Problem 15: Find the charge on 5 µF capacitor in the circuit shown:

Solution

Let C_{1} = 3 µF C_{2} = 2 µF and C_{3} = 5 µF. From the figure, C_{2} and C_{3} are in parallel. Therefore, by the formula for the equivalent capacitance of capacitors in parallel, C_{2,3} = 2 + 5 = 7 µF

Now this capacitor is in series with C_{1} and by the formula for the equivalent capacitance for series combination

So the equivalent capacitance of the circuit is

To calculate charge on the equivalent capacitance of the circuit, C_{e}, Q = 2.1 × 6 = 12.6 µF The voltage drop across the capacitance

This is the voltage across the parallel combination of C_{2} and C_{3}. Now we can calculate charge on C_{3}. Q_{3} = 5 × 1.8 = 9.0 µF Answer

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