Question

A 2.81 μF capacitor is charged to 1220 V and a 6.61 μF capacitor is charged to 560 V. These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each?

Answer 1

756.88 Volts will be the potential difference across each capacitor.

Step-by-step explanation:

`Q=C* V`

Q = Charge on capacitor

C = Capacitance

V = Voltage across capacitor

Capacitance of first capacitor =
`C_1=2.81 \mu F=2.81* 10^(-6) F`

Charge of first capacitor =
`Q_1`

Voltage across first capacitor =
`V_1=1220 V`

`Q_1=C_1V_1`

`Q_1=2.81* 10^(-6) F* 1220 V=0.0034282 C`

Capacitance of first capacitor =
`C_2=6.61\mu F=6.61* 10^(-6) F`

Charge of second capacitor =
`Q_2`

Voltage across first capacitor =
`V_2=560 V`

`Q_2=C_2V_2`

`Q_1=6.61* 10^(-6) F* 560 V=0.0037016 C`

Both the capacitors are disconnected and positive plates are now connected to each other and the negative plates are connected to each other. These capacitors are connected in parallel combination.

Total charge = Q

`Q =Q_1+Q_2=0.0034282 C+ 0.0037016 C=0.0071298 C`

Total capacitance in parallel combination:

`C_p=C_1+C_2=2.81* 10^(-6) F+6.61* 10^(-6) F=9.42* 10^(-6) F`

Potential across both capacitors = V

`V=(Q)/(C)=(0.0071298 C)/(9.42* 10^(-6) F)=756.88 V`

756.88 Volts will be the potential difference across each capacitor.