Question

A potentiometer wire has resistance 40 ohm and its length is 10m it is connected to a resistance of 760 ohm in series if emf of battery is 2 volt then potential gradient is

Answer 1

Given in the question :-

Refer to the attachment

Wire resistance = 40 ohm

length = 10 m.

R = 760 ohm

v = 2

We know the formula ,

i = v/R

i = 2/(760 +40)

i = 2/800 A

i = 1/400 A .

Now, Potential drop at wire = i × 40

= 1/10 v.

Potential gradient = Potential drop / length

Potential gradient  = (1/10) / 10 v/m

Potential gradient  = 0.01 v/m.

Hope it helps .

Answer 2

Answer:

0.01V/m

Explanation:

As we know,

K(potential gradient) =voltage of potentiometer wire /length of potentiometer wire

Voltage of wire=I×40

Net resistance= 760+40=800

I=net Emf/net resistance=2/800

Voltage=2×40/800=0.1

K=0.1/100=0.001V/cm or 0.01V/m

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