| A 100 W bulb \[{{B}_{1}}\], and two 60 W bulbs \[{{B}_{2}}\]and \[{{B}_{3}}\], are connected to a 250 V source, as shown in the figure. |
| Now, \[{{W}_{1}}\], \[{{W}_{2}}\] and \[{{W}_{3}}\]are the output powers of the bulbs \[{{B}_{1}}\], \[{{B}_{2}}\]and \[{{B}_{3}}\] respectively. Then: |
 |
A) \[{{W}_{1}}>{{W}_{2}}={{W}_{3}}\]
B) \[{{W}_{1}}>{{W}_{2}}>{{W}_{3}}\]
C) \[{{W}_{1}}<{{W}_{2}}={{W}_{3}}\]
D) \[{{W}_{1}}<{{W}_{2}}<{{W}_{3}}\]
Correct Answer: D
Solution :
| \[p=\frac{{{V}^{2}}}{R};\] so, \[R=\frac{{{V}^{2}}}{P}\] |
| \[\therefore \,\,\,\,\,\,{{R}_{1}}=\frac{{{V}^{2}}}{100}\] and \[{{R}_{2}}={{R}_{3}}=\frac{{{V}^{2}}}{60}\] |
| Now, \[{{W}_{1}}=\frac{{{\left( 250 \right)}^{2}}}{{{\left( {{R}_{1}}+{{R}_{2}} \right)}^{2}}}.{{R}_{1}}\] |
| \[{{W}_{2}}=\frac{{{\left( 250 \right)}^{2}}}{{{\left( {{R}_{1}}+{{R}_{2}} \right)}^{2}}}.{{R}_{2}}\] and \[{{W}_{3}}=\frac{{{\left( 250 \right)}^{2}}}{{{R}_{3}}}\] |
| \[{{W}_{1}}:{{W}_{2}};\,\,\,\,\,{{W}_{3}}-15:25:64\]or \[{{W}_{1}}<{{W}_{2}}<{{W}_{3}}\] |
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