Question

The value of the currents $I_1, \,I_2,$ 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. $I_1 = 1A, I_2 = -3A, I_3 = -2A$
• E. $I_1 = 2A, I_2 = -1 A, I_3 = -3A$

Answer 1

Answer: (B)

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