A) 4.5 V
B) 1.5 V
C) 2.67 V
D) 13.5 V
Correct Answer: D
Solution :
Current in the bulb \[=\frac{P}{V}=\frac{4.5}{1.5}=3A\] Current in 1 W resistance \[=\frac{1.5}{1}=1.5A\] Hence total current from the cell \[i=3+1.5=4.5A\] By using \[E=V+ir\]Þ \[E=1.5+4.5\times (2.67)=13.5V\]
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