ is the voltage across the resistor and $R = 2k\Omega$.

F. Draw the $T$ vs $I_{R=2k}$ and $T$ vs $V_{R=2k}$ graphs:

The graphs for current $I_{R=2k}(t)$ and voltage $V_{R=2k}(t)$ will follow an exponential charging curve.

G. Draw the $T$ vs $V_{AB}$ graph from $0$ seconds to $12$ seconds:

This graph will show the behavior of the total voltage $V_{AB}$ as the capacitors charge or discharge across the time span.

I'll now calculate the detailed values for these equations and graphs.