A) 200 W, 150 W
B) 50 W, 200 W
C) 50 W, 100 W
D) 100 W, 50 W
Correct Answer: B
Solution :
Let, \[{{P}_{1}}=100\,W,\,\,{{P}_{2}}=100\,W,\,V=220\,volt\] \[{{P}_{1}}=\frac{{{V}^{2}}}{{{R}_{1}}}\,and\,\,{{P}_{2}}=\frac{{{V}^{2}}}{{{R}_{2}}}\] \[\therefore {{R}_{1}}=\frac{{{V}^{2}}}{{{P}_{1}}}=\frac{{{(220)}^{2}}}{100}=\frac{220\times 220}{100}\Omega \] and \[{{R}_{2}}=\frac{{{V}^{2}}}{{{P}_{2}}}=\frac{{{(220)}^{2}}}{100}=\frac{220\times 220}{100}\Omega \] Case I: When two bulbs are connected in series. In series, \[{{R}_{eq}}={{R}_{1}}+{{R}_{2}}\] \[=\left( \frac{220\times 220}{100} \right)\times 2\] Hence, \[{{P}_{eq}}=\frac{{{V}^{2}}}{{{R}_{eq}}}=\frac{220\times 220}{\left( \frac{220\times 220}{100}\times 2 \right)}\] \[=\frac{100}{2}=50\,W\] Case II: When two bulbs are connected in parallel. In parallel, \[{{R}_{eq}}=\frac{{{R}_{1}}\,{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\] \[=\frac{{{\left( \frac{220\times 220}{100} \right)}^{2}}}{\frac{220\times 220}{100}\times 2}\] \[{{R}_{eq}}=\frac{220\times 220}{100}\times \frac{1}{2}\] Hence, \[{{P}_{eq}}=\frac{{{V}^{2}}}{{{R}_{eq}}}=\frac{220\times 220}{\frac{220\times 220}{100}\times \frac{1}{2}}\] = 200 W Alternative: For series \[{{P}_{eq}}=\frac{{{P}_{1}}{{P}_{2}}}{{{P}_{1}}+{{P}_{2}}}=\frac{100\times 100}{200}=50\,W\] For parallel \[{{P}_{eq}}={{P}_{1}}+{{P}_{2}}=100+100=200\,W\]You need to login to perform this action.
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