A wire is connected to a battery and has a certain current flowing through it. If the radius of a wire were doubled, what would happen to the current?
The correct answer and explanation is:
If the radius of a wire is doubled while connected to the same battery, the current flowing through the wire will increase. Specifically, the current will increase by a factor of four.
Here is the explanation:
The current II in a wire depends on the voltage VV supplied by the battery and the resistance RR of the wire, according to Ohm’s Law: I=VRI = \frac{V}{R}
The resistance RR of a wire is determined by the resistivity ρ\rho of the material, the length LL of the wire, and the cross-sectional area AA of the wire: R=ρLAR = \frac{\rho L}{A}
For a wire with a circular cross-section, the area AA is: A=πr2A = \pi r^2
where rr is the radius of the wire.
If the radius of the wire is doubled, the new radius becomes 2r2r. The new cross-sectional area A′A’ is: A′=π(2r)2=π×4r2=4×πr2=4AA’ = \pi (2r)^2 = \pi \times 4r^2 = 4 \times \pi r^2 = 4A
Because resistance is inversely proportional to the area, the new resistance R′R’ is: R′=ρLA′=ρL4A=R4R’ = \frac{\rho L}{A’} = \frac{\rho L}{4A} = \frac{R}{4}
With the resistance reduced to one quarter of its original value and the battery voltage unchanged, the current increases by a factor of four: I′=VR′=VR/4=4×VR=4II’ = \frac{V}{R’} = \frac{V}{R/4} = 4 \times \frac{V}{R} = 4I
Therefore, doubling the radius of the wire causes the current to increase four times, assuming the battery voltage and wire length remain constant. This is because a thicker wire has less resistance, allowing more current to flow through it under the same voltage.