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.
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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