Question 2 of 7: Thévenin Equivalent and Maximum Power Transfer

Question 2: Thévenin Equivalent and Maximum Power Transfer (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The network of Figure 2, with the ideal current source $I_s$ wired in series in the top rail (between the $R_1$–$R_4$ block and $R_5$–$R_6$–$R_7$ block); $R_L$ is the load across $R_7$.

Find. $V_{th}$, $R_{th}$, $R_{L,opt}$, $P_{max}$, and $P_{R_L}$ for $R_L=100\,\Omega$.

Approach. Because $I_s$ sits in series in the top rail, it is the only connection between the left block ($V_{s1},V_{s2},R_1$–$R_4$) and the right block ($R_5,R_6,R_7,R_L$). Opening $I_s$ for the resistance calculation therefore completely isolates the left block; only $R_5,R_6,R_7$ set $R_{th}$. For $V_{th}$, all sources stay active and $I_s$ injects its full 30 A into the right block.

1. b) Thévenin resistance. Deactivate $V_{s1},V_{s2}$ (short) and $I_s$ (open). Opening $I_s$ strands the left block entirely (it has no other path to the right block), so it contributes nothing. Looking into the $R_L$ terminals: $R_7$ is directly in parallel with the series combination $R_5+R_6$ (the top rail node beyond the open $I_s$ connects to $R_5$ down to ground and, via $R_6$, to $R_7$). $$R_{th} = R_7 \parallel (R_5+R_6) = 80 \parallel (100+20) = \frac{80\times120}{200} = \boxed{48\,\Omega}$$
2. a) Thévenin voltage. With $R_L$ removed, all 30 A of $I_s$ must flow out of node N3 (after $I_s$) through $R_5$ (to ground) and $R_6$ (to node N4), and then out of N4 through $R_7$ (to ground). Two KCL equations in the unknown node voltages $V_3,V_4$ (ground at the bottom rail): $$30 = \frac{V_3}{R_5}+\frac{V_3-V_4}{R_6}\ ,\qquad \frac{V_3-V_4}{R_6}=\frac{V_4}{R_7}$$ Solving simultaneously gives $V_3=1500$ V and $$V_{th}=V_4=\boxed{1200\text{ V}}$$ (The left block only fixes internal node voltages $V_{s1}=20$ V, $V_{s1}-V_{s2}=15$ V there but never reaches N3/N4, confirming it is a distractor for this particular load.)
3. c) Maximum power transfer. Maximum power is delivered when $R_L=R_{th}$. $$R_{L,opt}=R_{th}=\boxed{48\,\Omega}\ ,\qquad P_{max}=\frac{V_{th}^2}{4R_{th}}=\frac{1200^2}{4(48)}=\boxed{7500\text{ W}}$$
4. d) Power at $R_L=100\,\Omega$. $$I_{R_L}=\frac{V_{th}}{R_{th}+R_L}=\frac{1200}{48+100}=8.108\text{ A}\ ,\qquad P_{R_L}=I_{R_L}^2R_L=(8.108)^2(100)=\boxed{6574\text{ W}}$$