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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

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Pankaj

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