Current Electricity Charging Discharging Of Capacitors Question 629
Question: A battery having e.m.f. $ 5V $ and internal resistance 0.5 W is connected with a resistance of 4.5 W then the voltage at the terminals of battery is
[RPMT 2000]
Options:
A) 4.5 V
B) 4 V
C) 0 V
D) 2 V
Show Answer
Answer:
Correct Answer: A
Solution:
$ i=\frac{E}{R+r}=\frac{5}{4.5+0.5}=1A $
$ V=E-ir=5-1\times 0.5=4.5,\text{Volt} $
BETA
