Given three capacitors, C1C1C_1 = 2.0 μFμF, C2C2C_2 = 1.5 μFμF, and C3C3C_3 = 3.0 μFμF, what arrangement of parallel and series connections with a 12-VV battery will give the minimum voltage drop across the 2.0- μFμF capacitor?

Explanation:

If joined in parallel ,each will have potential of 12 volt

so on 2 μF capacitor , voltage will be 12 V

When joined in series , if C be equivalent capacitor

1/C = 1/2 + 1/1.5 + 1/3

= .5 + .667 + .333

= 1.5

C = .666 μF

Charge on each capacitor

= CV

= .666 x 12

= 8 μC

Potential on 2 μF capacitor

= charge / capacitance

= 8 μC / 2 x μF

= 4 V

So in series , potential at 2μF capacitor is 4 V which is less than 12 V in parallel combination.

harmonic law for those who use odyssey

Explanation: