A 2.50- F capacitor is charged to 760 V and a 6.80- F capacitor is charged to 57

Question

A 2.50-F capacitor is charged to 760 V and a 6.80-F capacitor is charged to 575 V . These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? [Hint: Charge is conserved.]

Part A

Determine the potential difference across the first capacitor.

Express your answer using three significant figures and include the appropriate units.

Part B

Determine the potential difference across the second capacitor.

Express your answer using three significant figures and include the appropriate units.

Part C

Determine the charge across the first capacitor.

Express your answer using three significant figures and include the appropriate units.

Part D

Determine the charge across the second capacitor.

Express your answer using three significant figures and include the appropriate units.

Explanation / Answer

C1 = Capacitance of first capacitor = 2.50 uF

C2 = Capacitance of second capacitor = 6.80 uF

V1 = Voltage across first capacitor = 760 v

V2 = Voltage across second capacitor = 575 v

Q1 = charge on first capacitor = C1 V1 = 2.50 x 10-6 x 760 = 1.9 x 10-3 C

Q2 = charge on second capacitor = C2 V2 = 6.80 x 10-6 x 575 = 3.91 x 10-3 C

After connecting the positive and negative plates Together, the capacitors are in Parallel combination and their Parallel combination is given as

Cp = C1 + C2 = 2.50 + 6.80 = 9.30 x 10-6 F

Total Charge = Q = Q1+ Q2 = 1.9 x 10-3 + 3.91 x 10-3 = 5.81 x 10-3 C

a)

So Voltage across first Capacitor = V1' = Q/Cp = 5.81 x 10-3 / (9.30 x 10-6) = 624.73 volts

b)

Voltage across fsecond Capacitor = V2' = Q/Cp = 5.81 x 10-3 / (9.30 x 10-6) = 624.73 volts

c)

Charge on first Capacitor = Q1' = C1 V1' = 2.50 x 10-6 x 624.73 = 1.56 x 10-3 C

d)

Charge on second Capacitor = Q2' = C2 V2' = 6.80 x 10-6 x 624.73 = 4.25 x 10-3 C

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