A) 100 W bulb will be fused
B) 200 W bulb will be fused
C) both bulbs will be fused
D) No bulb will be fused
Correct Answer: A
Solution :
Resistance of bulb having 100 W and 250 V \[\Rightarrow \] \[{{R}_{1}}=\frac{{{V}^{2}}}{P}\] \[\Rightarrow \] \[{{R}_{1}}=\frac{{{(250)}^{2}}}{100}\] \[=625\,\Omega \] Resistance of bulb having 200 W and 250 V \[\Rightarrow \] \[{{R}_{2}}=\frac{{{V}^{2}}}{P}\] \[=\frac{{{(250)}^{2}}}{200}\] \[=\frac{625}{2}\Omega =312.5\,\Omega \] Since, both are connected in series. \[\therefore \] \[{{R}_{eq}}=625+312.5\] = 937.5 Therefore, current in the circuit when connected across 500 V \[I=\frac{500}{937.5}\] \[\therefore \] Potential difference across 100 W bulb is more because of which it get fused.
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