Do you still remember the Resistance
Electricity and Magnetism
Level
3
Two wires of the same material and equal length are joined in parallel combination. If one of them has half the thickness of the other and the thinner wire has a resistance of 8 ohms, what is the resistance of the parallel combination (in Ohms)?
The answer is 1.6.
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1 solution
Two resistances in parallel have an equivalent resistance equal to:
1/R1 + 1/R2 = 1/R => R1*R2 / (R1 + R2) = R (i).
If the smaller and the larger wires have cross-sectional areas equal to A1 and A2 respectively, then we have:
R1 = k L/A1, R2 = k L/A2
where k is the resistivity of the common material and L is the overall wire length. We can combine these two expressions to obtain:
R2 = (A1/A2)*R1 (ii)
where R1 = 8 ohms. If the larger wire has diameter = d, then:
R2 = [pi/4 (d/2)^2 / pi/4 (d^2)] * 8 = 8/4 = 2 ohms.
Finally, the equivalent parallel resistance computes to:
R = (8*2) / (8+2) = 1.6 ohms.