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: $I_s$, $R_1$, $R_3$, $R_2$ form a dangling chain that only reaches the rest of the circuit at the node where the ideal source $V_{s1}$ sits directly in parallel to ground; from there $R_4$ and $C_4$ also shunt to ground at that same node; $V_{s2}$ is in series in the top rail beyond it; $R_5$ and $R_{LOAD}$ are both in parallel between the far node and a second ground rail, reached from the first ground rail only through the series resistor $R_6$.

Find. $R_{th}$, $V_{th}$ at the load terminals; $R_{L,opt}$ and $P_{max}$; and $P_{R_L}$ for $R_L=72\,\Omega$.

Approach. The ideal source $V_{s1}$ is wired directly in parallel to ground, so it pins that node's voltage absolutely regardless of any current drawn through it — which makes $I_s$, $R_1$, $R_2$, $R_3$ (upstream of it) and $R_4$, $C_4$ (shunting the same fixed node) all electrically irrelevant to the load terminals. The only elements that matter are $V_{s1}$, $V_{s2}$ (in series, both ideal), and the $R_5$–$R_6$ path back to the (first) ground rail. In DC steady state $C_4$ carries no current and drops out entirely.

1. a) Thévenin resistance. Deactivate all independent sources: $I_s\to$ open (stranding the $R_1$-$R_2$-$R_3$ chain completely), $V_{s1}\to$ short (collapsing its node to ground — taking $R_4$, $C_4$ down with it, contributing nothing), $V_{s2}\to$ short (so the far node collapses to the SAME ground too). Looking into the load terminals, $R_5$ and $R_6$ are then both wired directly between the same two terminals. $$R_{th}=R_5\parallel R_6=\frac{(40\text{k}\Omega)(10\text{k}\Omega)}{40\text{k}\Omega+10\text{k}\Omega}=\boxed{8\,000\ \Omega}$$
2. b) Thévenin voltage. With the load open, no current can flow into it, so the only closed path is the loop formed by $V_{s1}$, $V_{s2}$, $R_5$, and $R_6$ in series (both sources add, since $V_{s2}$'s $+$ terminal faces the $R_5$ side). $$I_{loop}=\frac{V_{s1}+V_{s2}}{R_5+R_6}=\frac{10+40}{40\,000+10\,000}=1.000\text{ mA}\ ,\qquad V_{th}=I_{loop}R_5=(1.0\text{ mA})(40\text{k}\Omega)=\boxed{40\text{ V}}$$
3. c) Maximum power transfer. Maximum power is delivered when $R_L=R_{th}$. $$R_{L,opt}=R_{th}=\boxed{8\,000\ \Omega}\ ,\qquad P_{max}=\frac{V_{th}^2}{4R_{th}}=\frac{40^2}{4(8\,000)}=\boxed{0.0500\text{ W}}=50.0\text{ mW}$$
4. d) Power at $R_L=72\,\Omega$. With $R_L\ll R_{th}$, most of the Thévenin voltage drops across $R_{th}$ and only a small fraction reaches the load. $$P_{R_L}=\frac{V_{th}^2R_L}{(R_{th}+R_L)^2}=\frac{40^2(72)}{(8\,000+72)^2}=\boxed{1.768\text{ mW}}$$