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A 10 m long wire of resistance 20Ω is connected in series with batter…

A 10 m long wire of resistance $20\Omega$ is connected in series with battery of e.m.f. 3V (negligible internal resistance) and a resistance of $10 \Omega$. The potential gradient along the wire in volt per metre is

Answer & explanation

Correct answer: option 3

The correct answer is Option 3: 0.2

The circuit contains:

Potentiometer wire = 20 Ω
External resistor = 10 Ω

So the total resistance is:

Using Ohm's Law ( $V = I R$ ):

I = V/ = 3 V /30 Ω = 0.1 A

V_{w i re} = I \times R_{w i re} = 0.1 A \times 20 Ω = 2 V The potential gradient is the potential drop per unit length of the wire (L = 10 m): Potential Gradient = V (wire) / L = 2 V/ 10 m = 0.2 v/m