Two capacitors of capacitances 6 µF and 12 µF are connected in series with 12 V battery. Find the equivalent capacitance of the combination. Find the charge and the potential difference across each capacitor. Ans. (4 µF, 48 µC, 8 V, 4 V)




Data Given:


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 = Cₑ = ?  
 
(2) Charge on each Capacitor = Q = ?

(3) Potential Difference (Voltage) across capacitor 1 = V₁ = ? 

(4) Potential Difference (Voltage) across capacitor 2 = V₂ = ?   




Solution:


(1) Equivalent Capacitor = Cₑ = ?  

We have the formula Equivalent capacitor connected in series as 

1Cₑ = 1C₁ + 1C₂

Now by putting the values

1Cₑ = 16ｘ10⁻⁶F + 112ｘ10⁻⁶F

1Cₑ = 2+112ｘ10⁻⁶ F

1Cₑ = 312ｘ10⁻⁶ F

or

Cₑ = 12ｘ10⁻⁶3 F

Cₑ = 4 𝐱 10⁻⁶ F = 4 µF ---------------Ans. 1


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(2) Charge on each Capacitor = Q = ?


We know that

Q = CV

Q = 4 𝐱 10⁻⁶ F ｘ 12 V

Q = 48 ｘ 10⁻⁶ C

Q = 48 µC    -------------------------Ans. 2



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(3) Potential Difference (Voltage) across capacitor 1 = V₁ = ?


We have

Q = C₁V₁

or

V₁ = QC₁

by putting values

V₁ = 48ｘ10⁻⁶C6ｘ10⁻⁶F

V₁ = 8 V  -----------------------------Ans. 3


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(4) Potential Difference (Voltage) across capacitor 2 = V₂ = ?  

We have

Q = C₂V₂

or

V₂ = QC₂

by putting values

V₂ = 48ｘ10⁻⁶C12ｘ10⁻⁶F

V₂ = 4 V  ----------------------Ans. 4

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