Problem 5.3-6

The two sources of voltage are transformed into sources of current.

The new currents are i₁ = 24/4 = 6A and i₂ = 12/12 = 1A

So the two parallel transistors give 1/R_e = 1/12 + 1/4 = 4/12 = 1/3 so then R_e = 3 Ω

We can apply the node’s law for the middle current source and we get 1 + i = i₃, so as i = 2A is the aim, we get i₃ = 3A.

Finally we have (on the right hand side), V = 5 + 2*RL and on the left side (remember we have taken the equivalent resistor for the two resistors in the middle) V = 3 (i₁ – i₃) = 3*(6 – 3) = 9 V

Problem 5.3-6

Thus finally we get 5 + 2*RL = 9 V thus RL = 2Ω

Thus finally we get 5 + 2*R_L = 9 V thus R_L = 2Ω