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(Please answer to the fourth decimal place - i.e 14.3225)
![## Analysis of the Given Circuit
Consider the circuit below. The capacitor has 32 nF capacitance. R1 has 125 kΩ of resistance. The battery outputs 20 V. The value of R2 is irrelevant. The switch has been in the right position for a long time (long enough to fully charge the capacitor). At t = 0 s, the switch is flipped into the left position. Calculate the energy stored in the capacitor at t = 6.2 ms, in units of nJ.
### Circuit Diagram:
- **R1**: 125 kΩ resistor
- **C1**: 32 nF capacitor
- **V1**: 20 V DC power source
- **Switch (S1)**: Can be toggled between two positions
- **R2**: (Irrelevant in this problem)
The circuit consists of:
1. A resistor (R1) connected in series with a capacitor (C1) and a switch (S1).
2. A power source (V1) that charges the capacitor (C1) through a resistor (R1).
### Circuit Description:
- Initially, the switch (S1) has been in the right position for a long time such that the capacitor (C1) is fully charged to the battery voltage.
- At t = 0 s, the switch is moved to the left position, allowing the capacitor to start discharging through the resistor (R1).
### Calculation of Energy Stored in the Capacitor:
1. **Initial Voltage Across Capacitor (V0)**:
Since the capacitor is fully charged before the switch is flipped, the initial voltage V0 across the capacitor is equal to the battery voltage:
\[ V_0 = 20 \text{ V} \]
2. **Discharging Equation**:
The voltage Vc(t) across the capacitor at any time t during discharging is given by:
\[ V_c(t) = V_0 \cdot e^{-\frac{t}{R_1 C_1}} \]
Where:
- \( R_1 = 125 \times 10^3 \ \Omega \)
- \( C_1 = 32 \times 10^{-9} \ \text{F} \)
- \( t = 6.2 \ \text{ms} = 6.2 \times 10^{-](https://content.bartleby.com/qna-images/question/021aca7b-3c6d-4d3c-bfc7-736350e47cc8/10b45ddc-254f-4f7f-a384-c95e07aa975f/ht9qq94_processed.png)
Transcribed Image Text:## Analysis of the Given Circuit
Consider the circuit below. The capacitor has 32 nF capacitance. R1 has 125 kΩ of resistance. The battery outputs 20 V. The value of R2 is irrelevant. The switch has been in the right position for a long time (long enough to fully charge the capacitor). At t = 0 s, the switch is flipped into the left position. Calculate the energy stored in the capacitor at t = 6.2 ms, in units of nJ.
### Circuit Diagram:
- **R1**: 125 kΩ resistor
- **C1**: 32 nF capacitor
- **V1**: 20 V DC power source
- **Switch (S1)**: Can be toggled between two positions
- **R2**: (Irrelevant in this problem)
The circuit consists of:
1. A resistor (R1) connected in series with a capacitor (C1) and a switch (S1).
2. A power source (V1) that charges the capacitor (C1) through a resistor (R1).
### Circuit Description:
- Initially, the switch (S1) has been in the right position for a long time such that the capacitor (C1) is fully charged to the battery voltage.
- At t = 0 s, the switch is moved to the left position, allowing the capacitor to start discharging through the resistor (R1).
### Calculation of Energy Stored in the Capacitor:
1. **Initial Voltage Across Capacitor (V0)**:
Since the capacitor is fully charged before the switch is flipped, the initial voltage V0 across the capacitor is equal to the battery voltage:
\[ V_0 = 20 \text{ V} \]
2. **Discharging Equation**:
The voltage Vc(t) across the capacitor at any time t during discharging is given by:
\[ V_c(t) = V_0 \cdot e^{-\frac{t}{R_1 C_1}} \]
Where:
- \( R_1 = 125 \times 10^3 \ \Omega \)
- \( C_1 = 32 \times 10^{-9} \ \text{F} \)
- \( t = 6.2 \ \text{ms} = 6.2 \times 10^{-
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