A) \[450\,\,\Omega \]
B) \[360\,\,\Omega \]
C) \[120\,\,\Omega \]
D) \[13.33\,\,\Omega \]
Correct Answer: B
Solution :
Power of electric bulb is given by \[p=\frac{{{V}^{2}}}{R}\] But from Ohms law, \[V=iR\] So, \[p=\frac{{{(iR)}^{2}}}{R}={{i}^{2}}R\] \[\therefore \] \[R=\frac{p}{{{i}^{2}}}\] Here, \[p=40\,\,W,\,\,i=\frac{1}{3}\,A\] Hence, \[R=\frac{40}{{{(1/3)}^{2}}}=40\times 9=360\,\,\Omega \]
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