Current Electricity Charging Discharging Of Capacitors Question 578
Question: The magnitude and direction of the current in the circuit shown will be
[CPMT 1986, 88]
Options:
A) $ \frac{7}{3} $ A from a to b through e
B) $ \frac{7}{3} $ A from b to a through e
C) 1A from b to a through e
D) 1A from a to b through e
Show Answer
Answer:
Correct Answer: D
Solution:
Since $ E _{1}(10,V)>E _{2}(4V) $ So current in the circuit will be clockwise. Applying Kirchoff’s voltage law $ -,1\times i+10-4-2\times i-3i=0 $
Therefore $ i=1A(a,\text{to }b\text{ via }e) $ \ Current $ =\frac{V}{R}=\frac{10-4}{6}=1.0,ampere $
BETA
