Question

A cell of internal resistance 3 ohm and emf 10 volt is connected to a uniform wire of length 500cm and resistance 3 ohm. What is the P.D between points A and B on the wire separated by a distance 100 cm​

Answer 1

Answer:

10mV/cm

Explanation:

Potential gradient =

$\frac{e.r}{r.r}l$

So = 10×3/(3+3)5=1V/m=10mV/cm

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Answer 2

Given: Internal resistance of a call = 3Ω

          Emf of a cell = 10 V

         Length of the wire  500cm

         External resistance = 3Ω

To find: Potential difference at any points on the wire separated by a distance of 100cm.

Solution:

Total resisatnce in the current = internal resiatnce + external resistance

                                                  = 3Ω + 3Ω = 6Ω

Current flowing through the circuit = V/R

                                                         = 10/6A = 5/3A

Resistance per unit length = 3/500

The Potential difference per unit length will be equal to the product of the current flowing through the circuit and the resistance per unit length.

dV/dx = current × resistance per unit length

          = (5/3) × (3/500)

           = 1/100 V/m

           = 1×10⁻³ V/cm

The Potential difference between the points on the wire separated by a distance of 100cm will be the product of The Potential difference per unit length and distance

V = dV/dx × 100

  = 1×10⁻³×100

  = 0.1 V

Therefore, the Potential difference between the points on the wire separated by a distance of 100cm is 0.1 V