A) \[100\times 0.8\,watt\]
B) \[100\times {{(0.8)}^{2}}watt\]
C) Between \[100\times 0.8\]watt and 100 watt
D) Between \[100\times {{(0.8)}^{2}}watt\] and \[100\times 0.8\]watt
Correct Answer: D
Solution :
\[{{P}_{1}}=\frac{{{(220)}^{2}}}{{{R}_{1}}}\] and \[{{P}_{2}}=\frac{{{(220\times 0.8)}^{2}}}{{{R}_{2}}}\] \[\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{{{(220\times 0.8)}^{2}}}{{{(220)}^{2}}}\times \frac{{{R}_{1}}}{{{R}_{2}}}\] Þ\[\frac{{{P}_{2}}}{{{P}_{1}}}={{(0.8)}^{2}}\times \frac{{{R}_{1}}}{{{R}_{2}}}\] Here R2 < R1 (because voltage decreases from 220 V ® 220 ´ 0.8 V It means heat produced ® decreases) So \[\frac{{{R}_{1}}}{{{R}_{2}}}>1\] Þ \[{{P}_{2}}>{{(0.8)}^{2}}{{P}_{1}}\] Þ \[{{P}_{2}}>{{(0.8)}^{2}}\times 100\,W\] Also \[\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{(220\times 0.8){{i}_{2}}}{220\,{{i}_{1}}},\] Since \[{{i}_{2}}<{{i}_{1}}\] (we expect) So \[\frac{{{P}_{2}}}{{{P}_{1}}}<0.8\] Þ \[{{P}_{2}}<(100\times 0.8)\] Hence the actual power would be between \[100\times {{(0.8)}^{2}}W\] and (100 ´ 0.8) W
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