A 2.50-mu F capacitor is charged to 747 V and a 6.80-mu F capacitor is charged t
ID: 1654722 • Letter: A
Question
A 2.50-mu F capacitor is charged to 747 V and a 6.80-mu 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? Determine the potential difference across the first capacitor. Express your answer using three significant figures and include the appropriate units. ANSWER: Determine the potential difference across the second capacitor. Express your answer using three significant figures and include the appropriate units. ANSWER: Determine the charge across the first capacitor. Express your answer using three significant figures and include the appropriate units. ANSWER: Determine the charge across the second capacitor. Express your answer using three significant figures and include the appropriate units. ANSWER:Explanation / Answer
We will call 2.50 uF capacitor as Capacitor 1 and 6.80 uF as Capacitor 2
Initially,
Q1 = 2.50*747 = 1867.5 uC
Q2 = 6.80*580 = 3944 uC
Total Charge is,
Q = Q1+Q2 = 5811.5 uC
Since charge is conserved, total charge will be 5811.5 uC on the combination.
Net Capacitance would be C1+C2 because of parallel Connection.
C = 2.50+6.80 = 9.3 uF
Since they are in parallel, Final voltages on them would be same,
V1 = V2 = V
Q=CV
5811.5 = 9.3*V
V = 624.89 volts
V1 = 624.89 Volts
V2 = 624.89 Volts
Q1 = C1V1 = 2.50*624.89
Q1 = 1562.23 uC
Q2 = C2V2 = 6.80*624.89
Q2 = 4249.27 uC
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