A) 33.3 W
B) 66.7 W
C) 300 W
D) 100 W
Correct Answer: B
Solution :
When two similar bulbs of different powers are connected in series, then \[\frac{1}{P}=\frac{1}{{{P}_{1}}}+\frac{1}{{{P}_{2}}}\] Given, \[{{P}_{1}}=200\text{ }W,\text{ }{{P}_{2}}=100\text{ }W\] \[\Rightarrow \] \[\frac{1}{P}=\frac{1+2}{200}\] \[\Rightarrow \] \[P=\frac{200}{3}=66.7\,W\]You need to login to perform this action.
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