Answer to Question #208325 in Electric Circuits for Thulani Silinda

Question

To make a potentiometer, a driver cell of e.m.f. 4.0 V is connected across a 1.00 m length of resistance wire.

1.What is the potential difference across each 1 cm length of the wire? What length of wire has a p.d. of 1.0 V across it?

2.A cell of unknown e.m.f. E is connected to the potentiometer and the balance point is found at a distance of 37.0 cm from the end of the wire to which the galvanometer is connected. Estimate the value of E. Explain why this can only be an estimate.

3.A standard cell of e.m.f. 1.230 V gives a balance length of 31.2 cm. Use this value to obtain a more accurate value for E.

Answer 1

Gives

Emf=4.0volt

l=1.00m

(1)

When V₂=?

$l_2=1cm=0.01m$

$\frac{E_2}{E_1}=\frac{l_2}{l_1}$

Put value

$E_2=\frac{0.01}{1}\times4=400V$

When potential V=1V

$l_2=\frac{1}{4}\times1=0.25m$

Part (2)

balence point (l)=37cm

$\frac{4}{E}=\frac{37}{100-l}$

$\frac{4}{E}=\frac{37}{100-37}$

$E=\frac{63}{37}\times4=6.82V$

Part(c)

V=1.230V

l=31.2cm

$\frac{1.230}{E}=\frac{31.2}{68.8}$

$E=2.71V$