Three capacitors 3 $μ$ F, 10 $μ$ F and 15 $μ$ F are connected in series to a voltage source of 100 V. The charge on 15 $μ$ F is

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500 $μ$ C

200 $μ$ C

100 $μ$ C

400 $μ$ C

answer is A.

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Detailed Solution

$C_{S} = \frac{C_{1} C_{2} C_{3}}{C_{1} C_{2} + C_{2} C_{3} + C_{3} C_{1}} i . e . \frac{3 \times 10 \times 15}{30 + 150 + 45} i . e 2 μF Q = CV i.e., 2 \times 100 = 200 μC$

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