First slide

potentiometer

Question

A potentiometer wire has length 4 m and resistance 8 $Ω$ . The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1mV per cm on the wire is

Moderate

A

44 $Ω$
B

48 $Ω$
C

32 $Ω$
D

40 $Ω$

Solution

$Required potential gradient = 1 {mVcm}^{- 1}$
$= \frac{1}{10} {Vm}^{- 1}$
$Length of potentiometer wire, l = 4 m$
So potential difference across potentiometer wire
$= \frac{1}{10} \times 4 = 0.4 V . . . . . (i)$
$In the circuit, potential difference across 8 Ω$
$= I \times 8 = \frac{2}{8 + R} \times 8 . . . . (i i)$
Using equation (i) and (ii), we get
$0.4 = \frac{2}{8 + R} \times 8$
$\frac{4}{10} = \frac{16}{8 + R}, 8 + R = 40$
$∴ R = 32 Ω$

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