Consider the following. (Let C1-9.80 F and C2-3.80 F.) 6.00 F 9.00 V (a) Find th

Question

Consider the following. (Let C1-9.80 F and C2-3.80 F.) 6.00 F 9.00 V (a) Find the equivalent capacitance of the capacitors in the figure. 9.8 Your response differs from the correct answer by more than 100%. F (b) Find the charge on each capacitor. on the right 9.80 F capacitor on the left 9.80 F capacitor on the 3.80 F capacitor on the 6.00 F capacitor uC uC (c) Find the potential difference across each capacitor. on the right 9.80 F capacitor on the left 9.80 F capacitor on the 3.80 F capacitor on the 6.00 uF capacitor

Explanation / Answer

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:

C2 and C3 are in parallel, So (C3 = 6 uF)

C23 = C2 + C3 = 3.80 + 6.0 = 9.8 uF

Now two C1 and C23 are in series, So

1/Ceq = 1/C1 + 1/C1 + 1/C23)

Ceq = (1/9.80 + 1/9.80 + 1/9.80)^-1

Ceq = 3.27 uF

Now

Qeq = Ceq*V = 3.27*9 = 29.43 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.

Charge in C1 = Charge in C23 = 29.43 uC

Q1 = Charge on right 9.80 uF = 29.43 uC

Q1 = Charge on left 9.80 uF = 29.43 uC

V1 = Q1/C1 = 29.43/9.80 = 3.00 V

V1 = Voltage on right 9.80 uF = 3.00 V

V1 = Voltage on left 9.80 uF = 3.00 V

Now, Q23 = 29.43 uC

V23 = V - V1 - V1 = 9 - 3.00 - 3.00 = 3.00 V

Since C2 and C3 are in parallel, So

V2 = V3 = V23

V2 = 3.00 V = Potential difference on 3.80 uF

V3 = 3.00 V = Potential difference on 6.00 uF

Q2 = C2*V2 = 3.80*3.00

Q2 = Charge on 3.80 uF = 11.4 uC

Q3 = C3*V3 = 6.00*3.00 = 18.0 uC

Q3 = Charge on 6.00 uF = 18.0 uC

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