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, \[R=\frac{40}{{{(1/3)}^{2}}}=40\times 9=360\,\Omega \]You need to login to perform this action.
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