https://file.aiquickdraw.com/mathbot/file/1731006691072z6hm4uc2.jpg

To solve this problem, let’s analyze the circuit and the given information step-by-step.

### Problem Summary
1. We have an RC (resistor-capacitor) circuit with a 12V DC power supply, a switch, a 50 kΩ resistor, and a 100 μF capacitor.
2. The problem states that the capacitor is initially discharged.
3. The switch is initially in position B, then moved to position A for 5 seconds, and finally returned to position B.
4. We need to calculate the voltage across the capacitor after the switch has been in:
   - Position A for 5 seconds.
   - Position B for 5 seconds after moving it back from position A.

### Step-by-Step Solution
#### Step 1: Voltage Across the Capacitor in Position A (Charging Phase)
When the switch is moved to position A, the capacitor starts charging through the resistor. The voltage across the capacitor \( V_C(t) \) during the charging phase is given by:

\[
V_C(t) = V_{\text{max}} \left(1 - e^{-\frac{t}{RC}}\right)
\]

where:
- \( V_{\text{max}} = 12 \, \text{V} \) (the power supply voltage),
- \( R = 50 \, \text{kΩ} = 50,000 \, \Omega \),
- \( C = 100 \, \mu\text{F} = 100 \times 10^{-6} \, \text{F} \),
- \( t = 5 \, \text{s} \).

Calculating \( RC \):
\[
RC = (50,000 \, \Omega)(100 \times 10^{-6} \, \text{F}) = 5 \, \text{s}
\]

Now, substitute \( t = 5 \, \text{s} \) into the equation:

\[
V_C(5) = 12 \, \text{V} \left(1 - e^{-\frac{5}{5}}\right)
\]
\[
V_C(5) = 12 \, \text{V} \left(1 - e^{-1}\right)
\]
Using \( e^{-1} \approx 0.3679 \):

\[
V_C(5) \approx 12 \, \text{V} \times (1 - 0.3679) \approx 12 \, \text{V} \times 0.6321 \approx 7.5852 \, \text{V}
\]

Thus, after 5 seconds in position A, the voltage across the capacitor is approximately **7.56 V**.

#### Step 2: Voltage Across the Capacitor in Position B (Discharging Phase)
When the switch is moved back to position B, the capacitor starts discharging through the resistor. The voltage across the capacitor during the discharging phase is given by:

\[
V_C(t) = V_{\text{initial}} e^{-\frac{t}{RC}}
\]

where:
- \( V_{\text{initial}} = 7.56 \, \text{V} \) (voltage at the end of the charging phase),
- \( RC = 5 \, \text{s} \),
- \( t = 5 \, \text{s} \).

Now, substitute these values into the equation:

\[
V_C(5) = 7.56 \, \text{V} \times e^{-\frac{5}{5}}
\]
\[
V_C(5) = 7.56 \, \text{V} \times e^{-1}
\]
\[
V_C(5) \approx 7.56 \, \text{V} \times 0.3679 \approx 2.78 \, \text{V}
\]

Thus, after 5 seconds in position B, the voltage across the capacitor is approximately **2.8 V**.

### Final Answers
1. After 5 seconds in position A: \( V_C \approx 7.56 \, \text{V} \).
2. After 5 seconds in position B: \( V_C \approx 2.8 \, \text{V} \).

Would you like more details or have any questions?

### Additional Related Questions
1. How does the time constant \( RC \) affect the charging and discharging rates in an RC circuit?
2. What happens to the final voltage across the capacitor if the resistance is doubled?
3. How would the charging and discharging curves look if plotted over time?
4. What is the significance of the exponential term in the equations for charging and discharging?
5. How would the circuit behave if a different capacitor value were used?

### Tip
In RC circuits, the time constant \( RC \) determines how quickly the capacitor charges or discharges. A larger \( RC \) means slower charging and discharging, while a smaller \( RC \) means faster responses.

Vea el circuito RC de abajo. Asumamos que el capacitor está inicialmente descargado y con el interruptor en la posición mostrada en la figura. El interruptor es movido a la posición A por exactamente 5 segundos y finalmente a su posición original. ¿Cuál es la diferencia de potencial a través del capacitor cuando: a) el interruptor estuvo en la posición A por 5 s? b) el interruptor ha estado en la posición B por 5 s?

Analyze the voltage across a capacitor in an RC circuit with a switch. This problem explores charging and discharging behaviors in a 12V RC circuit with a 50 kΩ resistor and a 100 μF capacitor, calculating the voltage after 5 seconds in both charging and discharging phases.

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