# Circuit LOSt

Circuit ....LOSt ## Homework Statement

A sandard cell or 1080V gives a balance length of 512 mm on a potentiometer. If the p.d across a standard 10.0 (ohm) resistor in a circuir gives a balance length of 784 mm, calculate the current in the circuit with the 10.0 (ohm) resistor. Draw a circuit sketch og the two circuit arrangements required in the above case.
I am stuck here..am i going right?.please help someone..

## The Attempt at a Solution

E = k * 512*10^-3 (512 times ten to the power of minus 3)

V = E - Ir
V = k * 784*10^-3

V = IR I = V/R = V/10

E = V + Ir

K* 784*10^-3 = k * 512*10^-3 - K*784*10^-3(r)/10

K's cancel

784*10^-3(r)/10 = 784*10^-3 - 512*10^-3
= 0.0784r = 0.272
= 3.46(ohms)

.....what else do i do from here?

## Answers and Replies

andrevdh
Homework Helper
The info on the standard cell (I assume you ment 1.080 V) gives the potential gradient of the potentiometer:

$$e = 1.080/0.512\ V/m$$

With this value you can subsequently calculate the potential over the standard resistor. Which enables you to calc the current through it via Ohm's law.

how do you calculate the potential? would it be V=0.784(1.080/0.512) ?

andrevdh
Homework Helper
Yes. The potential gradient over the wire of the potentiometer is constant (the diameter of the wire is constant).