Question

two capacitors of 2muf and 6muf are in series across a 20V source find the charge on these capacitors and the potential difference across each?
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Answer 1

Given :

Two capacitors of 2μF and 6μF are connected in series across a 20V source.

To Find :

Charge and potential difference across each capacitor.

Solution :

❖ In series combination, charge on each capacitor is same (equal to the charge supplied by battery) but potential differences across the capacitors may be different.

In series combination,

$\dag\:\underline{\boxed{\bf{\purple{\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}}}}}$

$\sf:\implies\:\dfrac{1}{C}=\dfrac{1}{2}+\dfrac{1}{6}$

$\sf:\implies\:\dfrac{1}{C}=\dfrac{4}{6}$

$\sf:\implies\:C=\dfrac{6}{4}$

$:\implies\:\underline{\boxed{\bf{\orange{C=1.5\:\mu F}}}}$

♦ Charge on each capacitor :

➙ Q = C × V

➙ Q = 1.5 × 20

➙ Q = 30 μC

♦ Pd across C₁ :

➙ V₁ = Q / C₁

➙ V₁ = 30/2

➙ V₁ = 15 volt

♦ Pd across C₂ :

➙ V₂ = V - V₁

➙ V₂ = 20 - 15

➙ V₂ = 5 volt

Answer 2

Given :

Two capacitors of 2μF and 6μF are connected in series across a 20V source.

To Find :

Charge and potential difference across each capacitor.

Solution :

❖ In series combination, charge on each capacitor is same (equal to the charge supplied by battery) but potential differences across the capacitors may be different.

In series combination,

1C = C 1/1 +C 21

⟹ 1 C= 21 +61

⟹ 1 C

= 4/5

⟹C= 6/4

⟹ C=1.5μF

♦ Charge on each capacitor :

➙ Q = C × V

➙ Q = 1.5 × 20

➙ Q = 30 μC

♦ Pd across C₁ :

➙ V₁ = Q / C₁

➙ V₁ = 30/2

➙ V₁ = 15 volt

♦ Pd across C₂ :

➙ V₂ = V - V₁

➙ V₂ = 20 - 15

➙ V₂ = 5 volt