A.The battery has 7.5 V output voltage and 19 Ohm internal resistance. The battery is used to boot a small electric motor with 155 Ohm resistance. Find the power delivered to the motor.
so my first attempt was 7.5/19 =.394737A
then .394737^2*155 = 24.1517W
that was wrong so i I'm thinking this but I have only one try left:
7.5/(19+155) = .043103A
.043103^2*155 = .287976W
B.Two batteries with 2.8 voltage and 9.5 Ohm internal resistance each are connected in series. This battery pack is used to boot a small electric motor with 270 Ohm resistance. Find the power delivered to the motor.
This second question i did the same as the first but only adding the voltage = 2.8 +2.8 = 5.6 and adding the internal resistance R = 19
the 5.6/19 = .294737
.294737^2*270 = 23.4548W
And I also have only one try left with this guess:
5.6/(19+270) = .019377
.01937^2*270 = .101378W