MHT CET · Physics · Capacitance
A series combination of \(n_1\) capacitors, each of value \(C_1\) is charged by a source of potential difference 6 V . Another parallel combination of \(\mathrm{n}_2\) capacitors, each of value \(\mathrm{C}_2\) is charged by a source of potential' difference 2 V . Total energy of both the combinations is same. The value of \(\mathrm{C}_2\) in terms of \(\mathrm{C}_1\) is
- A \(\frac{3 C_1}{n_1 n_2}\)
- B \(\frac{9 \mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1\)
- C \(\frac{3 \mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1\)
- D \(\frac{9 C_1}{n_1 n_2}\)
Answer & Solution
Correct Answer
(D) \(\frac{9 C_1}{n_1 n_2}\)
Step-by-step Solution
Detailed explanation
When connected in series,
\(\left(\mathrm{C}_{\mathrm{eq}}\right)_1=\frac{\mathrm{C}_1}{\mathrm{n}_1} ; \mathrm{V}_1=6 \cdot \mathrm{~V}\)
When connected in parallel,
\(\left(\mathrm{C}_{\mathrm{eq}}\right)_2=\mathrm{n}_2 \dot{\mathrm{C}}_2 ; \mathrm{V}_2=2 \mathrm{~V}\)
Total energy is the same for both connections, \(\mathrm{U}_1=\mathrm{U}_2\)
\(\therefore \frac{1}{2}\left(\mathrm{C}_{\mathrm{eq}}\right)_1 \mathrm{~V}_1^2=\frac{1}{2}\left(\mathrm{C}_{\mathrm{eq}}\right)_2 \mathrm{~V}_2^2 \quad \therefore(\because \mathrm{U}=\frac{1}{2}\) \(\mathrm{CV}^2) \)
\( \frac{1}{2} \frac{\mathrm{C}_1}{\mathrm{n}_1} 36=\frac{1}{2} \mathrm{n}_2 \mathrm{C}_2 4 \)
\( \mathrm{C}_2=\frac{9 \mathrm{C}_1}{\mathrm{n}_1 \mathrm{n}_2}\)
\(\left(\mathrm{C}_{\mathrm{eq}}\right)_1=\frac{\mathrm{C}_1}{\mathrm{n}_1} ; \mathrm{V}_1=6 \cdot \mathrm{~V}\)
When connected in parallel,
\(\left(\mathrm{C}_{\mathrm{eq}}\right)_2=\mathrm{n}_2 \dot{\mathrm{C}}_2 ; \mathrm{V}_2=2 \mathrm{~V}\)
Total energy is the same for both connections, \(\mathrm{U}_1=\mathrm{U}_2\)
\(\therefore \frac{1}{2}\left(\mathrm{C}_{\mathrm{eq}}\right)_1 \mathrm{~V}_1^2=\frac{1}{2}\left(\mathrm{C}_{\mathrm{eq}}\right)_2 \mathrm{~V}_2^2 \quad \therefore(\because \mathrm{U}=\frac{1}{2}\) \(\mathrm{CV}^2) \)
\( \frac{1}{2} \frac{\mathrm{C}_1}{\mathrm{n}_1} 36=\frac{1}{2} \mathrm{n}_2 \mathrm{C}_2 4 \)
\( \mathrm{C}_2=\frac{9 \mathrm{C}_1}{\mathrm{n}_1 \mathrm{n}_2}\)
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