Question:
Consider the combination of 2 capacitors \(C_1\) and \(C_2\) with \(C_2\gt C_1\), when connected in parallel, the equivalent capacitance is \(\frac{15}{4}\) times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, \(\frac{C_2}{C_1}\).
Consider the combination of 2 capacitors \(C_1\) and \(C_2\) with \(C_2\gt C_1\), when connected in parallel, the equivalent capacitance is \(\frac{15}{4}\) times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, \(\frac{C_2}{C_1}\).
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Form \(C_p=\frac{15}{4}C_s\) and check whether the resulting quadratic in \(C_2/C_1\) has real roots.
Updated On: Nov 4, 2026
- \(\frac{15}{11}\)
- No Solutions
- \(\frac{29}{15}\)
- \(\frac{15}{4}\)
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The Correct Option is B
Solution and Explanation
\(C_p=C_1+C_2\), \(C_s=\frac{C_1C_2}{C_1+C_2}\).
Given \(C_1+C_2=\frac{15}{4}\cdot\frac{C_1C_2}{C_1+C_2}\Rightarrow(C_1+C_2)^2=\frac{15}{4}C_1C_2\).
Let \(r=\frac{C_2}{C_1}\): \((1+r)^2=\frac{15}{4}r\Rightarrow4r^2+8r+4=15r\Rightarrow4r^2-7r+4=0\).
Discriminant \(=49-64=-15\lt0\), so no real \(r\) exists.
Answer: (2) No Solutions
Given \(C_1+C_2=\frac{15}{4}\cdot\frac{C_1C_2}{C_1+C_2}\Rightarrow(C_1+C_2)^2=\frac{15}{4}C_1C_2\).
Let \(r=\frac{C_2}{C_1}\): \((1+r)^2=\frac{15}{4}r\Rightarrow4r^2+8r+4=15r\Rightarrow4r^2-7r+4=0\).
Discriminant \(=49-64=-15\lt0\), so no real \(r\) exists.
Answer: (2) No Solutions
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