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 xx 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?

Options

A. `0.00145` V
B. 0.0145 V
C. `1.7 xx 10^(-6)` V
D. 0.117 V

Answer

D

Explanation

`R_(eq) = R_(1) +R_(2) = (rho_(1)l)/A + (rho_(2)l)/A`
`=1/(0.1 xx 10^(-4)) xx (1.7 xx 10^(-6) + 10 xx 10^(-6)) xx 10^(-2)`
`=0.117 Omega`
`I = V/R_(eq) rArr 1 = V/(11.7 xx 10^(-2)) = 0.117`Volt