Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electrostatic Potential and Capacitance

Question:

Options:

a

b

c

d

Correct Answer:

d

Explanation:

$ V = V_0 (1 - e^\frac{-t}{RC}) = \frac{V_0}{2}$

$\Rightarrow \frac{t}{RC} = ln2$

$\Rightarrow t = RC ln2$

$\text{For parallel arrangement } C_{eq} = 2C$

$\Rightarrow t_p= 2RC ln2$

$\text{For Series arrangemant } C_s = \frac{c}{2}$

$\Rightarrow t_s = \frac{RC}{2} ln2$

$\frac{t_s}{t_p} = \frac{1}{4}$

$\Rightarrow t_s = \frac{10}{4} = 2.5 Sec$