A) 0.83 A, Bulb B
B) 0.083 A, Bulb A
C) 0.83 A, Bulb A
D) 0.083 A, Bulb B
Correct Answer: C
Solution :
Resistance of bulb A is \[{{R}_{1}}=\frac{{{V}^{2}}}{{{P}_{1}}}=\frac{{{(120)}^{2}}}{100}=144\,\Omega \] Resistance of bulb B is \[{{R}_{2}}=\frac{{{V}^{2}}}{{{P}_{2}}}=\frac{{{(120)}^{2}}}{10}=1440\,\Omega \] As the bulbs are connected in parallel to the source, the voltage across each bulb is the same. l in bulb A is \[{{i}_{1}}=\frac{V}{{{R}_{1}}}=\frac{120}{144}=0.83A\] I in bulb B is \[{{i}_{2}}=\frac{V}{{{R}_{2}}}=\frac{120}{1440}=0.083A\] Power consumed by bulb A is 100 W and by bulb B is 10 W. Bulb A consumes more energy than bulb B.
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