Current Electricity Charging Discharging Of Capacitors Question 139
Question: A torch bulb rated as 4.5 W, 1.5 V is connected as shown in the figure. The e.m.f. of the cell needed to make the bulb glow at full intensity is
[MP PMT 2000]
Options:
A) 4.5 V
B) 1.5 V
C) 2.67 V
D) 13.5 V
Show Answer
Answer:
Correct Answer: D
Solution:
Current in the bulb $ =\frac{P}{V}=\frac{4.5}{1.5}=3A $ Current in 1 W resistance $ =\frac{V^2}{P}=\frac{1.5^2}{1}=2.25A $ Hence total current from the cell $ i=3+2.25=5.25A $ By using $ E=V+ir $
Therefore $ E=1.5+4.5\times (2.67)=13.5V $
BETA
