In a potentiometer, a wire of length 10 , m h | vedclass

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(A) Given: Length of wire $L = 10 \, m$, resistance of wire $R = 50 \, \Omega$, emf of primary battery $E_1 = 5 \, V$, series resistor $R_1 = 450 \, \

Vedclass · Accepted Answer

$0.225$

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(A) Given: Length of wire $L = 10 \, m$, resistance of wire $R = 50 \, \Omega$, emf of primary battery $E_1 = 5 \, V$, series resistor $R_1 = 450 \, \Omega$, balancing length $x = 450 \, cm = 4.5 \, m$.First, calculate the current $i$ flowing through the potentiometer wire:$i = \frac{E_1}{R + R_1} = \frac{5}{50 + 450} = \frac{5}{500} = 0.01 \, A$.The potential drop across the entire wire is $V = i 	imes R = 0.01 	imes 50 = 0.5 \, V$.The potential gradient $k$ of the wire is $k = \frac{V}{L} = \frac{0.5 \, V}{10 \, m} = 0.05 \, V/m$.The emf $E$ of the unknown battery is given by $E = k 	imes x = 0.05 \, V/m 	imes 4.5 \, m = 0.225 \, V$.

In a potentiometer, a wire of length $10 \, m$ having resistance $50 \, \Omega$ is used. $A$ battery of $5 \, V$ and a resistor of $450 \, \Omega$ are connected in series to the wire. If an unknown battery of emf $E$ balances the potentiometer at $450 \, cm$, then the value of $E$ is (in $V$)

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