Current Electricity Charging Discharging Of Capacitors Question 618
Question: Consider the circuit shown in the figure. The current $ I _{3} $ is equal to
[AMU 1995]
Options:
A) 5 amp
B) 3 amp
C) $ -3,amp $
D) $ -5/6,amp $
Show Answer
Answer:
Correct Answer: D
Solution:
Suppose current through different paths of the circuit is as follows. After applying KVL for loop (1) and loop (2) We get $ 28i _{1}=-6-8 $
Therefore $ i _{1}=-\frac{1}{2}A $ and $ 54i _{2}=-6-12 $
Therefore $ i _{2}=-\frac{1}{3}A $ Hence $ i _{3}=i _{1}+i _{2}=-\frac{5}{6}A $
BETA
