Solve this by mesh analysis.
Step one: identify the basic meshes.
#M=B_e -(N_e-1)#, #B_e = 3 #is number of essential branches, and
#N_e = 2#= number of essential nodes.
#M_e = 3-(2-1)= 2#
Here basic meshes are two. #
step two: assign a current to each mesh, say,# i_1 and i_2#
step three:apply KVL around each loop to get an equation in terms of loop current.
Consider loop 1:
#10i_1 +16+8(i_1-i_2) - 4=0#
#10i_1 +16+8i_1- 8i_2 - 4=0#
#18i_1 - 8i_2 + 12 =0#----divide by 2
#9i_1 -4i_2 =-6# ----------------------equation (1)
Consider loop2:
#4+8(i_2-i_1)+10+2i_2=0#
#4+8i_2- 8i_1+10+2i_2=0#
#10i_2- 8i_1+14 =0# ------divide by 2 and rearrange,
#-4i_1 + 5i_2 = -7# ------------------equation (2)
Solving equations(1) and(2) by solving linear equations method,
(1) #xx5# + (2) #xx 4# gives:
#45i_1-cancel(20i_2) -16i_1 +cancel(20i_2) = -30 -28#
#29i_1 =-58#
#=> i_1= -2# ---negative indicates assumed direction is reverse.
Substitute this value in equation (2) [ it can be substituted any equation]:
#-4i_1 + 5i_2 = -7#
#=> -4(-2) +5i_2 =-7#
#=> 5i_2 = -7-8#
#=> i_2 = -3# --negative indicates assumed direction is reverse.
Now let us consider the current through #8Omega# resistor as #i_0#
So #i_0 + i_2= i_1#-------according to KCL
#=> i_0 = i_1- i_2#
#=> i_0 = -2 -(-3) = -2+3 = 1 #
#therefore i_0 = 1 amp# ------ current through #8Omega# resistor.
#therefore i_ 1 = 2 amp # --------current through #10Omega# resistor, but the direction is opposite to that we have assumed.
#therefore i_2 = 3 amp# -----current through #2Omega# resistor, but the direction is opposite to that we have assumed.
Cross check :
Apply KVL to any one mesh with derived currents:
Mesh 1 :
#+16V +8 i_0 - 4V + 10i_1 #---- must sum upto #0#
#=> 16 +8 xx 1 - 4 +10 xx (-2) = 16+8 -4-20#
#=> 24 -24 =0# ------ hence values we achieved are correct.
Mesh 2 :
#+4V- 8i_0 +10V +2i_2# ---- must sum upto #0#
#4 - 8 xx (1) +10 + 2 xx (-3)#
#=> 4- 8 + 10 -6#
#14-14 =0# -------hence values we achieved are correct.