In the figure a potential difference V = 110 V is applied across a capacitor arr

Question

In the figure a potential difference V = 110 V is applied across a capacitor arrangement with capacitances C_1 = 14.1 mu F, C_2 = 4.22 mu F, and C_3 = 3-64 mu F. What are (a) charge q_3, (b) potential difference V_3, and (c) stored energy V_3 for capacitor 3, (d) q_1, (e) V_1, and (f) U_1 for capacitor 1, and (g q_2, (h) V_2, and (i) U_2 for capacitor 2? (a) Number Units (b) Number Units (c) Number Units (d) Number Units (e) Number Units (f) Number Units (g) Number Units (h) Number Units (i) Number Units

Explanation / Answer

The equivalent capacitance of the circuit needs to be calculated first.

C1 and C2 are in series and their combination is in parallel with C3

C12 = C1 x C2/ (C1 + C2) = 14.1 x 4.22/(14.1 + 4.22) = 3.25 uF

Ceq = 3.25 + 3.64 = 6.89 uF

Q = CV = 6.89 x 10^-6 x 110 V = 757.9 uC

a)Potential diff across C3 is C3 = 110 V

Q3 = C3 V3 = 3.64 x 110 = 400.4 uC

Hence, Q3 = 400.4 uC

b)V3 = 110 V ; since its in parallel.

c)U3 = 1/2 C3 V3^2

U3 = 0.5 x 3.64 x 10^-6 x 110^2 = 0.022022 J

Hence, U3 = 0.022022 J

d)The total voltage across C12 is 110 V

V1 + V2 = 110 V

same charge flows through the cap connected in series,

Q1 = Q2

C1V1 = C2V2

V2 = C1 V1/C2 = 14.1/4.22 V1 = 3.41 V1

V1 + 3.41V1 = 110 => V1 = 24.94

Q1 = C1V1 = 14.4 x 24.94 = 359.14 uC

Hence, Q1 = 359.14 uC

e)V1 = 24.94 V

f)U1 = 0.5 x 14.1 x 10^-6 x 24.94^2 = 0.0044 J

g)q2 = C2V2

Q2 = 3.41 V1

V1 = 3.41 x 24.94 = 85.06

Q2 = 4.22 x 85.06 = 358.95 uC

Hence, Q2 = 358.95 uC

h)V2 = 85.06 V

i)U2 = 0.5 x 4.22 x 10^-6 x 85.06^2 = 0.0153 J

Hence, U3 = 0.0153 J

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