Question 2 of 7: Thevenin Equivalent and Maximum Power Transfer

Question 2: Thevenin 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 ladder network of Figure 2 feeds an R_LOAD at its right-hand terminals. R6 sits in the lower rail, splitting it into two ground segments; Vs1 is an ideal source directly across R4; Vs2 is an ideal source in series in the top rail, its "+" facing the load side.

Find. R_th and V_th at the RLOAD terminals, the load resistance and power at maximum power transfer, and the power delivered to a 2 kΩ load.

Approach. Vs1 is an ideal source directly across the node shared with R2, R4 — it fixes that node's potential outright, so Is, R1, R2, R3 (everything upstream of Vs1) cannot affect anything measured at the RLOAD terminals. That leaves only Vs1, Vs2, R5, R6 in play for both Rth and Vth.

1. Collapse the distractor branch. Call the node where R2, Vs1, and R4 meet "n2." Because Vs1 is an ideal source directly between n2 and the ground rail, $V_{n2}=V_{s1}= 10\text{ V}$ regardless of Is, R1, R2, R3 — those elements only set the (irrelevant) current circulating in their own loop back to n2. R4, sitting in parallel with an ideal source, is likewise irrelevant to any node voltage.
2. Propagate through Vs2. Vs2 sits in series in the top rail with its "+" terminal facing the load side (call that node "n4"), so $$ V_{n4} = V_{n2} + V_{s2} = 10+40 = 50\text{ V}, $$ again independent of everything upstream of Vs1.
3. Thevenin resistance (part a). Deactivate all independent sources: Is→open (strands R1 as a dead-end, irrelevant), Vs1→short (ties n2 to ground, shorting out R4), Vs2→short (ties n4 directly to n2, i.e. to ground). With n4 tied to ground through zero resistance, R5 and R6 are now both simply connected between n4(=ground) and G2 — i.e. R5 and R6 are in parallel as seen from the load terminals: $$ R_{th} = R_5 \parallel R_6 = \frac{(40\text{k})(10\text{k})}{40\text{k}+10\text{k}} = \boxed{8\text{ k}\Omega}. $$
4. Thevenin voltage (part b). With RLOAD removed (open), the only closed path from n4 is through R5 then R6 back to ground, carrying a single loop current $$ I_x = \frac{V_{n4}}{R_5+R_6} = \frac{50}{50\text{k}} = 1\text{ mA}, \qquad V_{G2} = I_xR_6 = 10\text{ V}. $$ $$ V_{th} = V_{n4} - V_{G2} = 50-10 = \boxed{40\text{ V}}. $$
5. Maximum power transfer (part c). The load resistance for maximum power equals the source's Thevenin resistance: $$ R_{L,opt} = R_{th} = 8\text{ k}\Omega, \qquad P_{max} = \frac{V_{th}^2}{4R_{th}} = \frac{40^2}{4(8000)} = \boxed{0.050\text{ W} = 50\text{ mW}}. $$
6. Power at RL = 2 kΩ (part d). $$ I_{RL} = \frac{V_{th}}{R_{th}+R_L} = \frac{40}{8000+2000} = 4\text{ mA}, \qquad P_{RL} = I_{RL}^2R_L = (0.004)^2(2000) = \boxed{0.032\text{ W} = 32\text{ mW}}. $$