A 2.50-F capacitor is charged to 742 V and a 6.80-F capacitor is charged to 580

Question

A 2.50-F capacitor is charged to 742 V and a 6.80-F capacitor is charged to 580 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.

Part B Determine the potential difference across the second capacitor.

Part C Determine the charge across the first capacitor.

Part D Determine the charge across the second capacitor.

Explanation / Answer

Here, charge on the two caps is Q = CV
Q = 2.5x742 + 6.8x580 = 1855 + 3944 = 5799 µC

Now, as per the conditions, they are connected in parallel. So, equivalent capacitance,

C = 2.5 + 6.8 = 9.3 µF,

So, V = Q/C
V = 5799 µC / 9.3µF = 623.55 volts.

Part A -

Potential difference across the first capacitor = 623.55 V

Part B -

Potential difference across the second capacitor will also, 623.55 V, since the both are connected in parallel.

Part C -

Charge across the first capacitor, Q1 = 2.5x623.55 = 1558.87 uC

Part D -

Charge across the second capacitor, Q2 = 6.8x623.55 = 4240.14 uC

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.