First slide

Resistors

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 $2.7 \times 10^{- 8} Ω$ m and $1.0 \times 10^{- 7} Ω$ m, respectively. The electrical resistance between the two faces P and Q of the composite bar is

Moderate

A

$\frac{2475}{64} μ Ω$
B

$\frac{1875}{64} μ Ω$
C

$\frac{1875}{49} μ Ω$
D

$\frac{2475}{132} μ Ω$

Solution

$B o t h r o d s a r e i n p a r a l l e l, a s t h e y w i l l b e c o n n e c t e d t o s a m e p o t e n t i a l d i f f e r e n c e .$
$\begin{array}{l} \frac{1}{R} = \frac{1}{R_{A l}} + \frac{1}{R_{F e}} \\ \Rightarrow \frac{1}{R} = (\frac{A_{A l}}{ρ_{A l}} + \frac{A_{F e}}{ρ_{F e}}) \frac{1}{ℓ} \\ \Rightarrow \frac{1}{R} = [\frac{(7^{2} - 2^{2})}{2.7} + \frac{2^{2}}{10}] \frac{10^{- 6}}{10^{- 8}} \times \frac{1}{50 \times 10^{- 3}} \\ \Rightarrow \frac{1}{R} = [\frac{45}{2.7} + \frac{2}{5}] 2 \times 10^{3} = (\frac{450}{27} + \frac{2}{5}) 2 \times 10^{3} \\ \Rightarrow \frac{1}{R} = (\frac{50}{3} + \frac{2}{5}) 2 \times 10^{3} = \frac{256}{15} \times 2 \times 10^{3} \\ \Rightarrow R = \frac{15}{512} \times 10^{3} μ Ω = \frac{1875}{64} μ Ω \end{array}$

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