Question 153 of 949
The diagram above shows two capacitors P and Q of capacitance 5μF and 10μF. Find the charges stored in P and Q respectively
- A. 2μC and 4μC
- B. 4μC and 2μC
- C. 100μC and 200μC
- D. 200μC and 100μC
Correct Answer:
C
Explanation
To solve the problem of finding the charges stored in capacitors P and Q, we need to understand how capacitors work and how to calculate the charge stored in them.
### Step-by-Step Explanation
1. **Understanding Capacitance**:
- Capacitance (C) is defined as the ability of a capacitor to store charge (Q) per unit voltage (V). The relationship is given by the formula:
\[
C = \frac{Q}{V}
\]
- Rearranging this formula allows us to find the charge:
\[
Q = C \times V
\]
2. **Identifying the Values**:
- From the problem, we have two capacitors:
- Capacitor P has a capacitance of \( C_P = 5 \, \mu F \)
- Capacitor Q has a capacitance of \( C_Q = 10 \, \mu F \)
- We need to find the charges stored in these capacitors, denoted as \( Q_P \) and \( Q_Q \).
3. **Voltage Across the Capacitors**:
- The problem does not specify the voltage across the capacitors. However, for the sake of this explanation, let's assume they are connected in series to a voltage source \( V \).
- In a series connection, the charge on each capacitor is the same. Therefore, if we denote the total voltage across the series combination as \( V \), the voltage across each capacitor can be expressed as:
\[
V_P = \frac{Q}{C_P} \quad \text{and} \quad V_Q = \frac{Q}{C_Q}
\]
4. **Calculating Charge**:
- Since the charge is the same for both capacitors in series, we can express the total voltage as:
\[
V = V_P + V_Q = \frac{Q}{C_P} + \frac{Q}{C_Q}
\]
- Rearranging gives:
\[
V = Q \left( \frac{1}{C_P} + \frac{1}{C_Q} \right)^{-1}
\]
- However, without a specific voltage, we cannot calculate the exact charges.
5. **Assuming a Voltage**:
- If we assume a voltage \( V = 10V \) (for example), we can calculate the charges:
- For capacitor P:
\[
Q_P = C_P \times V = 5 \, \mu F \times 10 \, V = 50 \, \mu C
\]
- For capacitor Q:
\[
Q_Q = C_Q \times V = 10 \, \mu F \times 10 \, V = 100 \, \mu C
\]
6. **Final Charges**:
- Therefore, if we assume a voltage of 10V, the charges stored would be:
- \( Q_P = 50 \, \mu C \)
- \( Q_Q = 100 \, \mu C \)
### Evaluating the Options
Now, let's evaluate the provided options based on our calculations:
- **Option A: 2μC and 4μC** - This is incorrect because the calculated charges are much higher.
- **Option B: 4μC and 2μC** - This is also incorrect for the same reason as above.
- **Option C: 100μC and 200μC** - This is incorrect as well; we calculated 50μC and 100μC.
- **Option D: 200μC and 100μC** - This is incorrect; again, our calculations do not match.
### Conclusion
None of the options provided match the calculated charges based on the assumed voltage. The correct charges for the assumed voltage of 10V are 50μC for capacitor P and 100μC for capacitor Q.
### Revision Summary
- **Capacitance Formula**: \( C = \frac{Q}{V} \) and \( Q = C \times V \).
- **Series Connection**: In series, the charge is the same across all capacitors.
- **Assumed Voltage**: Without a specified voltage, we cannot determine exact charges.
- **Check Options**: Always verify calculated values against provided options.
If you have a specific voltage value or additional context, please provide it for a more accurate calculation!