17 Consider the following Let Ci 2420 muF and C2 1820 muF

#17

Consider the following. (Let Ci = 24.20 muF and C_2 = 18.20 muF.) Find the equivalent capacitance of the capacitors in the figure. Find the charge on each capacitor. Find the potential difference across each capacitor.

Solution

(b)

The equvalent capacitance, C_{E =} 8.07 F.

The total voltage V_T = 9 V.

The total charge stored is Q_T=C_{E *} V_T= 8.07 * 9 = 72.07 C

The potential difference across each capacitor

V₁ = Q₁/C₁

= 72.07 / 24.2 = 2.978 V

Therefore both the C₁s are having 2.978 V each

total 2.978 + 2.978 is split between both the C₁s.

i.e, 5.956 V

The rest voltage..i.e. 9 - 5.956=3.044 V is being equally taken by 18.2 and 6 F (voltage is the same in parallel connection)

The charges stored in each capacitor

Charge on C₁ is Q₁ = C₁ * V₁ = 24.2 * 2.978 = 72.06 C

Charge on 6 F capacitor is 6 * 3.044 = 18.264 C

Charge on 18.2 F capacitor is 18.2 * 3.044 = 55.04 C