(D) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.Initial energy $U_0 = \frac{1}{2}CV_0^2$,where $V_0 = 100\, V$.Final energy $U

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(D) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.Initial energy $U_0 = \frac{1}{2}CV_0^2$,where $V_0 = 100\, V$.Final energy $U = \frac{1}{2}CV^2$,where $V = 80\, V$.The energy lost is $\Delta U = U_0 - U = \frac{1}{2}CV_0^2 - \frac{1}{2}CV^2$.The fractional loss of energy is $\frac{\Delta U}{U_0} = \frac{\frac{1}{2}CV_0^2 - \frac{1}{2}CV^2}{\frac{1}{2}CV_0^2} = \frac{V_0^2 - V^2}{V_0^2}$.Substituting the values: $\frac{\Delta U}{U_0} = \frac{100^2 - 80^2}{100^2} = \frac{(100 - 80)(100 + 80)}{10000} = \frac{20 \times 180}{10000} = \frac{3600}{10000} = \frac{36}{100} = \frac{9}{25}$.

Class 12 Physics Electric Potential and Capacitance Charging and Discharging of Capacitance and RC circuit (DC)

Difficult

$A$ capacitor is charged to a potential difference of $100\, V$ and then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After $1\, s$,the potential difference across the capacitor plates is $80\, V$. What is the fractional loss of the stored energy?

A
$11/9$
B
$12/25$
C
$25/9$
D
$9/25$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Electric Potential and Capacitance

Subtopic

Charging and Discharging of Capacitance and RC circuit (DC)

$A$ capacitor of capacitance $5\,\mu F$ is connected to a source of constant $emf$ of $200\,V$ for a long time. Then,the switch is shifted from contact $1$ to contact $2$. The total amount of heat generated in the $500\,\Omega$ resistance thereafter is: (in $/32\,J$)

Medium

View Solution

$A$ capacitor of capacitance $C = 1 \, \mu F$ is suddenly connected to a battery of $100 \, V$ through a resistance $R = 100 \, \Omega$. The time taken for the capacitor to be charged to $50 \, V$ is $.... \times 10^{-4} \, s$. (Take $\ln 2 = 0.69$)

MediumJEE MAIN 2021

View Solution

In the following circuit,the switch $S$ is closed at $t = 0.$ The charge on the capacitor $C_1$ as a function of time will be given by $\left( {{C_{eq}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right).$

DifficultJEE MAIN 2018

View Solution

In an $RC$ circuit while charging,the graph of $\log i$ versus time is as shown by the dotted line in the diagram,where $i$ is the current. When the value of the resistance is doubled,which of the solid curves best represents the variation of $\log i$ versus time?

MediumIIT 2004

View Solution

$A$ capacitor is connected to a $20\, V$ battery through a resistance of $10\, \Omega$. It is found that the potential difference across the capacitor rises to $2\, V$ in $1\, \mu s$. The capacitance of the capacitor is $....\, \mu F$. Given: $\ln(10/9) = 0.105$.

DifficultJEE MAIN 2021

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

$A$ capacitor of capacitance $5\,\mu F$ is connected to a source of constant $emf$ of $200\,V$ for a long time. Then,the switch is shifted from contact $1$ to contact $2$. The total amount of heat generated in the $500\,\Omega$ resistance thereafter is: (in $/32\,J$)

$A$ capacitor of capacitance $C = 1 \, \mu F$ is suddenly connected to a battery of $100 \, V$ through a resistance $R = 100 \, \Omega$. The time taken for the capacitor to be charged to $50 \, V$ is $.... \times 10^{-4} \, s$. (Take $\ln 2 = 0.69$)

In the following circuit,the switch $S$ is closed at $t = 0.$ The charge on the capacitor $C_1$ as a function of time will be given by $\left( {{C_{eq}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right).$

In an $RC$ circuit while charging,the graph of $\log i$ versus time is as shown by the dotted line in the diagram,where $i$ is the current. When the value of the resistance is doubled,which of the solid curves best represents the variation of $\log i$ versus time?

$A$ capacitor is connected to a $20\, V$ battery through a resistance of $10\, \Omega$. It is found that the potential difference across the capacitor rises to $2\, V$ in $1\, \mu s$. The capacitance of the capacitor is $....\, \mu F$. Given: $\ln(10/9) = 0.105$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module