First slide

potentiometer

Question

Resistance of 100 cm long potentiometer wire is $10 Ω$ . It is connected to a battery (2 volt) and a resistance R in series. A source of 10 mV gives null point at 40 cm length, then external resistance R is

Easy

A

$490 Ω$
B

$790 Ω$
C

$590 Ω$
D

$990 Ω$

Solution

$\begin{array}{l} E = \frac{e}{(R + R_{h} + r)} \frac{R}{L} \times l \\ 10 \times 10^{- 3} = \frac{2}{(10 + R + 0)} \times \frac{10}{1} \times 0 .4 \Rightarrow R = 790 Ω \end{array}$

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