A) \[10\,\Omega \]
B) \[20\,\Omega \]
C) \[30\,\Omega \]
D) \[40\,\Omega \]
Correct Answer: D
Solution :
Given \[{{A}_{1}}:{{A}_{2}}=3:1,\] \[{{R}_{1}}=10\Omega \] Resistance \[(R)=\frac{\rho l}{A}\] \[\therefore \] \[R\propto \frac{1}{A}\] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{A}_{2}}}{{{A}_{1}}}\] \[\frac{10}{{{R}_{2}}}=\frac{1}{3}\] \[\therefore \] \[{{R}_{2}}=30\Omega \] In series, \[R={{R}_{1}}+{{R}_{2}}\] \[=10+30=40\Omega \]
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