First slide

potentiometer

Question

A 6 V battery is connected to the terminals of a three metre long wire of uniform thickness and resistance of $100 Ω$ . The difference of potential between two points on the wire separated by a distance or 50 cm, will be

Easy

A

2 V
B

3 V
C

1 V
D

1.5 V

Solution

Total current drawn from the battery
$i = \frac{E}{R + r} = \frac{6}{100 + 0} = 0 .06 A$
Resistance of 5o cm wire is $R^{'} = \frac{{ρl}^{'}}{A} = (\frac{ρ}{A}) l^{'}$
$\begin{array}{l} = (\frac{R}{l}) l^{'} (∵ R = \frac{ρl}{A}) \\ = \frac{100}{300} \times 50 \\ so, R^{'} = \frac{50}{30} Ω \end{array}$
Hence, the potential difference between two points on the wire separated by a distance l is
$V = {iR}^{'} = 0 .06 \times \frac{50}{3} = 1 V$

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