Question
A C1= 2.50uF capacitor is charged to 862 V and a C2= 6.80uF capacitor is charged to 660 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? [Hint: charge is conserved.]
What will be the charge on each?
What will be the potential difference across each? [Hint: charge is conserved.]
What will be the charge on each?
Answers
C = Q/V (That is the meaning of C)
how much charge on 2.5*10^-6 at 862
Q1 = C V = 2.5*10^-6 * 862 = 2155*10^-6 coulombs
how much charge on 6.8*10^-6 at 660 ?
Q2 = 6.8*10^-6 * 660 = 4488*10^-6
Total charge = Q1+Q2 = 6643*10^-6
Total Capacitance = (2.5+6.8)10^-6 = 9.30*10^-6 Farads
V = same on each now = 6643/9.30 = 714.3 volts
now just do
Q1 = C1 (714)
Q2 = C2 (714
for the charge on each
how much charge on 2.5*10^-6 at 862
Q1 = C V = 2.5*10^-6 * 862 = 2155*10^-6 coulombs
how much charge on 6.8*10^-6 at 660 ?
Q2 = 6.8*10^-6 * 660 = 4488*10^-6
Total charge = Q1+Q2 = 6643*10^-6
Total Capacitance = (2.5+6.8)10^-6 = 9.30*10^-6 Farads
V = same on each now = 6643/9.30 = 714.3 volts
now just do
Q1 = C1 (714)
Q2 = C2 (714
for the charge on each
Related Questions
A parallel plate capacitor with plate separation d is connected to a battery. The capacitor is fully...
A 600W flashlamp fitted to a camera is operated by discharging a 10 000 uF capacitor through the lam...
A 10uf capacitor is charged from a 30v supply and then connected across an uncharged 50uf capacitor....