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## Homework Statement

How do I find the answer for part C?

3) A battery with an Emf of 12 volts is hooked up to a 125 ohm resistor and an ammeter. The ammeter measures a current of 75 mA.

a) Calculate the internal resistance of the battery and the actual voltage across the battery.

b) Calculate the power dissipated in the resistor and the power generated by the battery.

c) The battery is now disconnected from the 125 ohm resistor and connected to a 60 watt light bulb.

What will be the actual power output of the light bulb?

## Homework Equations

V

_{battery}= Emf - IR

_{internal}

V = IR = I

^{2}R

P = iV

## The Attempt at a Solution

a) V

_{battery}= .075(125) = 9.375

R

_{internal}= (9.375 - 12) / -0.075 = 35ohms

b) P

_{resistor}= (0.075)

^{2}125 = 0.703 watts

P

_{battery}= 0.075(9.375) = 0.703 watts

c)I thought the power supplied to the lightbulb would be 0.703 watts, but it isn't... can anyone explain why?