In an aluminium (Al) bar of square cross-sect | vedclass

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(B) The composite bar acts as two resistors in parallel,one of $Fe$ and one of $Al$.Length $L = 50 	imes 10^{-3} \ m$.Area of $Fe$ core $A_{Fe} = (2

Vedclass · Accepted Answer

$\frac{1875}{64} \mu \Omega$

Vedclass · Answer

(B) The composite bar acts as two resistors in parallel,one of $Fe$ and one of $Al$.Length $L = 50 	imes 10^{-3} \ m$.Area of $Fe$ core $A_{Fe} = (2 	imes 10^{-3} \ m)^2 = 4 	imes 10^{-6} \ m^2$.Area of $Al$ part $A_{Al} = (7 	imes 10^{-3} \ m)^2 - (2 	imes 10^{-3} \ m)^2 = (49 - 4) 	imes 10^{-6} \ m^2 = 45 	imes 10^{-6} \ m^2$.Resistance of $Fe$ part: $R_{Fe} = \frac{ho_{Fe} L}{A_{Fe}} = \frac{1.0 	imes 10^{-7} 	imes 50 	imes 10^{-3}}{4 	imes 10^{-6}} = 1.25 	imes 10^{-3} \ \Omega = 1250 \ \mu \Omega$.Resistance of $Al$ part: $R_{Al} = \frac{ho_{Al} L}{A_{Al}} = \frac{2.7 	imes 10^{-8} 	imes 50 	imes 10^{-3}}{45 	imes 10^{-6}} = 0.03 	imes 10^{-3} \ \Omega = 30 \ \mu \Omega$.Since they are in parallel,the equivalent resistance $R_{eq} = \frac{R_{Fe} 	imes R_{Al}}{R_{Fe} + R_{Al}} = \frac{1250 	imes 30}{1250 + 30} = \frac{37500}{1280} \ \mu \Omega = \frac{3750}{128} \ \mu \Omega = \frac{1875}{64} \ \mu \Omega$.

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