KCET · Physics · Capacitance
In this diagram, the PD between \(A\) and \(B\) is \(60 \mathrm{~V}\), The PD across \(6 \mu \mathrm{F}\) capacitor is
- A \(4 \mathrm{~V}\)
- B \(10 \mathrm{~V}\)
- C \(5 \mathrm{~V}\)
- D \(20 \mathrm{~V}\)
Answer & Solution
Correct Answer
(B) \(10 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Let us first calculate equivalent capacitance between \(A\) and \(B\).
\(\mathrm{C}_{\mathrm{e}}=1 \mu \mathrm{F} \quad \text { (In series combination) }\)
\(\therefore\) Charge on the combination,
\(\mathrm{q}=\mathrm{CV}=1 \times 60=60 \mu \mathrm{C}\)
As \(6 \mu \mathrm{F}\) capacitor is in series with all other capacitors, hence, charge on \(6 \mu \mathrm{F}\) capacitance is also \(60 \mu \mathrm{C}\),
\(\therefore\) Potential difference across \(6 \mu \mathrm{F}\) capacitor
\(\mathrm{V}_{1}=\frac{\mathrm{q}}{\mathrm{C}_{1}}=\frac{60}{6}=10 \mathrm{~V}\)
\(\mathrm{C}_{\mathrm{e}}=1 \mu \mathrm{F} \quad \text { (In series combination) }\)
\(\therefore\) Charge on the combination,
\(\mathrm{q}=\mathrm{CV}=1 \times 60=60 \mu \mathrm{C}\)
As \(6 \mu \mathrm{F}\) capacitor is in series with all other capacitors, hence, charge on \(6 \mu \mathrm{F}\) capacitance is also \(60 \mu \mathrm{C}\),
\(\therefore\) Potential difference across \(6 \mu \mathrm{F}\) capacitor
\(\mathrm{V}_{1}=\frac{\mathrm{q}}{\mathrm{C}_{1}}=\frac{60}{6}=10 \mathrm{~V}\)
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