# Finding pd given r, R, and emf?

ugodzilla

## Homework Statement

What potential difference is measured across
a 23 ohm load resistor when it is connected
across a battery of emf 7.99 V and internal
resistance 0.555 ohms

V=IR
V=emf - Ir

## The Attempt at a Solution

V=IR
emf-Ir=IR
(emf/I)-r=R
emf/I=R-r
I=(emf/(R-r))
I=(7.99/((23+.555)-.555)))
I=0.34739

V=IR
V=(0.34739)(23.555)
V=8.182802174
INCORRECT

Staff Emeritus
Homework Helper
Welcome to Physics Forums.

## Homework Statement

What potential difference is measured across
a 23 ohm load resistor when it is connected
across a battery of emf 7.99 V and internal
resistance 0.555 ohms

V=IR
V=emf - Ir

## The Attempt at a Solution

V=IR
emf-Ir=IR
(emf/I)-r=R
emf/I=R-r
That should be R+r in that last equation, since you are adding +r to both sides of the equation before it.
I=(emf/(R-r))
I=(7.99/((23+.555)-.555)))
R is simply 23 ohms, not 23+.555
I=0.34739

V=IR
V=(0.34739)(23.555)
V=8.182802174
INCORRECT

ugodzilla
so,
I=7.99/22.445
I=0.355598
then,
V=(0.355598)(23) ?
or V=(0.355598)(23.555)

I don't understand why I'm not supposed to use the total resistance as opposed to just the external resistance.

ugodzilla
I tried out both ways, and each answer was deemed incorrect.
8.385139229
and
8.187569615

ugodzilla
I finally got! I didn't see the comment you made about the mistake in my simplifying! Thanks for the help

Staff Emeritus
Homework Helper
so,
I=7.99/22.445
I don't understand why I'm not supposed to use the total resistance as opposed to just the external resistance.
When you said that "V=emf - Ir", that V is the voltage across the 23 ohm resistor.

And you also said "V=IR" -- the same V here. Since V is the voltage across the 23 ohm resistor, R must be referring to that resistor.

I finally got! I didn't see the comment you made about the mistake in my simplifying! Thanks for the help
You're welcome, glad it worked out.