The current in the primary circuit of a potentiometer wire is 0-5 A- - for the wire is 4 - 10-7- -m and area of cross-section of wire is 8 - 10-6 m2- The potential gradient in the wire would be

The resistivity of a potential meter wire is if the area of cross section of the wire is $4 c m^{2}$ . The current flowing in the circuit is 1A, the potential gradient is 7.5 v/m

Q. The specific resistance and cross-section area of potentiometer wire is

ρ

and

A

respectively. If a current

I

is passed through the wire, the potential gradient of the wire will be :

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The current in the primary circuit of a potentiometer wire is 0.5 A, ρ for the wire is 4×10−7Ω−m and area of cross-section of wire is 8×10−6m2. The potential gradient in the wire would be

The correct option is B 25 mV/meterPotential gradient of wire =Vl=(ρA)×I where l & A are the length and cross-section of wire so Vl=4×10−78×10−6×0.5=25 mV/meter

The current in the primary circuit of a potentiometer wire is 0.5 A, $ρ$ for the wire is $4 \times 10^{- 7} Ω - m$ and area of cross-section of wire is $8 \times 10^{- 6} m^{2}$ . The potential gradient in the wire would be

The correct option is B 25 mV/meter
Potential gradient of wire $= \frac{V}{l} = (\frac{ρ}{A}) \times I$ where $l & A$ are the length and cross-section of wire so $\frac{V}{l} = \frac{4 \times 10^{- 7}}{8 \times 10^{- 6}} \times 0.5 = 25 m V / m e t e r$