A) 100 W bulb fuses
B) 25 W bulb fuses
C) both the bulbs fuse
D) neither of the bulb fuses
Correct Answer: B
Solution :
Resistance of a bulb \[R=\frac{{{V}^{2}}}{P}\] \[\therefore \] \[{{R}_{1}}=\frac{{{(220)}^{2}}}{25}=1936\,\,\Omega \] \[{{R}_{2}}=\frac{{{(220)}^{2}}}{100}=484\,\Omega \] Since, \[{{R}_{1}}\] and \[{{R}_{2}}\] are in series \[{{R}_{net}}={{R}_{1}}+{{R}_{2}}\] \[=1936+484=2420\,\,\Omega \] Hence, current \[I=\frac{V}{R}\] \[=\frac{440}{2420}=\frac{2}{1}\,A\] Potential difference across 25 W bulb \[{{V}_{1}}=I{{R}_{!}}\] \[=\frac{2}{11}\times 1936\] = 352V Potential difference across 100 W bulb \[{{V}_{2}}=I{{R}_{2}}\] \[=\frac{2}{11}\times 484\] = 88V Potential difference across 25 W bulb in this combination is 352 V, but it can tolerate only 220 V. Hence, it will fuse.
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