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 \[I=\frac{V}{R}\] \[=\frac{440}{2420}=\frac{2}{11}A\] Potential difference across 25 W bulb \[{{V}_{1}}=I{{R}_{1}}\] \[=\frac{2}{11}\times 1936\] \[=352V\] 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.You need to login to perform this action.
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