R5, tell me.

As shown in the figure, according to KCL law, i3=i 1-i2 and i5 = I2-I4;

For loop 1, KVL law gives 20=R 1i 1+R3i3.

For loop 2, KVL's law is i3R3=i2R5+i5R4.

According to KVL's law, loop 3 has i5R4=i4R2+40.

Solve? I 1 =- 1.06 mA

I2 =-2.33 mA

I3 = 1.27 mA

I4 =-3.33 mA

i5= 1mA

You'd better do the calculation again. I think I'm right. A negative sign in the result indicates that the current direction is opposite to that of the logo.