A potentiometer wire of length 300   cm is connected in series with a resistance 780   Ω and a standard cell of emf 4   V . A constant current flows through potentiometer wire. The length of the null point for cell of emf 20 mV is found to be 60   cm . The resistance of…

Question

A potentiometer wire of length 300 &#160; cm is connected in series with a resistance 780 &#160; &#937; and a standard cell of emf 4 &#160; V . A constant current flows through potentiometer wire. The length of the null point for cell of emf 20 mV is found to be 60 &#160; cm . The resistance of the potentiometer wire is _____ &#937; .

Aicharya · Accepted Answer

Let resistance of potentiometers wire is $R$ , then the current through $A B$
$i_{A B} = \frac{4}{R + 780}$
Potential difference across $A B$
$V_{A B} = \frac{4 R}{R + 780}$
Potential difference across $AC$
$V_{A C} = V_{A B} \frac{l}{L} = \frac{4 R}{(R + 780)} \times \frac{60}{300} = \frac{4 R}{5 (R + 780)}$
This should be equal to $20 mV$
$\Rightarrow \frac{4 R}{5 (R + 780)} = 20 \times 10^{- 3} = 2 \times 10^{- 2}$
$\Rightarrow 4 R = 10^{- 1} (R + 780)$
$\Rightarrow 4 R - \frac{R}{10} = 78$
$\Rightarrow \frac{39 R}{10} = 78$
$\Rightarrow R = 20 Ω$

A potentiometer wire of length 300   cm is connected in series with a resistance 780   Ω and…

Solution

A potentiometer wire of length 300 &#160; cm is connected in series with a resistance 780 &#160; &#937; and…

Solution

A potentiometer wire of length 300 cm is connected in series with a resistance 780 Ω and…