Question

Q18.Two capacitors of capacitances C1 = 6 uF and C2 = 3uF are connected in series across a cell of emf 18 V. Calculate
(i) The equivalent capacitance
(ii) the potential difference across each capacitor
(iii) the charge on each capacitor
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Answer 1

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Answer 2

Answer:

(i) $C_{eq} =$2 µF

(ii) $V_{1}=6\ V$ and $V_{2}=12\ V$

(iii) q=36 µC on each capacitor

Explanation:

(i) Given, $C_{1}$= 6 µF and $C_{2}$= 3 µF

    $\frac{1}{C_{eq} } =\frac{1}{C_{1} }+\frac{1}{C_{2} }$

 ⇒ $\frac{1}{C_{eq} } =\frac{1}{6 }+\frac{1}{3 }$

 ⇒ $\frac{1}{C_{eq} } =\frac{1+2}{6 }$

 ⇒ $\frac{1}{C_{eq} } =\frac{3}{6 }$

 ∴ $C_{eq} =$2 µF

(ii) Since, the capacitors are connected in series to calculate the potential difference across each capacitors we need to calculate the charge flowing through them.

  q=$C_{eq} V$

⇒q= 2*18

∴  q=36 µC

  $V_{1}=\frac{q}{C_{1} }$

⇒$V_{1}=\frac{36}{6 }$

∴ $V_{1}=6\ V$

Similarly,

  $V_{2}=\frac{q}{C_{2} }$

⇒ $V_{2}=\frac{36}{3}$

∴ $V_{2}=12\ V$

(iii) As calculated above charge on each capacitor will be 36 µC.

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