Question

The value of the current $I_{1}, \, I_{2}$ and $I_{3}$ flowing through the circuit given below is

• A. $I_{1}=-3A, \, I_{2}=2A, \, I_{3}=-1A$
• B. $I_{1}=2A, \, I_{2}=-3A, \, I_{3}=-1A \, $
• C. $I_{1}=3A, \, I_{2}=-1A, \, I_{3}=-2A$
• D. $I_{1}=1A, \, I_{2}=-3A, \, I_{3}=-2A$

Answer 1

Answer: (B)

Applying Kirchoff's law
$10-6I_{B}+14-4I_{A}=0$ ...(loop 1)
$24=6I_{B}+4I_{A}$
$12=3I_{B}+2I_{A}$ .....(i)
$I_{A}=I_{B}+I_{C}$ .....(junction)
$10-6I_{B}+2I_{C}=0$ .....(loop 2)
$10=6I_{B}-2I_{C}$
$10=6I_{B}-2\left(I_{A} - I_{B}\right)$ .....(ii)
On solving Equations (i) and (ii), we get
$I_{B}=2A, \, I_{A}=3A, \, I_{C}=1A$
Now as per given circuit,
$I_{1}=I_{B}=2A, \, I_{2}=-I_{A}=-3A$
$I_{3}=-I_{C}= \, -1A$