A) 25 W bulb
B) 100 W bulb
C) Both will have equal incandescence
D) Neither 25 W nor 100 W bulb will give light
Correct Answer: A
Solution :
Since power P is given by \[P={{V}^{2}}/R\], so \[R={{V}^{2}}/P\] For the first bulb, \[{{R}_{1}}=\left( \frac{{{V}^{2}}}{{{P}_{1}}} \right)=\left[ \frac{{{(220)}^{2}}}{25} \right]=1936\Omega \] For the second bulb, \[{{R}_{2}}=\left( \frac{{{V}^{2}}}{{{P}_{2}}} \right)=\left[ \frac{{{(220)}^{2}}}{100} \right]=484\Omega \] Current in series combination is the same in the two bulbs and current i is given by \[i=\frac{V}{{{R}_{1}}+{{R}_{2}}}=\frac{220}{1936+484}\] \[=\frac{220}{2420}=\frac{1}{11}A\] If the actual powers in the two bulbs be \[{{P}_{1}}\] and \[{{P}_{2}}\] then \[P_{1}^{}={{i}^{2}}{{R}_{1}}={{\left( \frac{1}{11} \right)}^{2}}\times 1936=16W\] and \[P_{2}^{}={{i}^{2}}{{R}_{2}}={{\left( \frac{1}{11} \right)}^{2}}\times 484=4W\] Since \[P_{1}^{}>P_{2}^{},25W\] bulb will glow more brightly.
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