Question:
Two capacitors \( C_1 \) and \( C_2 \) in a circuit are joined as shown in the figure. The potentials of points A and B are \( V_1 \) and \( V_2 \), respectively. Then the potential of point D will be
Two capacitors \( C_1 \) and \( C_2 \) in a circuit are joined as shown in the figure. The potentials of points A and B are \( V_1 \) and \( V_2 \), respectively. Then the potential of point D will be
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For capacitors in parallel, the potential across them is the same and can be calculated as a weighted average based on their capacitances.
Updated On: Jan 12, 2026
- \( \frac{V_1 + V_2}{2} \)
- \( C_1 V_1 + C_2 V_2 \)
- \( \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \)
- \( C_2 V_1 + C_1 V_2 \)
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The Correct Option is C
Solution and Explanation
Step 1: Use the formula for the potential in a parallel combination of capacitors.
In a parallel combination of capacitors, the total potential at point D is a weighted average of the potentials at points A and B, where the weights are the capacitances of the capacitors.
Step 2: Calculate the potential at point D.
The potential at point D is given by the formula: \[ V_D = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \]
Final Answer: \[ \boxed{\frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}} \]
In a parallel combination of capacitors, the total potential at point D is a weighted average of the potentials at points A and B, where the weights are the capacitances of the capacitors.
Step 2: Calculate the potential at point D.
The potential at point D is given by the formula: \[ V_D = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \]
Final Answer: \[ \boxed{\frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}} \]
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