Question 719: Current Electricity

Question

In an aluminium (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 $$	imes$$ 10
$$-$$8
$$\Omega$$ m and 1.0 $$	imes$$ 10
$$-$$7
$$\Omega$$ m, respectively. The electrical resistance between the two faces P and Q of the composite bar is

Accepted Answer

Resistance of a wire, $$R = {{ho l} \over A}$$
For iron (Fe) bar,
$$ho$$ = 10
$$-$$7
$$\Omega$$m, l = 50 mm = 50 $$	imes$$ 10
$$-$$3
m
A = (2 mm) $$	imes$$ (2 mm) = 4 mm
2
= 4 $$	imes$$ 10
$$-$$6
m
2
$${R_1} = {{{{10}^{ - 7}} 	imes 50 	imes {{10}^{ - 3}}} \over {4 	imes {{10}^{ - 6}}}} = 1250 	imes {10^{ - 6}}$$ $$\Omega$$ = 1250 $$\mu$$W
For aluminium (Al) bar,
$$ho$$ = 2.7 $$	imes$$ 10
$$-$$8
$$\Omega$$ m, l = 50 mm = 50 $$	imes$$ 10
$$-$$3
m
A = (7
2
$$-$$ 2
2
) mm
2
= 45 mm
2
= 45 $$	imes$$ 10
$$-$$6
m
2
$$	herefore$$ $${R_2} = {{2.7 	imes {{10}^{ - 8}} 	imes 50 	imes {{10}^{ - 3}}} \over {45 	imes {{10}^{ - 6}}}}$$
$$ = {{27 	imes 50} \over {45}} 	imes {10^{ - 6}} = 30 	imes {10^{ - 6}}$$ $$\Omega$$ = 30 $$\mu$$$$\Omega$$
Potential difference across both bars (resistors) is same so they are in parallel combination.
Equivalent resistance between P and Q is given by
$$R = {{{R_1}{R_2}} \over {{R_1} + {R_2}}} = {{1250 	imes 30} \over {1250 + 30}} = {{125 	imes 30} \over {128}} = {{1875} \over {64}}\mu \Omega $$.

Answer

$${{2475} \over {64}}\mu \Omega $$

Answer

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

Answer

$${{1875} \over {49}}\mu \Omega $$

Answer

$${{2475} \over {132}}\mu \Omega $$

Current Electricity

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