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GUJCET · Physics · Electrostatic Potential and Capacitance
Two capacitors of capacitance \(2 \mu F\) and \(4 \mu F\) respectively are connected in series. The combination is connected across a potential difference of 10 V . The ratio of energies stored by capacitors will be __________ .
- A \(2: 1\)
- B \(4: 1\)
- C \(1: \sqrt{2}\)
- D \(1: 4\)
Answer & Solution
Correct Answer
(A) \(2: 1\)
Step-by-step Solution
Detailed explanation
(A)
as charge is constant in series connection,
\(\begin{aligned} U _1=\frac{ Q ^2}{2 C _1} & \text { and } U _2=\frac{ Q ^2}{2 C _2} \\ \therefore \quad \frac{ U _1}{ U _2} & =\frac{ C _2}{ C _1} \\ \frac{ U _1}{ U _2} & =\frac{4}{2} \\ \frac{ U _1^2}{ U _2} & =\frac{2}{1}\end{aligned}\)
as charge is constant in series connection,
\(\begin{aligned} U _1=\frac{ Q ^2}{2 C _1} & \text { and } U _2=\frac{ Q ^2}{2 C _2} \\ \therefore \quad \frac{ U _1}{ U _2} & =\frac{ C _2}{ C _1} \\ \frac{ U _1}{ U _2} & =\frac{4}{2} \\ \frac{ U _1^2}{ U _2} & =\frac{2}{1}\end{aligned}\)
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