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?
![](https://neetcare-image.cdn.examgoal.net/1qpqahkk7u23iw0/975b88f8-5e9d-429a-b3fe-dffc8559a84c/0fbaa600-674b-11ea-aedc-f1c8d8e0194b/file-1qpqahkk7u23iw1-277w.jpg)

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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