Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electrostatic Potential and Capacitance

Question:

Two capacitors of capacitances 8 μF and 20 μF are connected in series with a battery. The voltage across the 8 μF capacitor is 5 V. The total battery voltage is :

Options:

5 V

10 V

7 V

12.5 V

Correct Answer:

7 V

Explanation:

When two capacitor $C_1$ and $C_2$ re Connected in series then voltage across $C_1$ is 

$ V_1 = \frac{C_2}{C_1+C_2} V $ 

$\text{Here } C_1 = 8\mu F , C_2 = 20\mu F , V_1 = 5V$

$\Rightarrow 5 = \frac{20}{28} \times V $

$\Rightarrow V = 7 V$