# Potentiometer and resistance

## Homework Statement

The slide wire of a simple potentiometer has a resistance of 5.0 ohm . The current source for the potentiometer is a cell of emf of 2.0V and of negligible internal resistance . Draw a labelled circuit diagram showing how the potentiometer might be used to measure an emf of up to about 2.5 mV from a copper-constantan thermocouple . Calculate a suitable value for the resistor which should be placed in series in the slide wire .

## The Attempt at a Solution

I have drawn a diagram (attached) . Let resistance of the resistor be R .

the pd across AJ = pd of thermocouple

$$\frac{2}{R+5}(R)=2.5\times 10^{-3}$$

maybe i made a mistake in calculating the resistance of AJ , do i have to add the resistance of the wire as well , or the resistance is simply R .

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The potentiometer is used to yield a variable potential difference between one end of the slide wire and the sliding contact. The ends of the wire are connected into a circuit, so that current I flows through it. The variable potential difference is then

U=IRx

where Rx is the resistance between the terminal and the sliding contact. If the resistance of the slide wire is Rs and R is the constant resistance of a series resistor,

I=Emf(battery)/ (Rs+R).

Rx changes from zero to Rs.

The thermometer is connected in series with a very sensitive ammeter between the end of the slide wire and the sliding contact, and zero current is set by the slide. The unknown emf of the thermocouple is equal to

Emf(unknown)=IRx=Emf(battery) Rx/(Rs+R).

Try to draw the circuit. You need an output voltage that varies between 0 to 25 mV. The Emf of the battery is 2 V, Rs=5 ohm, what should be the series resistance R?

ehild