A) 40 W
B) 60 W
C) 90 W
D) 180 W
Correct Answer: A
Solution :
Resistance of each bulbs, \[R=\frac{{{\operatorname{V}}^{2}}}{P}=\frac{{{\left( 120 \right)}^{2}}}{60}\,\,=\,\,240\,\Omega \] Net resistance of circuit \[=\,\,R+\frac{R\times R}{R+R}\] \[=\,\,240+\frac{240\times 240}{240+240}\] \[=\,\,\,360\,\Omega \] \[\therefore \,Current in circuit i = R+\frac{120}{360}=\frac{1}{3}A\] Now, total power consumed \[=\,\,{{i}^{2}}\,\left( R+\frac{R\times R}{R+R} \right)\] \[=\,\,\frac{1}{9}\times 360\,\,=\,\,40\,\,W\]
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