Respuesta :
Answer:
756.88 Volts will be the potential difference across each capacitor.
Explanation:
[tex]Q=C\times V[/tex]
Q = Charge on capacitor
C = Capacitance
V = Voltage across capacitor
Capacitance of first capacitor = [tex]C_1=2.81 \mu F=2.81\times 10^{-6} F[/tex]
Charge of first capacitor = [tex]Q_1[/tex]
Voltage across first capacitor = [tex]V_1=1220 V[/tex]
[tex]Q_1=C_1V_1[/tex]
[tex]Q_1=2.81\times 10^{-6} F\times 1220 V=0.0034282 C[/tex]
Capacitance of first capacitor = [tex]C_2=6.61\mu F=6.61\times 10^{-6} F[/tex]
Charge of second capacitor = [tex]Q_2[/tex]
Voltage across first capacitor = [tex]V_2=560 V[/tex]
[tex]Q_2=C_2V_2[/tex]
[tex]Q_1=6.61\times 10^{-6} F\times 560 V=0.0037016 C[/tex]
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
[tex]Q =Q_1+Q_2=0.0034282 C+ 0.0037016 C=0.0071298 C[/tex]
Total capacitance in parallel combination:
[tex]C_p=C_1+C_2=2.81\times 10^{-6} F+6.61\times 10^{-6} F=9.42\times 10^{-6} F[/tex]
Potential across both capacitors = V
[tex]V=\frac{Q}{C}=\frac{0.0071298 C}{9.42\times 10^{-6} F}=756.88 V[/tex]
756.88 Volts will be the potential difference across each capacitor.
