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The problem you've shared consists of a circuit involving capacitors, resistors, and a DC source, with various tasks like finding voltages, currents, and drawing graphs. I'll walk through how to approach each part, using the following setup from your image:

### Circuit Overview:
1. **Two capacitors**: \(10 \, \mu F\) each.
2. **Three resistors**: Two \(1 \, k\Omega\) resistors in parallel, and one \(2 \, k\Omega\) resistor in series with the battery and the parallel combination.
3. **DC Voltage source**: \(1.5 \, V\).
4. The circuit is closed at \(t = 9 \, sec\), meaning that the capacitors start charging at that moment.

### Now, let's break down the questions:

#### **D. Voltage across capacitors at \( t = 9.5 \, sec \)**
- When the circuit is closed at \(t = 9 \, sec\), the capacitors start charging through the resistors.
- We use the charging formula for a capacitor:
  \[
  V_C(t) = V_{\text{max}} \left(1 - e^{-t/RC}\right)
  \]
  where \(V_{\text{max}} = 1.5V\), \(R\) is the equivalent resistance seen by the capacitor, and \(C = 10 \, \mu F\).
- The equivalent resistance across the charging path will be the series combination of the \(2 \, k\Omega\) resistor and the parallel combination of the two \(1 \, k\Omega\) resistors.

Let me calculate the voltage at \(t = 9.5 \, sec\).The voltage across the capacitors at \( t = 9.5 \, \text{sec} \) is approximately \( 1.5 \, \text{V} \).

#### **E. Expression for the current across the 2kΩ resistor**
The current through the 2kΩ resistor as a function of time can be expressed using Ohm's law and the capacitor charging formula. The current \( I(t) \) is given by:
\[
I(t) = \frac{V_{\text{max}}}{R_{\text{total}}} \cdot e^{-t / (R_{\text{total}} C)}
\]
where:
- \(V_{\text{max}} = 1.5V\),
- \(R_{\text{total}} = R_{\parallel} + R_2\),
- \(C = 10 \, \mu F\).

#### **F. Drawing \(T\) vs \(I_{R=2k}\) and \(T\) vs \(V_{R=2k}\) graphs**
1. For \( I_{R=2k} \), the graph will show an exponential decay.
2. For \( V_{R=2k} \), we can calculate the voltage across the 2kΩ resistor using \(V = I \times R\).

#### **G. Drawing \(T\) vs \(V_{AB}\) graph**
- The voltage \( V_{AB} \) starts from zero at \(t = 9 \, sec\) and follows the same charging curve as the capacitors, reaching 1.5V asymptotically over time.

Would you like detailed calculations or graph sketches? Also, here are some related questions to explore:

1. How does the initial current change when the switch is closed at \(t=9\) sec?
2. How would the time constant change if the capacitor value was doubled?
3. What would happen if the resistors were swapped in value?
4. How long does it take for the capacitors to charge up to 99% of the source voltage?
5. What is the energy stored in the capacitors at \(t = 12\) seconds?

**Tip**: Always pay attention to the equivalent resistance when multiple resistors are involved in series and parallel configurations, as it affects charging and discharging behavior.

The problem involves analyzing a circuit with two 10μF capacitors, three resistors (1KΩ, 1KΩ, 2KΩ), and a 1.5V DC source. The tasks include calculating the voltage across the capacitors, expressing the current through the 2KΩ resistor, and drawing time vs. voltage/current graphs.

Analyze a circuit with 10μF capacitors, 1KΩ, 2KΩ resistors, and a 1.5V source. Calculate the voltage across capacitors and the current through the 2KΩ resistor. Explore time vs voltage and current graphs for exponential decay in the RC circuit.

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