A) 100 W lamp will fuse
B) 40 W lamp will fuse
C) Both lamps will fuse
D) Neither lamp will fuse
Correct Answer: B
Solution :
When current \[i\] flows, across potential\[V\], then power\[={{v}_{i}}\], and \[V=i\,\,R\](Ohm?s law)
The currents required by the two lamps for their normal brightness are \[{{i}_{1}}=\frac{{{P}_{1}}}{{{V}_{1}}}=\frac{40}{220}\] \[=\frac{2}{11}=0.18\,A\] \[{{i}_{2}}=\frac{{{P}_{2}}}{{{V}_{2}}}=\frac{100}{200}\] \[=0.45\,\,A\] The resistance of the filaments are \[{{R}_{1}}=\frac{V}{{{i}_{1}}}\] \[=\frac{220\times 11}{2}=1210\Omega \] \[{{R}_{2}}=\frac{V}{{{i}_{2}}}\] \[=\frac{220\times 22}{10}=484\,\Omega \] The current in each lamp when connected in series with a \[40\,V\] supply is \[i=\frac{V}{{{R}_{1}}+{{R}_{2}}}\] \[=\frac{40}{1694}\] \[=0.0236\]. Thus, \[i>{{i}_{1}}\,and\,i<{{i}_{2}}\] Thus, the 40 W lamp will fuse, while the 100 W lamp will light dim. Alternative: Resistance of 40 W lamps is \[R=\frac{{{V}^{2}}}{P}=\frac{{{\left( 220 \right)}^{2}}}{40}=1210\,\Omega \] Resistance of 100 W lamp will be \[R=\frac{{{V}^{2}}}{100}=\frac{{{\left( 220 \right)}^{2}}}{100}=484\,\,\Omega \] Now the current through series combination is given by \[=\frac{440}{1210+484}=0.26\,A\] Potential drop across 40 W lamp will be given by \[=0.26\times 1210\] \[=314.6\,V\] And across 100 W lamp \[=0.26\times 484\] \[=125.84\,V\] Therefore, 40 W bulb will fuse because the lamp can tolerate only 220 V.
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