PART 1 A 2.50- ? F capacitor is charged to 747V and a6.80- ? F capacitor is char

Question

PART 1

A 2.50-?F capacitor is charged to 747V and a6.80-?F capacitor is charged to 580V . 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.]

1) Determine the potential difference across the first capacitor.

2) Determine the potential difference across the second capacitor.

3) Determine the charge across the first capacitor.

4) Determine the charge across the second capacitor.

PART 2

A 7.4-?F capacitor is charged by a 175-V battery (see(Figure 1) a) and then is disconnected from the battery. When this capacitor (C1) is then connected (see (Figure 1) b) to a second (initially uncharged) capacitor, C2, the final voltage on each capacitor is12V .

1) What is the value of C2? [Hint: charge is conserved.]

Explanation / Answer

the common potential of the two capacitors is gven by

V = (C1V1+C2V2)/(C1+C2) = 624.9V

a) potential difference across the first capacitor. = 624.9 V

b) potential difference across the second capacitor. = 624.9 V

c) charge on first capacitor = C1V = 1.562 x 10-3C

d) charge on second capacitor = C2V = 4.25 x 10-3C

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