Question

Two rods are joined end to end, as shown. Both have a cross-sectional area of $0.01c{m}^2$.

Each is $1$ meter long. One rod is of copper with a resistivity of $1.7\times 10^{-6}$ 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?

• A. $0.00145V$
• B. $0.0145V$
• C. $1.7\times 10^{-6}V$
• D. $0.117V$

Answer 1

Answer: (D)

Here,
Length of each rod, $I=1m$
Area of cross-section of each rod,
$A=0.01c{m}^2=0.01\times 10^{-4}{m}^2$
Resistivity of copper rod,
${\rho }_{cu}=1.7\times 10^{-6}\Omega cm$
$=1.7\times 10^{-6}\times 10^{-2}\Omega m=1.7\times 10^{-8}\Omega m$
Resistivity of iron rod,
${\rho }_{Fe}=10^{-5}\Omega cm$
$=10^{-5}\times 10^{-2}\Omega m=10^{-7}\Omega cm$
$\therefore$ Resistance of copper rod,
$R_{cu}={\rho }_{cu}\frac{l}{A}$
and resistance of iron rod,
$R_{Fe}={\rho }_{Fe}\frac{l}{A}$
As copper and iron rods are connected in series, therefore equivalent resistance is
$R=R_{cu}+R_{Fe}={\rho }_{cu}\frac{l}{A}+{\rho }_{Fe}\frac{l}{A}=\left({\rho }_{cu}+{\rho }_{Fe}\right)\frac{l}{A}$

Voltage required to produce $1A$ current in the rods is
$V=IR=(1)\left(R_{cu}+R_{Fe}\right)$
$=\left({\rho }_{cu}+{\rho }_{Fe}\right)\left(\frac{l}{A}\right)$
$=\left(1.7\times 10^{-8}+10^{-7}\right)\left(\frac{1}{0.01\times 10^{-4}}\right)V$
$=10^{-7}(0.17+1)\left(10^6\right)V$
$=1.17\times 10^{-1}V=0.117V$