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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 |
0.02 0.1 0.2 1.2 |
0.2 |
The correct answer is Option 3: 0.2 The circuit contains:
So the total resistance is: R (total) =20+10=30 Ω Using Ohm's Law (V=IR): I=V/R (total) =3 V /30 Ω = 0.1 A
Vwire=I×Rwire=0.1A×20Ω=2V
The potential gradient is the potential drop per unit length of the wire (L=10m): Potential Gradient=V (wire) / L = 2 V/ 10 m = 0.2 v/m
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