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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? Opens at t-Os 100V 6052 ww 10 SE 4052 245 a) The fact that the switch has been closed for a very long time means that the capacitor in the scheme is fully charged, and it cre- ates a physical barrier to the current. Therefore, the current through the branch of the circuit with the 10 S2 resistor and the capacitor is zero (if the capacitor was in the process of charging, there would be a non-zero current in that branch carrying the charge to the capacitor). Find the current flowing throm the other two resistors. FIG. 1: The scheme for Problem

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? Opens at t-Os 100V 6052 ww 10 SE 4052 245 a) The fact that the switch has been closed for a very long time means that the capacitor in the scheme is fully charged, and it cre- ates a physical barrier to the current. Therefore, the current through the branch of the circuit with the 10 S2 resistor and the capacitor is zero (if the capacitor was in the process of charging, there would be a non-zero current in that branch carrying the charge to the capacitor). Find the current flowing throm the other two resistors. FIG. 1: The scheme for Problem

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# Problem 1: Capacitor Discharge **Introduction** In the given circuit diagram (Fig. 1), a switch has been closed for a prolonged period. We need to investigate the charge on the capacitor under various conditions. **Diagram Description** - The circuit consists of a battery (100V), three resistors (60Ω, 40Ω, 10Ω), and a capacitor (2µF). - The switch is initially closed and opens at \( t = 0 \). --- **Questions and Solutions** 1. **What is the charge on the capacitor when the switch is closed for a long time?** - **Explanation:** With the switch closed for a long time, the capacitor gets fully charged and no current flows through the branch containing the 10Ω resistor and the capacitor. The branch acts as an open circuit. 2. **At what time does the charge on the capacitor decrease to 10% of its initial value after the switch is opened?** a) **Initial Condition Analysis:** - The fully charged capacitor initially creates a barrier to the current in the branch with the 10Ω resistor, resulting in zero current flow. b) **Kirchhoff’s Loop Law Application:** - Analyze the rightmost loop, which includes the 10Ω resistor, 40Ω resistor, and the capacitor. - As you traverse the loop from the negative to the positive plate of the capacitor, the potential increases by \( \Delta V_c \). Conversely, moving from positive to negative results in potential loss. - Use Kirchhoff's loop law to deduce the charge on the capacitor from the potential difference and capacitance. c) **Post-Switch Opening:** - Once the switch opens, the system behaves as an RC circuit. - **Determine the Equivalent Resistance:** Combine the resistances (10Ω and 40Ω) in series for the RC circuit setup. - **Calculate the Discharge Time:** Using the formula for exponential decay in an RC circuit, calculate the time required for the capacitor’s charge to reach 10% of its initial value. **Answer Calculation:** - The calculated time is \( 2.3 \times 10^{-4} \) seconds. --- This exercise demonstrates the application of Kirchhoff’s laws and RC circuit analysis to understand how charge dissipates in electrical circuits.
Transcribed Image Text:# Problem 1: Capacitor Discharge **Introduction** In the given circuit diagram (Fig. 1), a switch has been closed for a prolonged period. We need to investigate the charge on the capacitor under various conditions. **Diagram Description** - The circuit consists of a battery (100V), three resistors (60Ω, 40Ω, 10Ω), and a capacitor (2µF). - The switch is initially closed and opens at \( t = 0 \). --- **Questions and Solutions** 1. **What is the charge on the capacitor when the switch is closed for a long time?** - **Explanation:** With the switch closed for a long time, the capacitor gets fully charged and no current flows through the branch containing the 10Ω resistor and the capacitor. The branch acts as an open circuit. 2. **At what time does the charge on the capacitor decrease to 10% of its initial value after the switch is opened?** a) **Initial Condition Analysis:** - The fully charged capacitor initially creates a barrier to the current in the branch with the 10Ω resistor, resulting in zero current flow. b) **Kirchhoff’s Loop Law Application:** - Analyze the rightmost loop, which includes the 10Ω resistor, 40Ω resistor, and the capacitor. - As you traverse the loop from the negative to the positive plate of the capacitor, the potential increases by \( \Delta V_c \). Conversely, moving from positive to negative results in potential loss. - Use Kirchhoff's loop law to deduce the charge on the capacitor from the potential difference and capacitance. c) **Post-Switch Opening:** - Once the switch opens, the system behaves as an RC circuit. - **Determine the Equivalent Resistance:** Combine the resistances (10Ω and 40Ω) in series for the RC circuit setup. - **Calculate the Discharge Time:** Using the formula for exponential decay in an RC circuit, calculate the time required for the capacitor’s charge to reach 10% of its initial value. **Answer Calculation:** - The calculated time is \( 2.3 \times 10^{-4} \) seconds. --- This exercise demonstrates the application of Kirchhoff’s laws and RC circuit analysis to understand how charge dissipates in electrical circuits.
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