SATHEE: Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance - Result Question 26

28.

A capacitor of $2 μ F$ is charged as shown in the diagram. When the switch $S$ is turned to position 2 , the percentage of its stored energy dissipated is :

(a) $0$

(b) $20$

(d) $80$

[2016]

Show Answer

Answer:

Correct Answer: 28. (d)

Solution:

(d) When $S$ and 1 are connected

The $2 μ F$ capacitor gets charged. The potential difference across its plates will be $V$ .

The potential energy stored in $2 μ F$ capacitor

$U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} \times 2 \times V^{2} = V^{2}$

When $S$ and 2 are connected

The $8 μ F$ capacitor also gets charged. During this charging process current flows in the wire and some amount of energy is dissipated as heat. The energy loss is

$Δ U = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} (V_{1} - V_{2})^{2}$

Here, $C_{1} = 2 μ F, C_{2} = 8 μ F, V_{1} = V, V_{2} = 0$

$∴ Δ U = \frac{1}{2} \times \frac{2 \times 8}{2 + 8} (V - 0)^{2} = \frac{4}{5} V^{2}$

The percentage of the energy dissipated

$= \frac{Δ U}{U_{i}} \times 100 = \frac{\frac{4}{5} V^{2}}{V^{2}} \times 100 = 80$

Electrostatic Potential and Capacitance - Result Question 26

28.

Answer:

When $S$ and 2 are connected

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Electrostatic Potential and Capacitance - Result Question 26

28.

Answer:

When S and 2 are connected

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

When $S$ and 2 are connected