2.3 1 solution: Let the currents in the three grids from left to right be I 1, I2 and I3 respectively, all clockwise. According to KVL, the voltage equations of the three grids are as follows:

Grid1:(2+4+6) I1-6i2 = 32-48+16, simplified as: 2I 1=I2.

Grid 2: (6+3+8)I2-6I 1-8I3=48, which is simplified as: 17I2-6I 1-8I3=48.

Grid 3: (8+5+3)I3-8I2=0, simplified as: 2I3=I2.

Solving equations, I 1=2.4, I2=4.8, I3=2.4.

So: I = i3 = 2.4a ..

2.32 Solution: Let the currents in the left, upper and right grids be I 1, I2 and I3 respectively, clockwise. It is obvious that I2 =1a. According to KVL, the battery voltage equation is obtained:

Grid left: (5+5+30)I 1-5I2-30I3=30, simplified as: 8I 1-6I3=7.

Grid right: (30+20)I3-20I2-30I 1=-5, simplified as: 10I3-6I 1=3.

Solve equations, I 1=2, I3= 1.5.

According to KCl: I = I1-i3 = 2-1.5 = 0.5 (a).