Three capacitors \({C}_{1}=2\, \mu {F}, {C}_{2}=6 \, \mu {F}\) and \({C}_{3}=12 \, \mu {F}\) are connected as shown in figure. Find the ratio of the charges on capacitors \({C}_{1}, {C}_{2}\) and \({C}_{3}\) respectively:

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Answer & Solution

Correct Answer

(C) \(1: 2: 2\)

Detailed explanation

\(\left(V_{D}-V\right) C_{2}+\left(V_{D}-0\right) C_{3}=0\)

\(\left(V_{D}-V\right) 6+\left(V_{D}-0\right) 12=0\)

\(V_{D}-V+2 V_{D}=0\)

\(V_{D}=\frac{V}{3}\)

\({q}_{2}=\left({V}-{V}_{{D}}\right) {C}_{2}=\left({V}-\frac{{V}}{3}\right)(6 \mu {F})\)

\({q}_{2}=(4 {V}) \mu {F}\)

\({q}_{3}=\left({V}_{{D}}-0\right) {C}_{3}=\frac{{V}}{3} \times 12 \mu {F}=4 {V} \mu {F}\)

\({q}_{1}=({V}-0) {C}_{1}={V}(2 \mu {F})\)

\({q}_{1}: {q}_{2}: {q}_{3}=2: 4: 4\)

\({q}_{1}: {q}_{2}: {q}_{3}=1: 2: 2\)

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