Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A capacitor of capacitance 0.1  F is connected to a battery of emf 8V as shown in the fig. Under steady state condi

A capacitor of capacitance 0.1 \mu F is connected to a battery of emf 8V as shown in the fig. Under steady state condition.  

Option: 1

Charge on the capacitor is 0.6 \mu C.


Option: 2

Charge on the capacitor is 0.2 \mu C.    


Option: 3

Current in the resistor(R) between points A & B is 0.1 A.  


Option: 4

Current in the resistor(R) between point A & B is 0.2 A.


Answers (1)

best_answer

 

Charging Of Capacitors -

Q=Q_{0}\left ( 1-e^{\frac{-t}{Rc}} \right )

 

 At steady state capacitor acts as open circuit

The equivalent circuit is as shown in figure (b)

    

In the steady  state the Potentil difference across AB is 4 volts .

\therefore charge on capacitor in steady state is

q= CV= 0.4 \,\mu C

Current through resistor  R is I = \frac{V}{R}= \frac{4}{20}= 0.2A

Posted by

Pankaj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: 2nd attempt? Mistakes to avoid, tips to improve score, percentile
anam-khan

JEE Main 2025: What to expect on exam day? Section-wise paper pattern, marking scheme, last-minute tips
anam-khan

JEE Main 2025 Live: NTA JEE session 2 exams from tomorrow; updates on admit card, exam pattern, passing marks

Latest Question