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## Homework Statement

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?

## Homework Equations

Potential difference between the terminals = emf - Ir (r = internal resistance)

(P.D. = emf - Ir)

V = IR

## The Attempt at a Solution

Part a:

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.

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