Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electrostatic Potential and Capacitance

Question:

Two parallel plate capacitors of capacitances 2 µF and 3 µF are joined in series and the combination is connected to a battery of V volts. The values of potential across the two capacitors $V_1$ and $V_2$ and energy stored in the two capacitors $U_1$ and $U_2$ respectively are related as ________.

Fill in the blank with the correct answer from the options given below.

Options:

$\frac{V_1}{V_2}=\frac{U_1}{U_2}=\frac{3}{2}$

$\frac{V_1}{V_2}=\frac{U_1}{U_2}=\frac{2}{3}$

$\frac{V_1}{V_2}=\frac{3}{2}$ and $\frac{U_1}{U_2}=\frac{2}{3}$

$\frac{V_1}{V_2}=\frac{2}{3}$ and $\frac{U_1}{U_2}=\frac{3}{2}$

Correct Answer:

$\frac{V_1}{V_2}=\frac{U_1}{U_2}=\frac{3}{2}$

Explanation:

$ V_1 = \frac{C_2 V}{C_1 + C_2} = \frac{3V}{5}$

$ V_2 = \frac{C_1 V}{C_1 + C_2} = \frac{2V}{5}$

$\Rightarrow \frac{V_1}{V_2} = \frac{3}{2}$

$ U_1 = \frac {q^2}{2C_1}, U_2 = \frac{q^2}{2C_2} \Rightarrow \frac{U_1}{U_2} = \frac{C_2}{C_1} = \frac{3}{2}$

The correct answer is Option (1) → $\frac{V_1}{V_2}=\frac{U_1}{U_2}=\frac{3}{2}$