A) both
B) 100W
C) 25W
D) neither
Correct Answer: C
Solution :
[c] |
As \[{{R}_{1}}=\frac{220}{25}\times 200\] and \[{{R}_{2}}=\frac{220}{100}\times 220\] |
\[R={{R}_{1}}+{{R}_{2}}\] |
\[=220\times 220\left( \frac{1}{25}+\frac{1}{100} \right)\] |
\[=220\times 220\frac{1}{20}\] |
\[\therefore \,{{I}_{live}}=\frac{440}{\frac{220\times 220}{20}}=\frac{40}{220}A\] |
\[\therefore \] 1st bulb (25 W) will fuse only |
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