Three capacitors with capacitances C 1 = 6.3 F, C 2 = 1.9 F, and C 3 = 4.7 F are

Question

Three capacitors with capacitances C1 = 6.3 F, C2 = 1.9 F, and C3 = 4.7 F are connected in a circuit as shown in the figure, with an applied potential of V. After the charges on the capacitors have reached their equilibrium values, the charge Q2 on the second capacitor is found to be 36. C.

a) What is the charge, Q1, on capacitor C1?

b) What is the charge, Q3, on capacitor C3?

c) How much voltage, V, was applied across the capacitors?

Incorrect.

If you can write out everystep with the numbers that would be great. Helps me follow it. Also have no idea what units to put into LONCAPA for it

Explanation / Answer

given three capacitors, C1= 6.3 micro F

C2 = 1.9 micro F

C3 = 4.7 micro F

potential difference applied = V

charge on C2 = Q2 = 36 micro C

a. Now charge on C1 = Charge on C2 ( both are in series)

hence Q1 = Q2 = 36 micro C

b. voltage drop across C2 = V2 = Q2/C2 = 18.947 V

voltage drop across C1 = V1 = Q1/C1 = 5.7142 V

hence net voltage drop across C3 = V3 = V2 + v1 = 24.6612 V

so charge on C3 = Q3 = C3V3 = 115.908 micro C

c. Net voltage applied, V = V1 + V2 = 24.6612 V

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