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 25 W, 220 V and 100 W, 200 V are connected in series across a 220 V source. The 25 W and 100 W bulbs now draw powers ?i and P^ respectively, then: (1) \[{{P}_{1}}=16W\] (2) \[{{P}_{1}}=4\,W\] (3) \[{{P}_{2}}=16\,W\] (4) \[{{P}_{2}}=4\,W\]A) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 are correct
D) 1, 2 and 3 are correct
Correct Answer: C
Solution :
Resistance of first bulb. \[{{R}_{1}}=\frac{V_{1}^{2}}{{{P}_{1}}}=\frac{{{(220)}^{2}}}{25}\] Resistance of second bulb \[{{R}_{2}}=\frac{V_{1}^{2}}{{{P}_{2}}}=\frac{{{(220)}^{2}}}{100}\] Net resistance in series combination will be \[R={{R}_{1}}+{{R}_{2}}=\] \[\frac{{{(220)}^{2}}}{25}+\frac{{{(220)}^{2}}}{100}=\frac{{{(220)}^{2}}\times 5}{100}=\frac{{{(220)}^{2}}}{20}\] Current through the circuit, \[I=\frac{V}{R}=\frac{220}{{{(220)}^{2}}/20}=\frac{20}{220}=\frac{1}{11}A\] Power in 25W bulb, \[{{P}_{1}}={{I}^{2}}{{R}_{2}}={{\left( \frac{1}{11} \right)}^{2}}\times \frac{{{(220)}^{2}}}{25}=16W\] Power in 100 W bulb, \[{{P}_{2}}={{I}^{2}}{{R}_{2}}={{\left( \frac{1}{11} \right)}^{2}}\times \frac{{{(220)}^{2}}}{100}=4W\]
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