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Question: A 10 m long wire of resistance 20 Ω is connected in series with a battery of emf 3V (negligible inte...

A 10 m long wire of resistance 20 Ω is connected in series with a battery of emf 3V (negligible internal resistance) and a resistance of 10 Ω. Find the potential gradient along the wire–

A

3 V/m

B

0.2 V/m

C

0.1 V/m

D

0.3 V/m

Answer

0.2 V/m

Explanation

Solution

= (Rwire) × Il\frac{I}{\mathcal{l}}

I = ε(Rwire+Rexternal)\frac{\varepsilon}{(R_{wire} + R_{external})} = 3(20+10)\frac{3}{(20 + 10)} = 330\frac{3}{30} = 0.1A

φ = 20×0.110\frac{20 \times 0.1}{10} = 0.2 V/m