A) 5/2 \[\Omega \]
B) 40/3 \[\Omega \]
C) 40 \[\Omega \]
D) 100 \[\Omega \]
Correct Answer: C
Solution :
Given: Ratio of cross-sectional areas of the wires =3 : 1 and resistance of thick wire \[{{R}_{1}}=10\,\Omega \]. We know that resistance (R)\[=\rho \frac{1}{A}\propto \frac{1}{A}.\] Therefore, \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{1}{3}\] or \[{{R}_{2}}=3{{R}_{1}}=3\times 10=30\,\Omega \] and equivalent resistance of these two series in parallel combination\[={{R}_{1}}+{{R}_{2}}=30+10=40\,\Omega .\]
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