Three capacitors (C1 = 71 nF, C2 = 173 nF, and C3 = 398 nF) are connected to a b

Question

Three capacitors (C1 = 71 nF, C2 = 173 nF, and C3 = 398 nF) are connected to a battery as shown in the picture. The battery creates potential difference V = 7.5 V.

(a) Find the equivalent capacitance of this set of capacitors.

Equivalent capacitance = 63.15 nF

(b) Find charge on each capacitor.

Charge on the capacitor C1 = ____?C

Charge on the capacitor C2 = ____ ?C

Charge on the capacitor C3 = ____?C

I posted this before but the answers were wrong. Part a is correct. I really don't understand how to get the charges when the capicators are in parallel.

Explanation / Answer

a) C = C1(C1 + C2)/(C1 + C2 + C3)

C = 71(173+398)/(71+173+398)

C = 63.147 nF

b) Q1 = C1V1

= 71 x V1

now V1 = (C1 + C2)V/(c1 + c2 + c3)

V1 = 6.67 V

hence,

Q1 = 71 x 6.67

Q1 = 473.6098 nC

Q2 = C2V2

= 173 x (7.5-6.67)

= 143.59 nC

Q3 = 330.34 nC

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