# Circuit LOSt

1. Feb 27, 2007

### natural

Circuit ....LOSt

1. The problem statement, all variables and given/known data
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.
3. 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?

2. Feb 28, 2007

### andrevdh

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.

3. Feb 28, 2007

### n00b

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

4. Mar 1, 2007

### andrevdh

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