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.You need to login to perform this action.
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