A motor as an internal resistance of 0.5 ohms. It draws 20 amperes at full load from a 120-V supply.
a) What is the emf of the motor?
b) What is the load resistance?
Potential difference between the terminals = emf - Ir (r = internal resistance)
(P.D. = emf - Ir)
V = IR
The Attempt at a Solution
P.D. = emf - Ir
120 = emf - (20)(0.5)
120 = emf - 10
emf = 130 volts
I'm not sure I understand what the 120-V supply is referring to. I assumed it means the potential difference between the terminals when the current is 20 amperes?
My hesitation is that it says "120-V supply" so I was thinking that maybe 120-V is the emf, and in that case the potential difference would be 120 - 10 = 110 Volts? This would also change the answer in part b.
Here I assume if we have a circuit with a 120-V source and a 0.5 ohms internal resistance, the load resistance is a second resistance (R) on the circuit.
So I apply:
V = IR (where R is the load resistance)
R = V / I
R = 120 / 20 = 6 ohms