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**Problem 1:** The switch in Fig. 1 has been closed for a very long time. What is the charge on the capacitor? The switch is opened at \( t = 0 \) s. At what time has the charge on the capacitor decreased to 10% of its initial value? **Diagram Description:** - **Fig. 1: The scheme for Problem 1** includes a closed loop circuit with: - A 100 V battery - A 60 Ω resistor - A 40 Ω resistor - A 10 Ω resistor - A 2 μF capacitor The switch opens at \( t = 0 \) s. **a)** The fact that the switch has been closed for a very long time means that the capacitor in the scheme is fully charged, creating a physical barrier to the current. Therefore, the current through the branch of the circuit with the 10 Ω resistor and the capacitor is zero. **b)** Write down Kirchhoff's loop law for the rightmost loop in the circuit (passing through 10 Ω and 40 Ω resistors, and the capacitor). When moving along the loop from the negative plate to the positive plate of the capacitor, you gain the potential \( \Delta V_c \). Moving from the positive to the negative plate, you lose the potential. Determine the charges on the plates using Kirchhoff's loop law. **c)** After the switch is opened, calculate the equivalent resistance in this RC circuit. Compute the time at which the charge on the capacitor decreases to 10% of its initial value. **Answer:** \( 2.3 \times 10^{-4} \, \text{s} \).

**Problem 1:** The switch in Fig. 1 has been closed for a very long time. What is the charge on the capacitor? The switch is opened at \( t = 0 \) s. At what time has the charge on the capacitor decreased to 10% of its initial value? **Diagram Description:** - **Fig. 1: The scheme for Problem 1** includes a closed loop circuit with: - A 100 V battery - A 60 Ω resistor - A 40 Ω resistor - A 10 Ω resistor - A 2 μF capacitor The switch opens at \( t = 0 \) s. **a)** The fact that the switch has been closed for a very long time means that the capacitor in the scheme is fully charged, creating a physical barrier to the current. Therefore, the current through the branch of the circuit with the 10 Ω resistor and the capacitor is zero. **b)** Write down Kirchhoff's loop law for the rightmost loop in the circuit (passing through 10 Ω and 40 Ω resistors, and the capacitor). When moving along the loop from the negative plate to the positive plate of the capacitor, you gain the potential \( \Delta V_c \). Moving from the positive to the negative plate, you lose the potential. Determine the charges on the plates using Kirchhoff's loop law. **c)** After the switch is opened, calculate the equivalent resistance in this RC circuit. Compute the time at which the charge on the capacitor decreases to 10% of its initial value. **Answer:** \( 2.3 \times 10^{-4} \, \text{s} \).

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Hello, I really truly need help with part A, Part B and Part C, I keep getting the wrong answer and I don't know how to do the problem is there any chance you can help me with part A, Part B, and Part C please thank you so much.

Also, can you label which one is part A, Part B and Part C. Thank you

**Problem 1:** The switch in Fig. 1 has been closed for a very long time. What is the charge on the capacitor? The switch is opened at \( t = 0 \) s. At what time has the charge on the capacitor decreased to 10% of its initial value? **Diagram Description:** - **Fig. 1: The scheme for Problem 1** includes a closed loop circuit with: - A 100 V battery - A 60 Ω resistor - A 40 Ω resistor - A 10 Ω resistor - A 2 μF capacitor The switch opens at \( t = 0 \) s. **a)** The fact that the switch has been closed for a very long time means that the capacitor in the scheme is fully charged, creating a physical barrier to the current. Therefore, the current through the branch of the circuit with the 10 Ω resistor and the capacitor is zero. **b)** Write down Kirchhoff's loop law for the rightmost loop in the circuit (passing through 10 Ω and 40 Ω resistors, and the capacitor). When moving along the loop from the negative plate to the positive plate of the capacitor, you gain the potential \( \Delta V_c \). Moving from the positive to the negative plate, you lose the potential. Determine the charges on the plates using Kirchhoff's loop law. **c)** After the switch is opened, calculate the equivalent resistance in this RC circuit. Compute the time at which the charge on the capacitor decreases to 10% of its initial value. **Answer:** \( 2.3 \times 10^{-4} \, \text{s} \).
Transcribed Image Text:**Problem 1:** The switch in Fig. 1 has been closed for a very long time. What is the charge on the capacitor? The switch is opened at \( t = 0 \) s. At what time has the charge on the capacitor decreased to 10% of its initial value? **Diagram Description:** - **Fig. 1: The scheme for Problem 1** includes a closed loop circuit with: - A 100 V battery - A 60 Ω resistor - A 40 Ω resistor - A 10 Ω resistor - A 2 μF capacitor The switch opens at \( t = 0 \) s. **a)** The fact that the switch has been closed for a very long time means that the capacitor in the scheme is fully charged, creating a physical barrier to the current. Therefore, the current through the branch of the circuit with the 10 Ω resistor and the capacitor is zero. **b)** Write down Kirchhoff's loop law for the rightmost loop in the circuit (passing through 10 Ω and 40 Ω resistors, and the capacitor). When moving along the loop from the negative plate to the positive plate of the capacitor, you gain the potential \( \Delta V_c \). Moving from the positive to the negative plate, you lose the potential. Determine the charges on the plates using Kirchhoff's loop law. **c)** After the switch is opened, calculate the equivalent resistance in this RC circuit. Compute the time at which the charge on the capacitor decreases to 10% of its initial value. **Answer:** \( 2.3 \times 10^{-4} \, \text{s} \).
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