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  • Two identical capacitors of 10 pF each are connected in turn (i) in series, and (ii) in parallel across a 20 V battery. Calculate the potential difference across each capacitor in the first case and charge acquired by each capacitor in the second case.

Two identical capacitors of 10 pF each are connected in turn (i) in series, and (ii) in parallel across a 20 V battery. Calculate the potential difference across each capacitor in the first case and charge acquired by each capacitor in the second case.

 

 
 
 
 
 

Answers (1)

(i) given,

 capacitance of capacitor = 10pF

  Voltage V = 20 V

Arrangement is 

  

In series connection charges are same

        Q =Q_{1}=Q_{2}

Know, Q =CV

            V=\frac{Q}{C}

Here, capacitance is equal so potential difference across each capacitor is equal which is 10V.

(ii)

 

In parallel connection potential difference across the capacitor are same.

That is, V=V_{1}=V_{2}=20V

Therefore, Q=CV

                    =10pF\times 20V=200pC

Charge acquired by each capacitor is 200 pC

Posted by

Safeer PP

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