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In an experiment with a potentiometer, V_{B}=10V R is adjusted to be 50 \Omega (Figure). A student wanting to measure voltage E1 of a battery (approx. 8V) finds no null point possible. He then diminishes R to 10 \Omega and is able to locate the null point on the last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

 

Answers (1)

Req=50+R

Veq=10V

I=\frac{10}{50+R'}

The potential difference across the potentiometer

 V'=IR'=\frac{10R'}{50+R'}

V'<8

10R'<8(50+R' )

R'<200

\frac{10R'}{10+R'}>8

R'>40

At balance point:

\frac{10\times \frac{3}{}4R'}{10+R'}<8

R'>-160

As R’ can only be positive, R'>160

k\times 400cm>8V

k>2 V/m

As balance, point isn’t at 3m

K\times 3<8

k<2\frac{2}{}3\frac{V}{}m

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