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

Answer 1

Answer: (D)

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

Voltage required to produce $1 A$ current in the rods is
$V=I R=(1)\left(R_{c u}+R_{F e}\right)$
$=\left(\rho_{c u}+\rho_{F e}\right)\left(\frac{l}{A}\right) $
$=\left(1.7 \times 10^{-8}+10^{-7}\right)\left(\frac{1}{0.01 \times 10^{-4}}\right) V$
$=10^{-7}(0.17+1)\left(10^{6}\right) V$
$=1.17 \times 10^{-1} V=0.117 \,V$