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

To determine the potential at point D in the circuit with capacitors \(C_1\) and \(C_2\), we can use the principle of charge conservation on capacitors. Since these capacitors are connected in a combination involving points A and B, we consider the equivalent single capacitor between A and B and calculate its potential.

The charge \(Q\) on a capacitor is given by \(Q = CV\), where \(C\) is the capacitance and \(V\) is the potential across its plates. For this system in equilibrium, the sum of the charges on both capacitors should be equal:

\(Q_1 = C_1(V_1 - V_D)\)
\(Q_2 = C_2(V_D - V_2)\)

Since no charge is lost, \(Q_1 = Q_2\). Hence,

\(C_1(V_1 - V_D) = C_2(V_D - V_2)\)

Expanding and rearranging gives:

\(C_1V_1 - C_1V_D = C_2V_D - C_2V_2\)

Bringing terms involving \(V_D\) on one side:

\(C_1V_D + C_2V_D = C_1V_1 + C_2V_2\)

Factoring out \(V_D\):

\(V_D(C_1 + C_2) = C_1V_1 + C_2V_2\)

Solving for \(V_D\):

\(V_D = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}\)

Thus, the potential at point D is \( \frac{C_1V_1 + C_2V_2}{C_1 + C_2} \).

Answer 2

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).