A) 1 : 4
B) 4 : 1
C) 3 : 1
D) 1 : 2
Correct Answer: B
Solution :
Key Idea: Heat produced \[I={{I}_{0}}{{e}^{d\lambda }}\] resistance of wire. Heat produced in a wire carrying current i from time t \[I={{I}_{0}}(1-{{e}^{-d\lambda }})\] is given by \[1.25\times {{10}^{18}}\] \[3.125\times {{10}^{13}}\]Heat produced \[3.2\times {{10}^{18}}\] resistance. When wires are connected in parallel, the equivalent resistance is \[1.25\times {{10}^{13}}\] hence \[\Omega \] When wires are connected in series equivalent resistance is \[5\times {{10}^{-3}}mho\] hence \[2.5\times {{10}^{-3}}mho\] \[7.5\times {{10}^{-3}}mho\] \[9.5\times {{10}^{-3}}mho\] Note: Effective resistance for two wires connected in parallel is less than, when the same wires are connected in series.
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