Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Two electric bulbs rated at 25 W, 220 V and 100 W, 220 V are connected in series across a 220 V source. The 25 W and 100 W bulbs now draw powers\[{{p}_{1}}\]and\[{{p}_{2}}\]respectively. Then (1) \[{{P}_{1}}=4W\] (2) \[{{P}_{1}}=16W\] (3) \[{{P}_{2}}=16W\] (4) \[{{P}_{2}}=4W\]A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: C
Solution :
Resistance of a bulb of\[25\text{ }W=\frac{{{V}^{2}}}{P}\] \[{{R}_{1}}=\frac{{{(220)}^{2}}}{25}\] \[=1936\,\Omega \] Resistance of a bulb of \[100\text{ }W=\frac{{{(220)}^{2}}}{100}\] \[{{R}_{2}}=484\,\Omega \] Total resistance\[R={{R}_{1}}+{{R}_{2}}\] \[=1936+484=2420\] Current, \[i=\frac{V}{R}\] \[=\frac{220}{2420}=\frac{1}{11}A\] Power in bulb of 25 W \[{{P}_{1}}={{i}_{2}}{{R}_{1}}\] \[={{\left( \frac{1}{11} \right)}^{2}}\times 1936=16W\] Power in bulb of 100 W \[{{p}_{2}}={{I}^{2}}{{R}_{2}}\] \[={{\left( \frac{1}{11} \right)}^{2}}\times 484=4\,W\] Hence, option is correct.
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