Question

Two condensers \( C_1 \) & \( C_2 \) in a circuit are joined as shown in the figure. The potential of point A is \( V_1 \) and that of point B is \( V_2 \). The potential at point D will be:

• A. \( \frac{1}{2} (V_1 + V_2) \)
• B. \( \)
• C. \( \frac{C_1V_1 + C_2V_2}{C_1 + C_2} \)
• D. \( \frac{C_2V_1 - C_1V_2}{C_1 + C_2} \)

Hint:

For capacitors in series, the charge remains constant, and the potential at any point is determined using the weighted average formula based on capacitance values.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept of Potential Division in a Capacitor Network In a series capacitor network, the charge on each capacitor remains the same. However, the voltage is divided across the capacitors in proportion to their capacitances. The potential at a point between two capacitors is given by the formula: \[ V_D = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \] where: - \( C_1 \) and \( C_2 \) are the capacitances of the two capacitors, - \( V_1 \) and \( V_2 \) are the potentials at points A and B respectively. 
Step 2: Explanation of the Formula Since charge conservation applies, the potential at the junction \( D \) can be derived using the principle of charge distribution, which leads to the formula: \[ V_D = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \] 
Step 3: Conclusion Thus, the potential at point D is given by: \[ V_D = \frac{C_1V_1 + C_2V_2}{C_1 + C_2} \] which matches option (3).