Question

5. In a potentiometer experiment, a wire of length 10m having resistance 2 m is connected with acell of 5 V, 1 Q and an unknown resistor RIf potential gradient of wire is 2 mV/crn, then valueof R is(1) 302(2) 30(3) 299(4) 500* ohronometer shows nul​

Answer 1

Answer:

The value of resistance R is 497 ohm

Explanation:

As we know that that wire is connected with unknown resistance R and a battery of given EMF

so here we can say that current flowing in the circuit is

$i = \frac{5}{2 + 1 + R}$

so the voltage drop in the wire is given as

$\Delta V = i R$

$\Delta V = \frac{5}{3 + R}(2)$

$\Delta V = \frac{10}{3 + R}$

now potential gradient of the wire is given as

$\phi = \frac{V}{L}$

$\frac{2}{1000} = \frac{1}{3 + R}$

$3 + R = 500$

$R = 497 ohm$

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Topic : Potentiometer

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