MHT CET · Physics · Capacitance
A parallel combination of two capacitors of capacities ' \(2 \mathrm{C}\) ' and ' \(\mathrm{C}\) ' is connected across \(5 \mathrm{~V}\) battery. When they are fully charged, the charges and energies stored in them be ' \(\mathrm{Q}_1\) ', ' \(\mathrm{Q}_2\) ' and ' \(\mathrm{E}_1\) ', ' \(\mathrm{E}_2\) ' respectively. Then \(\frac{\mathrm{E}_1-\mathrm{E}_2}{\mathrm{Q}_1-\mathrm{Q}_2}\) in \(\mathrm{J} / \mathrm{C}\) is (capacity is in Farad, charge in Coulomb and energy in \(J\) )
- A \(\frac{5}{4}\)
- B \(\frac{4}{5}\)
- C \(\frac{5}{2}\)
- D \(\frac{2}{5}\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
We know, \(\mathrm{Q}=\mathrm{C} . \mathrm{V}\)
\(\therefore \quad \mathrm{Q}_1=10 \mathrm{C} \text { and } \mathrm{Q}_2=5 \mathrm{C}\)
Energy stored, \(\mathrm{E}=\frac{1}{2} \mathrm{CV}^2\)
\(\begin{aligned}
\therefore \quad \mathrm{E}_1=\frac{1}{2} \mathrm{C}_1 \mathrm{~V}^2 & =\frac{1}{2} \times 2 \mathrm{C} \times 25 \\
& =25 \mathrm{~J}
\end{aligned}\)
Similarly, \(\mathrm{E}_2=\frac{1}{2} \mathrm{C}_2 \mathrm{~V}^2=\frac{1}{2} \times \mathrm{C} \times 25\) \(=12.5 \mathrm{~J}\)
\(\therefore \quad \frac{\mathrm{E}_1-\mathrm{E}_2}{\mathrm{Q}_1-\mathrm{Q}_2}=\frac{12.5}{5}=\frac{5}{2}\)
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