Current Electricity Charging Discharging Of Capacitors Question 606
Question: A torch battery consisting of two cells of 1.45 volts and an internal resistance $ 0.15,\Omega $ , each cell sending currents through the filament of the lamps having resistance 1.5ohms. The value of current will be
[MP PET 1994]
Options:
A) 16.11 amp
B) 1.611 amp
C) 0.1611 amp
D) 2.6 amp
Show Answer
Answer:
Correct Answer: B
Solution:
Here two cells are in series.
Therefore total emf = 2E. Total resistance = R + 2r \ $ i=\frac{2E}{R+2r}=\frac{2\times 1.45}{1.5+2\times 0.15}=\frac{2.9}{1.8}=\frac{29}{18}=1.611,amp $
BETA
