Draw a circuit sketch og the two circuit arrangements

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SUMMARY

The discussion centers on calculating the current in a circuit using a potentiometer with a standard cell of 1.080V and a 10.0-ohm resistor. The balance lengths measured were 512 mm for the standard cell and 784 mm for the resistor. The potential gradient is determined to be 2.109375 V/m, allowing for the calculation of potential across the resistor using the formula V = 0.784 * (1.080 / 0.512), which leads to the application of Ohm's law to find the current.

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