A potentiometer wire has length 10, m and res | vedclass

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(A) The potential gradient $x$ is defined as the potential drop per unit length of the wire.First,calculate the total current $I$ flowing through the

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$5 	imes 10^{-5}\,V/mm$

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(A) The potential gradient $x$ is defined as the potential drop per unit length of the wire.First,calculate the total current $I$ flowing through the circuit:$I = \frac{V}{R_{total}} = \frac{2.5\,V}{20\,\Omega + 80\,\Omega} = \frac{2.5}{100} = 0.025\,A$.The potential drop across the potentiometer wire of resistance $R = 20\,\Omega$ is:$V_{wire} = I 	imes R = 0.025\,A 	imes 20\,\Omega = 0.5\,V$.The potential gradient $x$ is given by:$x = \frac{V_{wire}}{L} = \frac{0.5\,V}{10\,m} = 0.05\,V/m$.To convert this to $V/mm$,we know $1\,m = 1000\,mm$:$x = \frac{0.05\,V}{1000\,mm} = 5 	imes 10^{-5}\,V/mm$.

$A$ potentiometer wire has length $10\, m$ and resistance $20\,\Omega$. $A$ $2.5\, V$ battery of negligible internal resistance is connected across the wire with an $80\,\Omega$ series resistance. The potential gradient on the wire will be

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