A) 4
B) 6
C) 8
D) 11
Correct Answer: A
Solution :
[a] Resistance of bulb \[A=\frac{{{v}^{2}}}{P}=\frac{4}{10}=0.4\] Resistance of bulb \[B=\frac{{{v}^{2}}}{P}=0.2\] \[emf=\frac{d\phi }{dt}=\frac{d}{dt}({{\mu }_{0}}nI\times A)\] \[={{\mu }_{0}}n\times A\times \frac{dI}{dt}={{10}^{-7}}\times 4\pi \times 1000\times \pi {{(1)}^{2}}\times 9\] \[v=36\times {{10}^{-3}}\] \[I=\frac{v}{{{\operatorname{R}}_{eq}}}=\frac{36\times {{10}^{-3}}}{0.6}=6\times {{10}^{-2}}\,A\] Power dissipated through bulb \[B={{I}^{2}}R\] \[=36\times {{10}^{-4}}\times 0.2=7.2\times {{10}^{-4}}\,watt\]You need to login to perform this action.
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