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Question:

In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are 2.7×10⁻⁸ Ωm and 1.0×10⁻⁷ Ωm, respectively. The electrical resistance between the two faces P and Q of the composite bar is?

187549μΩ

247564μΩ

2475132μΩ

187564μΩ

Solution:

RAl and RFe are parallel to each other.
R = ρl/A
Al: Area = 7² - 2² = 45mm²
ρ = 2.7×10⁻⁸ Ωm
l = 50mm
Fe: Area = 2² = 4mm²
ρ = 1.0×10⁻⁷ Ωm
l = 50mm
Substituting values:
RAl = 30μΩ
RFe = 1250μΩ
Rtotal = RAlRFe/(RAl + RFe) = 30 × 1250 / (30 + 1250) = 187564μΩ