MHT CET · Physics · Capacitance
A parallel combination of three condensers of capacities ' \(2 \mathrm{C}\), ' \(\mathrm{C}\) ' and ' \(\mathrm{C} / 2\) ' is connected
Across \(10 \mathrm{~V}\) battery. All the condensers are fully charged to charges ' \(\mathrm{Q}_1\) ', ' \(\mathrm{Q}_2\) ' and ' \(\mathrm{Q}\) 3' respectively. The ratio \(\mathrm{Q}_1: \mathrm{Q}_2: \mathrm{Q}_3\) is
- A \(4: 1: 2\)
- B \(1: 4: 2\)
- C \(1: 2: 4\)
- D \(4: 2: 1\)
Answer & Solution
Correct Answer
(D) \(4: 2: 1\)
Step-by-step Solution
Detailed explanation
In parallel combination the potential drop across each capacitor is same.
\(\therefore \Delta V=\frac{Q_1}{C_1}=\frac{Q_2}{C_2}=\frac{Q_3}{C_3} \)
\( \Rightarrow \Delta V=\frac{Q_1}{2 C}=\frac{Q_2}{C}=\frac{Q_3}{(C / 2)} \)
\( \Rightarrow(C \Delta V)=\frac{Q_1}{2}=\frac{Q_2}{1}=\frac{Q_3}{1 / 2} \)
\( \therefore Q_1=2 \times(C \Delta V), Q_2=1 \times(C \Delta V) \text { and }\)\(Q_3=\frac{1}{2} \times(C \Delta V) \)
\( \therefore Q_1: Q_2: Q_3=2: 1: 0.5 \)
\( \Rightarrow Q_1: Q_2: Q_2=4: 2: 1\)
\(\therefore \Delta V=\frac{Q_1}{C_1}=\frac{Q_2}{C_2}=\frac{Q_3}{C_3} \)
\( \Rightarrow \Delta V=\frac{Q_1}{2 C}=\frac{Q_2}{C}=\frac{Q_3}{(C / 2)} \)
\( \Rightarrow(C \Delta V)=\frac{Q_1}{2}=\frac{Q_2}{1}=\frac{Q_3}{1 / 2} \)
\( \therefore Q_1=2 \times(C \Delta V), Q_2=1 \times(C \Delta V) \text { and }\)\(Q_3=\frac{1}{2} \times(C \Delta V) \)
\( \therefore Q_1: Q_2: Q_3=2: 1: 0.5 \)
\( \Rightarrow Q_1: Q_2: Q_2=4: 2: 1\)
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