A copper rod of length 20 cm and cross-sectional area 2 mm2 is joined with an identical aluminium rod as shown in the figure. The resistance of the combination between the two ends (A and B) is approximately
(Given: Resistivity of Cu = 1.7 × 10−8 Ω m and resistivity of Al = 2.6 × 10−8 Ω m)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1 mΩ
In the figure the copper rod and the aluminium rod both run from A to B, so between A and B they are in parallel. Each resistance follows from \( R = \rho l/A \), with every length in metres.
\( l = 20\,\mathrm{cm} = 0.20\,\mathrm{m} \)
\( A = 2\,\mathrm{mm^{2}} = 2\times10^{-6}\,\mathrm{m^{2}} \)
\( \displaystyle R_{\mathrm{Cu}} = \frac{(1.7\times10^{-8}\,\Omega\,\mathrm{m}) \times (0.20\,\mathrm{m})}{2\times10^{-6}\,\mathrm{m^{2}}} = 1.7\times10^{-3}\,\Omega \)
\( \displaystyle R_{\mathrm{Al}} = \frac{(2.6\times10^{-8}\,\Omega\,\mathrm{m}) \times (0.20\,\mathrm{m})}{2\times10^{-6}\,\mathrm{m^{2}}} = 2.6\times10^{-3}\,\Omega \)
\( \displaystyle R = \frac{R_{\mathrm{Cu}}R_{\mathrm{Al}}}{R_{\mathrm{Cu}} + R_{\mathrm{Al}}} = \frac{1.7 \times 2.6}{1.7 + 2.6}\times10^{-3}\,\Omega \)
\( R \approx 1.03\times10^{-3}\,\Omega \approx 1\,\mathrm{m}\Omega \)