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

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(A) The current $I$ flowing through the potentiometer wire is given by the Ohm's law: $I = \frac{V_{total}}{R_{total}} = \frac{5 \ V}{50 \ \Omega + 45

Vedclass · Accepted Answer

$0.225$

Vedclass · Answer

(A) The current $I$ flowing through the potentiometer wire is given by the Ohm's law: $I = \frac{V_{total}}{R_{total}} = \frac{5 \ V}{50 \ \Omega + 450 \ \Omega} = \frac{5}{500} \ A = 0.01 \ A$.The potential drop across the entire wire is $V_{wire} = I 	imes R_{wire} = 0.01 \ A 	imes 50 \ \Omega = 0.5 \ V$.The potential gradient $k$ (potential drop per unit length) is $k = \frac{V_{wire}}{L} = \frac{0.5 \ V}{10 \ m} = 0.05 \ V/m$.The emf $E$ of the unknown battery is balanced at a length $l = 450 \ cm = 4.5 \ m$.Therefore,$E = k 	imes l = 0.05 \ V/m 	imes 4.5 \ m = 0.225 \ V$.

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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