Question

If length, thickness, specific resistance and current flowing in the potentiometer wire is doubled, then the potential gradient will be​

Answer 1

Please find below the solution to the asked query:

Potential gradient is nothing but the potential per unit length. So, Initial Potential gradient is,

(dVdl)i=IRL=IρLAL=IρA

Now, length, thickness, specific resistance (resistivity) and current are doubled. Therefore, the final potential gradient is,

(dVdl)i=IρA⇒(dVdl)f=(2I)(2ρ)(2A)=2(IρA)⇒(dVdl)f=2(dVdl)i

Therefore, the potential gradient gets doubled.

Hope this information will clear your doubts about the topic.

Regards

Answer 2

Answer:

If length, thickness, specific resistance and current flowing in the Potentiometer wire is doubled, then the potential gradient will be​ also doubled.

Explanation:

The wire has the length L, area A, current I, Specific resistance ρ.

Potential gradient  γ = ΔV / ΔL

γ = $\frac{V}{L}$

We know that V = IR

γ  = $\frac{I R}{L}$    .................................(1)

The resistance R can be written in the form of specific resistance.

Where, R = ρ L / A

Substitute the value of R in equation (1) we get

$Potential gradient = \frac{I L}{L A}$ ×  ρ

Potential gradient = Iρ / A    ......................(2)

To find the potential gradient, If the length, thickness, specific resistance and current flowing in the Potentiometer wire is doubled:

• L = 2 L
• A = 2 A
• I = 2 I
• ρ = 2 ρ

$Potential gradient = \frac{I L}{L A}$× ρ

Potential gradient = Iρ / A    

The Potential gradient = $\frac{2 I}{2 A}$ ×(2ρ)

Potential gradient = 2ρ$\frac{I}{A}$    ...........................(3)

From equation (2) and (3) It was found that,

If the length, thickness, specific resistance and current flowing in the Potentiometer wire is doubled, then the potential gradient will be doubled.

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