Question

Consider the network of capacitors in the figure below. If C = 1.90 nF, and the charge stored in one of the 3C capacitors is 36.8 nC, what is the potential difference (in V) across the points ab? Round your answer to 2 decimal places. C - 30. 30. a b 내 C

Answer 1

Here

C = 1.90 x10^-9 F

Charge in 3C capicitor = 36.8×10^-9 C

Now

both the 3C capacitor are in series and thus equivalent capacitance will be

Ce= (3×3)/(3+3)

Ce = 9/6

Ce= 1.5C

Now

Since in series connection charge remain same.

Therefore

Q = 36.8×10^-9C

Ce=1.5×1.90×10^-9 F

Now

V = (36.8×10^-9)/(2.85×10^-9)

V= 12.91V

Ce of 3C capacitance  will be in parallel with both capacitor with capacitance C, we know that voltage across parallel capacitor remains same.

Therefore potential across a and b is 12.91V