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Question
Find the charges on the four capacitors of capacitances 1 μF, 2 μF, 3 μF and 4 μF shown in the figure.
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Solution
When the capacitors are fully charged, they attain steady state and no current flows through them. Then, equivalent resistance of the circuit,
\[R_{eff} = \frac{3 \times 6}{3 + 6} = 2 \Omega\]
Current through the circuit,
\[i = \frac{6}{2} = 3 A\]
Current i is divided in the inverse ratio of the resistance in each branch. One branch has resistance of 3 Ω and the other branch has resistance of 6 Ω.
Current i' through the 3 Ω branch,
\[i' = \frac{6}{9}i = \frac{2}{3} \times 3 A = 2 A\]
Current i'' through the 6 Ω branch,
\[i'' = i - i' = 1 A\]
Voltage across the 1 Ω resistor = 2 A × 1 Ω = 2 V
Charge on the 1 μF capacitor = 2 × 1 μF = 2 μC
Voltage across the 2 Ω resistor = 2 Ω × 2 A = 4 V
Charge on the 2 μF capacitor = 4V × 2 μF = 8 μC
Voltage across each 3 Ω resistor = 3 Ω × 1 A = 3 V
Charge on the 4 μF capacitor = 3 × 4 μC = 12 μC
Charge on the 3 μF capacitor = 3 × 3 μC = 9 μC
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