- #1
aaronmilk3
- 12
- 0
A person runs a 9.0 V, 12.0 W radio for 1.33 hours. How many coulombs of charge pass through the wires in the radio during this time?
I used Power= pot difference/Resistance = 9v/120w=7.5e-2
I=current
Then I used I²=P/R= 120v/7.5e-2 = 1.6e3
I=40A
1.33h*60mins*60secs = 4.788e3
Q=Charge
Then Q=It = (40A)(4.788e3 secs) = 1.92e5 but this is not the right answer.
Any help would be great. Thanks!
I used Power= pot difference/Resistance = 9v/120w=7.5e-2
I=current
Then I used I²=P/R= 120v/7.5e-2 = 1.6e3
I=40A
1.33h*60mins*60secs = 4.788e3
Q=Charge
Then Q=It = (40A)(4.788e3 secs) = 1.92e5 but this is not the right answer.
Any help would be great. Thanks!