Two capacitors of capacitances 6 µF and 12 µF are connected in parallel with the 12V battery. Find the equivalent capacitance of the combination. Find the charge and the potential difference across each capacitor. Ans. (18 µF, 72 µC, 144 µC, 12 V)



Given Data:


Capacitance of capacitor 1 = C₁ =6 µF = 6 𝐱 10⁻⁶ F 

Capacitance of capacitor 2 = C₂ =12 µF = 12 𝐱 10⁻⁶ F 

Potential Difference (Voltage) = V = 12 V 



To Find:


(1) Equivalent Capacitor = Ceq = ?  
 
(2) Charge on capacitor 1 = Q₁ = ? 

(3) Charge on capacitor 2 = Q₂ = ? 

(4) Potential Difference (Voltage) across each capacitor = V = ? 



Solution:


(1) Equivalent Capacitor = Ceq = ?  


We have the formula Equivalent capacitor connected in parallel as

Ceq = C₁ + C₂

Ceq = 6 µF + 12 µF

Ceq = 18 µF



(2) Charge on capacitor 1 = Q₁ = ? 

 
We know that

Q₁ = C₁V

Q₁ = 6 µF ｘ 12 V

Q₁ = 72 ｘ 10⁻⁶ FV

Q₁ = 72 µC




(3) Charge on capacitor 2 = Q₂ = ? 

 
We know that

Q₂ = C₂V

Q₂ = 12 µF ｘ 12 V

Q₂ = 144 µ FV

Q₂ = 144 µC



(4) Potential Difference (Voltage) across each capacitor = V = ? 

 
For capacitors connected in a parallel combination, the potential difference across each capacitor remains the same as supplied by a battery i.e. 12 V.  

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