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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!
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