Question
In the circuit of series and parallel arrangement,c1=4microfarad,c2=3microfarad and c3=1microfarad,given that v=12volts,calculate the charge on each capacitor and voltage across each capacitor.
Answers
GPT-4o mini
To solve this problem, we need to clarify the arrangement of the capacitors (C1, C2, C3) in the circuit. Let’s calculate based on two common arrangements: all three capacitors in series and all three in parallel.
### Case 1: Capacitors in Series
1. **Total capacitance (C_total) for series capacitors:**
\[
\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}
\]
\[
\frac{1}{C_{total}} = \frac{1}{4\mu F} + \frac{1}{3\mu F} + \frac{1}{1\mu F}
\]
To proceed with this calculation, we convert these values to a common denominator (12):
\[
\frac{1}{C_{total}} = \frac{3}{12} + \frac{4}{12} + \frac{12}{12} = \frac{19}{12}
\]
Hence,
\[
C_{total} = \frac{12}{19}\mu F \approx 0.632 \mu F
\]
2. **Total charge (Q) stored in series capacitors:**
The charge (Q) is the same across all capacitors in series and can be calculated using:
\[
Q = C_{total} \times V
\]
\[
Q = 0.632 \mu F \times 12V = 7.584 \mu C
\]
3. **Voltage across each capacitor:**
- For \( C_1 (4 \mu F) \):
\[
V_1 = \frac{Q}{C_1} = \frac{7.584 \mu C}{4 \mu F} = 1.896 V
\]
- For \( C_2 (3 \mu F) \):
\[
V_2 = \frac{Q}{C_2} = \frac{7.584 \mu C}{3 \mu F} = 2.528 V
\]
- For \( C_3 (1 \mu F) \):
\[
V_3 = \frac{Q}{C_3} = \frac{7.584 \mu C}{1 \mu F} = 7.584 V
\]
### Summary for Series Arrangement:
- Charge on each capacitor: **7.584 µC**
- Voltage across \( C_1 \): **1.896 V**
- Voltage across \( C_2 \): **2.528 V**
- Voltage across \( C_3 \): **7.584 V**
### Case 2: Capacitors in Parallel
1. **Total capacitance (C_total) for parallel capacitors:**
\[
C_{total} = C_1 + C_2 + C_3 = 4 \mu F + 3 \mu F + 1 \mu F = 8 \mu F
\]
2. **Total charge (Q) stored in parallel capacitors:**
\[
Q = C_{total} \times V = 8 \mu F \times 12 V = 96 \mu C
\]
3. **Voltage across each capacitor:**
In parallel, the voltage across each capacitor is the same and equals the supply voltage:
\[
V_1 = V_2 = V_3 = 12 V
\]
- For \( C_1 (4 \mu F) \):
\[
Q_1 = C_1 \times V = 4 \mu F \times 12 V = 48 \mu C
\]
- For \( C_2 (3 \mu F) \):
\[
Q_2 = C_2 \times V = 3 \mu F \times 12 V = 36 \mu C
\]
- For \( C_3 (1 \mu F) \):
\[
Q_3 = C_3 \times V = 1 \mu F \times 12 V = 12 \mu C
\]
### Summary for Parallel Arrangement:
- Voltage across each capacitor: **12 V**
- Charge on \( C_1 \): **48 µC**
- Charge on \( C_2 \): **36 µC**
- Charge on \( C_3 \): **12 µC**
### Conclusion:
- In **series** arrangement:
- All capacitors have a charge of **7.584 µC** and voltages as calculated above.
- In **parallel** arrangement:
- All capacitors have the same voltage of **12 V** and the charges are as calculated above.
Please specify which arrangement you are interested in or provide additional details if needed!
### Case 1: Capacitors in Series
1. **Total capacitance (C_total) for series capacitors:**
\[
\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}
\]
\[
\frac{1}{C_{total}} = \frac{1}{4\mu F} + \frac{1}{3\mu F} + \frac{1}{1\mu F}
\]
To proceed with this calculation, we convert these values to a common denominator (12):
\[
\frac{1}{C_{total}} = \frac{3}{12} + \frac{4}{12} + \frac{12}{12} = \frac{19}{12}
\]
Hence,
\[
C_{total} = \frac{12}{19}\mu F \approx 0.632 \mu F
\]
2. **Total charge (Q) stored in series capacitors:**
The charge (Q) is the same across all capacitors in series and can be calculated using:
\[
Q = C_{total} \times V
\]
\[
Q = 0.632 \mu F \times 12V = 7.584 \mu C
\]
3. **Voltage across each capacitor:**
- For \( C_1 (4 \mu F) \):
\[
V_1 = \frac{Q}{C_1} = \frac{7.584 \mu C}{4 \mu F} = 1.896 V
\]
- For \( C_2 (3 \mu F) \):
\[
V_2 = \frac{Q}{C_2} = \frac{7.584 \mu C}{3 \mu F} = 2.528 V
\]
- For \( C_3 (1 \mu F) \):
\[
V_3 = \frac{Q}{C_3} = \frac{7.584 \mu C}{1 \mu F} = 7.584 V
\]
### Summary for Series Arrangement:
- Charge on each capacitor: **7.584 µC**
- Voltage across \( C_1 \): **1.896 V**
- Voltage across \( C_2 \): **2.528 V**
- Voltage across \( C_3 \): **7.584 V**
### Case 2: Capacitors in Parallel
1. **Total capacitance (C_total) for parallel capacitors:**
\[
C_{total} = C_1 + C_2 + C_3 = 4 \mu F + 3 \mu F + 1 \mu F = 8 \mu F
\]
2. **Total charge (Q) stored in parallel capacitors:**
\[
Q = C_{total} \times V = 8 \mu F \times 12 V = 96 \mu C
\]
3. **Voltage across each capacitor:**
In parallel, the voltage across each capacitor is the same and equals the supply voltage:
\[
V_1 = V_2 = V_3 = 12 V
\]
- For \( C_1 (4 \mu F) \):
\[
Q_1 = C_1 \times V = 4 \mu F \times 12 V = 48 \mu C
\]
- For \( C_2 (3 \mu F) \):
\[
Q_2 = C_2 \times V = 3 \mu F \times 12 V = 36 \mu C
\]
- For \( C_3 (1 \mu F) \):
\[
Q_3 = C_3 \times V = 1 \mu F \times 12 V = 12 \mu C
\]
### Summary for Parallel Arrangement:
- Voltage across each capacitor: **12 V**
- Charge on \( C_1 \): **48 µC**
- Charge on \( C_2 \): **36 µC**
- Charge on \( C_3 \): **12 µC**
### Conclusion:
- In **series** arrangement:
- All capacitors have a charge of **7.584 µC** and voltages as calculated above.
- In **parallel** arrangement:
- All capacitors have the same voltage of **12 V** and the charges are as calculated above.
Please specify which arrangement you are interested in or provide additional details if needed!