that the batteries shown have thier positive terminal on the long bar end so the 100 V unit has its positive terminal on the bottom and the 50 V unit has its positive terminal on top
sum of all voltages around any loop must be zero
looking at the left hand loop and working around the loop in the direction of the assumed current i1.
-100 - i1(75 + 25 + 100) + i2(25) + i3(100) = 0
looking at the top central loop in the direction of the assumed current i2
-50 - i2(25) + i1(25) = 0
looking at the right hand loop and working around the loop in the direction of the assumed current i3.
50 - i3(200 + 100) + i1(100) = 0
if any current results as a negative number, we know that the positive current is in the opposite direction than assumed for that loop.
Answers & Comments
Verified answer
Assume
i1 is a clockwise current in the left side loop
i2 is a clockwise current in the top central loop
i3 is a clockwise current in the right side loop
that the batteries shown have thier positive terminal on the long bar end so the 100 V unit has its positive terminal on the bottom and the 50 V unit has its positive terminal on top
sum of all voltages around any loop must be zero
looking at the left hand loop and working around the loop in the direction of the assumed current i1.
-100 - i1(75 + 25 + 100) + i2(25) + i3(100) = 0
looking at the top central loop in the direction of the assumed current i2
-50 - i2(25) + i1(25) = 0
looking at the right hand loop and working around the loop in the direction of the assumed current i3.
50 - i3(200 + 100) + i1(100) = 0
if any current results as a negative number, we know that the positive current is in the opposite direction than assumed for that loop.