Question 641: Current Electricity

Question

Two rods are joined end to end, as shown. Both have a cross-sectional area of 0.01 cm
2
. Each is 1 meter long. One rod is of copper with a resistivity of 1.7 $$ 	imes $$ 10
$$-$$6
ohm-centimeter, the other is of iron with a resistivity of 10
$$-$$5
ohm-centimeter.
How much voltage is required to produce a current of 1 ampere in the rods?

Accepted Answer

Copper rod and iron rod are joined in series.
$$ 	herefore R = {R_{Cu}} + {R_{Fe}} = \left( {{ho _1} + {ho _2}} ight){\ell  \over A}$$
$$\left(\because {R = ho {\ell  \over A}} ight)$$
From ohmÃ¢ÂÂs law
V = RI = (1.7 ÃÂ 10
Ã¢ÂÂ6
ÃÂ 10
Ã¢ÂÂ2
+ 10
Ã¢ÂÂ5
ÃÂ 10
Ã¢ÂÂ2
)
$$ \div $$ 0.01 ÃÂ 10
Ã¢ÂÂ4
volt
= 0.117 volt   ($$ \because $$ I = 1A)

Answer

0.00145 V

Answer

0.0145 V

Answer

1.7 $$ 	imes $$ 10
$$-$$6
V

Answer

0.117 V

Current Electricity

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