A) \[33.3\text{ }W\]
B) \[66.7\text{ }W\]
C) \[300W\]
D) \[100W\]
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, \[{{P}_{1}}=200W,\,\,\,\,\,\,{{P}_{2}}=100W\] \[\Rightarrow \] \[\frac{1}{P}=\frac{1+2}{200}\] \[\Rightarrow \] S\[P=\frac{200}{3}=66.7W\]You need to login to perform this action.
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