Question

Two capacitors, one a 4.0 μF capacitor, C₁, and the other a 7.0 μF capacitor, C₂, are connected in series. If a 90.0 V voltage source is applied to the capacitors, as shown in the figure, find the voltage drop across the 4.0 μF capacitor.​

36 V

54 V

9.0 V

60 V​

Answer 1

Solution:

Givne C1 and C2 are in series.

Equivalent capacitance = C1 x C2 / C1 + C2

=> C = 28 / 11 uF

Now Q = C * V

=> Q = 28 / 11 * 90 uC

=> Q = 2520 / 11 uC

potential drop across 4.0 uF capacitor

=> V = Q / C1

=> V = 2520 / 11 * 4

=> V = 57.27 V

Note the option is not matching , Acutally is think there should be 6.0 uF capacitor instead of 7.0 uF

Considering we have 6.0 uF capacitor if we solve the quesiton in same way

We will get V = 54 V ( option A as the Right answer)

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