Chapter 25, Problem 034 In the figure a potential difference V = 95.0 V is appli

Question

Chapter 25, Problem 034 In the figure a potential difference V = 95.0 V is applied across a capacitor arrangement with capacitances C1-11.4 uf C2-4.83 and C3 5.85 ?F. what are (a) charge q3, (b) potential difference V3, and (c) stored energy U3 for capacitor 3, (d) 91, (e) vi, and (f) U1 for capacitor 1, and (g) 92, (h) V2, and (i) U2 for capacitor 2? C1 Cs C2 (a) Number (b) Number (c) Number (d) Number (e) Number (f) Number (g) Number (h) Number (i) Number Click if you would like to Show Work for this question: Units Units Units Units Units Units Units Units Units Open Show Work

Explanation / Answer

Remember:

For parallel combination

Ceq = C1 + C2 + C3 +...............

for series combination

1/Ceq = 1/C1 + 1/C2 + 1/C3 + ............

for 2 capacitors in series it will be

Ceq = C1*C2/(C1+C2)

Using this Information:

C1 and C2 are in Series, So

C12 = 11.4*4.83/(11.4 + 4.83) = 3.39 uF

Now C12 and C3 are in paralel, So

Ceq = C12 + C3 = 3.39 + 5.85 = 9.24 uF

Qeq = Ceq*V

Qeq = 9.24*95 = 877.8 uC

Now remember in capacitors parallel combination voltage distribution in each part will be same and in series combination charge distribution in each capacitor will be same. So

C12 and C3 are in parallel, So

Voltage in C12 = Voltage in C3 = V

V12 = 95 V

V3 = 95 V

Q3 = C3*V3 = 5.85*95 = 555.75 uC

U3 = 0.5*C3*V3^2

U3 = 0.5*5.85*10^-6*95^2 = 0.0264 J

Now Charge on C12 will be

Q12 = C12*V12 = 3.39*95 = 322.05 uC

Since C1 and C2 are in series, So

Q1 = Q2 = Q12

Q1 = 322.05 uC

V1 = Q1/C1 = 322.05/11.4 = 28.25 V

U1 = 0.5*C1*V1^2

U1 = 0.5*11.4*10^-6*28.25^2 = 0.0045 J

Q2 = 322.05 uC

V2 = Q2/C2 = 322.05/4.83 = 66.68 V

U1 = 0.5*C1*V1^2

U1 = 0.5*4.83*10^-6*66.68^2 = 0.0107 J

To check

U = 0.5*Ceq*V^2 = 0.5*9.24*10^-6*95^2 = 0.0416

U = U1 + U2 + U3 = 0.0045 + 0.0107 + 0.0264 = 0.0416

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