Consider two wires, AB and CD, of lengths 2L and L and radius r and 2r, respectively. AB and CD are made of material of resistivity ρ and 2ρ, respectively. The wires are connected in parallel to a battery of EMF E of negligible internal resistance. The ratio of currents through AB and CD. (IAB/ICD) is:

1
2
2
4
3
\(\frac{1}{4}\)
4
\(\frac{1}{2}\)

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