Question

Two capacitors, one a 4.0 μF capacitor, *C _{1}*,
and the other a 7.0 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)**

please comment for Queries.

Please rate thanks.

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