A battery (ε= 10.0V, r = 0.50Ω) is connected to three light bulbs in parallels (R1= 15.0Ω, R2= 21.0Ω, R3= 24.0Ω). Calculate the current in R3.
3 answers
See previous post: 12-13-18. 5PM.
1/R = 1/15 + 1/21 + 1/24 = 0.156
so R = 6.41 Ohms
total resistance including internal of battery = 6.91 Oms
total current = V/6.91 = 10/6.91 = 1.45 amps
loss of voltage due to internal resistance = 1.45*0.50 = 0.723 volts
so useful voltage = 10 - 0.723 = 9.28 volts
current we want = 9.28/24 = 0.387 amps
so R = 6.41 Ohms
total resistance including internal of battery = 6.91 Oms
total current = V/6.91 = 10/6.91 = 1.45 amps
loss of voltage due to internal resistance = 1.45*0.50 = 0.723 volts
so useful voltage = 10 - 0.723 = 9.28 volts
current we want = 9.28/24 = 0.387 amps
but now they are in parallel :)
2 answers