Resistance of 100 , cm long potentiometer wir | vedclass

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(B) Let the resistance of the potentiometer wire be $R_w = 10 \, \Omega$ and its length be $L = 100 \, cm = 1 \, m$.The total resistance in the primar

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$790$

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(B) Let the resistance of the potentiometer wire be $R_w = 10 \, \Omega$ and its length be $L = 100 \, cm = 1 \, m$.The total resistance in the primary circuit is $R_{total} = R + R_w = R + 10$.The current in the primary circuit is $I = \frac{V}{R + R_w} = \frac{2}{R + 10}$.The potential drop across the potentiometer wire is $V_w = I 	imes R_w = \frac{2 	imes 10}{R + 10} = \frac{20}{R + 10}$.The potential gradient $k$ is given by $k = \frac{V_w}{L} = \frac{20}{R + 10} 	imes \frac{1}{100} = \frac{0.2}{R + 10} \, V/cm$.The null point is obtained at $l = 40 \, cm$ for a source of $E = 10 \, mV = 10 	imes 10^{-3} \, V$.Using the principle of the potentiometer,$E = k 	imes l$.$10 	imes 10^{-3} = \left( \frac{0.2}{R + 10} ight) 	imes 40$.$0.01 = \frac{8}{R + 10}$.$R + 10 = \frac{8}{0.01} = 800$.$R = 800 - 10 = 790 \, \Omega$.

Resistance of $100 \, cm$ long potentiometer wire is $10 \, \Omega$. It is connected to a battery of $2 \, V$ and a resistance $R$ in series. $A$ source of $10 \, mV$ gives a null point at $40 \, cm$ length. The value of external resistance $R$ is ........... $\Omega$.

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