Maximum Power Transfer

The source is already in Thévenin form, so no reduction is needed:

The matched load and the power it receives:

Check it directly. With \(R_L = 6\ \Omega\) the loop current is \(12/12 = 1\ \text{A}\), so \(P_L = 1^2 \times 6 = 6\ \text{W}\;\checkmark\)

Confirm no other load does better by tabulating the power curve \(P_L = V_{TH}^2R_L/(R_{TH}+R_L)^2\):

Why both ends give zero. A short carries the largest current but drops no voltage; an open holds the largest voltage but carries no current. Power is the product, so it vanishes at both extremes and peaks between them.

Notice the symmetry: \(R_L = 3\) and \(R_L = 12\) both give 5.33 W, as do \(4\) and \(9\). Loads related by \(R_LR_L' = R_{TH}^2\) always receive equal power — a fact Problem 5 puts to use.

With the network in Thévenin form, the load current and power are

Differentiate with respect to \(R_L\), treating \(V_{TH}\) and \(R_{TH}\) as constants. By the quotient rule:

The stationary point. The denominator is positive for positive resistances, so the derivative vanishes only when the numerator does:

It is a maximum, and the sign of the derivative shows it without any second differentiation:

Rising then falling — a genuine maximum, and the only one, since the numerator is linear in \(R_L\) and vanishes just once.

The value at the peak:

Every assumption used:

Open-circuit voltage. With no load current, all 3 A flows through the 4 Ω:

Thévenin resistance. Deactivating the current source opens it, leaving only the 4 Ω:

The power:

The Norton form gives the same answer more directly. The network is already \(I_N = 3\ \text{A}\) in parallel with \(R_N = 4\ \Omega\), so:

No conversion to Thévenin form was necessary — use whichever form the network is already in.

Check directly. With \(R_L = 4\ \Omega\), the 3 A divides equally between two 4 Ω branches, so the load takes 1.5 A and \(P_L = 1.5^2 \times 4 = 9\ \text{W}\;\checkmark\)

With \(R_L = R_{TH}\) the loop current is

The load power and the total power produced:

The general case. Writing \(k = R_L/R_{TH}\), the load and internal resistances share the same current, so the power splits in proportion to resistance:

Tabulating both quantities against \(k\):

The two quantities peak in different places, and neither optimum is a compromise for the other:

Efficiency rises monotonically with \(k\) and never has an interior maximum — you can always improve it by making the load larger, at the cost of the power delivered.

The symmetry. Write \(k = R_L/R_{TH}\) and normalise the power:

Replacing \(k\) by \(1/k\):

Identical — so \(k\) and \(1/k\) give the same power, which is the statement \(R_LR_L' = R_{TH}^2\). The curve is symmetric on a logarithmic axis, not a linear one.

The 90% range. Setting the normalised power to 0.9:

Note \(0.519 \times 1.925 = 1.000\), as the symmetry requires.

So anything within a factor of about two either way delivers 90% or more:

For Problem 1's 6 Ω source that is roughly 3 Ω to 12 Ω — and even a factor of four out, \(k = 4\) or \(0.25\), still gives 64%.

What this means practically. Precise matching is rarely worth pursuing. Real internal resistances drift with temperature, ageing and operating point by more than a few per cent, and the penalty for that drift is second-order: near the peak, \(dP/dR_L = 0\), so a small mismatch costs power only to second order in the error.

From Set 9, \(V_{TH} = V_{oc}\) and \(R_{TH} = V_{oc}/I_{sc}\). Substituting into the standard formula:

The geometric reading. Set 10, Problem 2 showed the terminal characteristic is the line joining \((V_{oc}, 0)\) to \((0, I_{sc})\). The power delivered at any operating point is the area of the rectangle \(V \times I\) under that point, and the largest rectangle inscribed under a straight line has exactly a quarter of the area of the triangle's bounding box — at the midpoint of the line:

The matched condition is simply "operate at half the open-circuit voltage and half the short-circuit current".

Applying it to the measured source:

Cross-check: \(V_{TH}^2/4R_{TH} = 400/20 = 20\ \text{W}\;\checkmark\)

Why it is the useful form in practice. Both \(V_{oc}\) and \(I_{sc}\) are directly measurable at the terminals, with no knowledge of what is inside. The formula converts two meter readings into the maximum power available and the load needed to extract it.

With the caution of Set 9, Problem 18. Measuring \(I_{sc}\) directly may damage the source, so it is usually inferred: measure \(V_{oc}\), then find the load that halves the terminal voltage — that load is \(R_{TH}\), and the source is then delivering \(P_{\max}\) by construction.

Open-circuit voltage. With \(R\) removed the 6 V branch carries no current, so the left part is a two-mesh network. Writing the mesh equations by inspection:

Mesh 1 totals \(1+2 = 3\ \Omega\); mesh 2 totals \(2+5+1 = 8\ \Omega\); they share the 2 Ω.

From the second equation \(i_1 = 4i_2\); substituting into the first:

That current flows in the right-hand 1 Ω, so walking from the reference up through it and through the 6 V source to the open terminal:

Thévenin resistance. Short both sources and look back from the load terminal. The 6 V short brings us to node \(S\), from which there are two paths to the reference — the 1 Ω directly, or the 5 Ω followed by \(2 \parallel 1\):

The matched load and its power:

Note the size of the answer. A 6.4 V source delivering 12 W implies a load current of \(6.4/1.7 = 3.76\ \text{A}\) — large, but consistent, because \(R_{TH}\) is well under an ohm. Low internal resistance means high available power, which is exactly what \(V_{TH}^2/4R_{TH}\) says.

The input stage is isolated — no current is drawn from it by the dependent source — so it is a plain divider:

Because \(V_0\) does not depend on anything in the output stage, the dependent source behaves as a fixed 200 V source — which is what makes deactivation-free analysis unnecessary here.

Open-circuit voltage. Remove \(R\); the output stage is then a divider of 10 kΩ and 40 kΩ:

Short-circuit current. Shorting the load terminals shorts out the 40 kΩ entirely, leaving the 200 V across the 10 kΩ alone:

Hence

Consistent with deactivation, which would give \(10\text{k} \parallel 40\text{k} = 8\ \text{k}\Omega\) — legitimate here only because the dependent source is effectively independent.

The power:

Or from Problem 6: \(V_{oc}I_{sc}/4 = 160 \times 0.02/4 = 0.8\ \text{W}\;\checkmark\)

The controlling variable depends on the circuit's own response, so deactivation is not permitted. Deactivate only the independent 100 V source — short it — and drive the load terminals with a test source \(V_1\) delivering \(I_1\).

Trace the test current. Of the \(I_1\) entering, the dependent source diverts \(V_x/4\), so the series 4 Ω carries the remainder:

KVL round the test loop, from the test terminal through the controlling element and the series resistor:

The controlling voltage cancels identically. That is not a coincidence of the numbers — the dependent source's transconductance is exactly \(1/4\ \text{S}\), matching the 4 Ω it works against.

Hence

Note what did not matter. The 100 V source plays no part in \(R_{TH}\), and neither does the value of \(V_x\). Only \(V_{TH}\) depends on the independent source — so if the question had asked for the power as well, a second calculation would be needed.

Why the test-source method was required. Deactivating the dependent source would leave the two 4 Ω resistors in some series–parallel arrangement and give the wrong answer; and \(V_{oc}/I_{sc}\) would need two full circuit solutions where one test loop sufficed.

Set 9, Problem 7 established the Thévenin equivalent at \(c\!-\!d\):

The matched load and the power available:

Compare with the actual 18 Ω arm. That load carries 83.3 mA (established four separate ways in earlier sets), so

Reading the result through Problem 5. Here \(k = 18/7.6 = 2.37\), and the formula gives

A load 2.4 times the matched value still collects five-sixths of the available power — the flat peak again.

A practical note on bridges. A measurement bridge is normally used near balance, where \(V_{TH} \to 0\) and the available power vanishes. Matching the detector is then pointless for power reasons; what matters is the detector's sensitivity, and a high-impedance detector loads the bridge less and disturbs the balance point less.

The load power has the same algebraic form, but now \(R_L\) is the constant and \(R_{TH}\) the variable:

Differentiate with respect to \(R_{TH}\) this time:

Strictly negative — there is no stationary point at all. The power decreases monotonically as \(R_{TH}\) grows.

So the optimum is at the boundary:

Tabulating the difference:

The "matched" design delivers a quarter of what the ideal source would, and wastes half of that again internally.

Why the answers differ. Both problems maximise the same expression, but with respect to different variables. Increasing \(R_L\) has two competing effects — less current, but a larger share of it usefully employed — and the competition produces an interior optimum. Increasing \(R_{TH}\) has only one effect: less current, all of it wasted more. Nothing competes, so nothing peaks.

The unconstrained optimum is \(R_L = 6\ \Omega\), which lies outside the permitted range. Problem 2 showed the derivative

is negative for every \(R_L > R_{TH}\). On the whole interval \([8, 20]\) the power is therefore strictly decreasing.

So the best is the smallest permitted value — the boundary nearest the unconstrained optimum:

Comparing with the unreachable ideal:

A 33% overshoot in resistance costs 2% of the power — Problem 5's flat peak, doing useful work. Here \(k = 8/6 = 1.33\) and \(4k/(1+k)^2 = 0.980\;\checkmark\)

The general rule for a constrained range \([R_{\min}, R_{\max}]\):

Because the function has a single peak and no other stationary points, the constrained optimum is always either the peak itself or whichever endpoint is closer to it.

A caution. "Closer" means closer on a logarithmic scale, by Problem 5's symmetry. If the range were \([2, 8]\) with \(R_{TH} = 6\), the peak lies inside and the answer is 6 Ω; but for a range like \([1, 3]\) the endpoint 3 wins over 1 because \(3/6 = 0.5\) beats \(1/6 = 0.167\).

The naive application gives \(R_L = R_{TH} = -4\ \Omega\) — not a resistor that can be bought. Worse, substituting into the power formula:

A negative "maximum" power, which is meaningless.

Look at the actual power curve for a positive load with \(R_{TH} = -4\ \Omega\) and, say, \(V_{TH} = 12\ \text{V}\):

The power is unbounded as \(R_L \to |R_{TH}|\). There is no maximum to find.

Where the derivation broke. Problem 2's derivative

assumed the denominator was positive and non-zero. With \(R_{TH} = -4\) it vanishes at \(R_L = 4\), and the "stationary point" at \(R_L = R_{TH} = -4\) sits on the far side of that pole — a minimum of a different branch of the function, not a maximum of anything physical.

What actually happens. The total loop resistance \(R_{TH}+R_L\) reaches zero at \(R_L = 4\ \Omega\), so the circuit has no net resistance to limit the current. A real circuit does not deliver infinite power — it oscillates, saturates, or destroys itself. The linear model has been pushed outside its range of validity.

The useful reading. A negative \(R_{TH}\) means the network supplies energy in response to load current rather than absorbing it, which is the condition for instability rather than for power transfer. This is exactly how oscillators are designed — a negative resistance is arranged to cancel a circuit's positive losses, and the resulting growth is limited by a non-linearity the linear model does not contain.

All three follow from the loop current \(I = V_{TH}/(R_{TH}+R_L)\):

(b) and (c) are both monotonically decreasing in \(R_L\), since each depends on \(R_L\) only through \(I\), which falls as \(R_L\) rises. Both are therefore maximised at \(R_L = 0\) — a dead short:

Identical, of course — with the load shorted, everything the source produces is burnt internally.

(a) alone has an interior optimum, at \(R_L = 6\ \Omega\), giving 6 W. Tabulating all three:

Every row satisfies \(P_S = P_L + P_{R_{TH}}\), as conservation demands.

Why they differ. Only the load power is a product of two competing factors. The source's output and the internal dissipation both track the current alone, and current is maximised by removing all resistance — so neither has anything to trade off against.

The practical consequence. "Maximum power" is ambiguous until you say whose. A question about a source's rating concerns \(P_S\) and is answered by the short-circuit condition; a question about a heat sink concerns \(P_{R_{TH}}\) and is answered the same way; only a question about what the load receives gives the matched condition.

Here the load is fixed and the source resistance is adjustable — Problem 11's situation, but with a subtlety: \(R_{TH}\) can be tuned continuously by choosing \(k\).

By Problem 11, the answer is not to match. With \(R_L\) fixed, the power falls monotonically with \(R_{TH}\), so the best choice is the smallest achievable \(R_{TH}\) — which means the most negative permissible \(k\):

Setting \(k = 0.9\) to "match" the 40 Ω load would be a serious error — it gives \(P_L = V_{TH}^2/160\), against \(V_{TH}^2/40\) for a near-zero \(R_{TH}\): four times worse.

The range of validity. The theorem requires \(R_{TH} > 0\) (Problem 13), so

The three regimes are genuinely different circuits. At \(k = 1\) the dependent source exactly replaces the current the 4 Ω would draw, so the terminals present an infinite resistance — the active-load principle of Problem 9 taken to its limit. Beyond that the source over-compensates and the network begins to supply energy in response to load current.

A caution about "maximum". Even within \(k < 1\), driving \(k\) very negative is not free: the dependent source must supply the extra current, and a real transistor or amplifier providing it has its own limits. The linear model says "smaller is always better" because it contains no such limit.

The operating point. At 400 kV delivering 500 MW, the line current and effective load are

The efficiency, from Problem 4:

The line loses \(I^2R_{TH} = 1250^2 \times 5 = 7.8\ \text{MW}\) — 1.5% of the total.

Now "match" it. Setting \(R_L = 5\ \Omega\) with the same source:

Sixteen times the delivered power — but half of it, 4.1 GW, would be dissipated in the line itself, against 7.8 MW at present. The line would vaporise.

The design rule. Which optimum to pursue depends entirely on whether the source energy is scarce:

The unifying statement. Match only when the source is fixed by physics and its wasted half costs nothing. Whenever the source is designed rather than given, Problem 11 applies and the answer is \(R_{TH}\) as small as possible.

The pad. A shunt \(R_p\) at the source side and a series \(R_s\) to the load, chosen so both ports look right. The standard minimum-loss values are

Verify both ports:

The source is now matched, so it delivers its full available power into the pad:

But most of it never reaches the load. The 6 V divides between the shunt and the series path:

Compare with connecting the 50 Ω directly, with no pad at all:

The pad is twelve times worse than doing nothing. Its 14.96 dB insertion loss more than cancels the benefit of matching.

Why this happens. A resistive network can transform an impedance only by dissipating. The pad presents the right resistance to each port, but it does so by burning the difference — and the power it burns exceeds the power the mismatch was costing.

The common feature of every genuine matching application is that the source cannot be improved. Its internal resistance is fixed by physics, not by a design decision, so Problem 11's escape route is closed.

The cases:

The second common feature: in every case the wasted half costs nothing that matters. Sunlight, radio waves and ambient heat arrive free; there is no fuel bill attached to the 50% dissipated internally, so efficiency in the Problem 4 sense is simply not the figure of merit.

The transmission-line case is different in kind. There, matching prevents reflections rather than maximising power — a mismatched line returns part of the signal towards the source, producing standing waves and, at high power, breakdown. The condition happens to be the same, but the reason is not.

A note on solar cells. These are matched dynamically, because the optimum load moves with irradiance and temperature. A maximum power point tracker continuously adjusts the effective load to sit at the peak — a direct, commercial application of Problem 5's flat peak, since the tracker need only get close.

The equivalent source. With \(s\) cells in series per string and \(m\) such strings in parallel:

Both scale with the arrangement, which is what makes this different from Problem 11 — here the source's voltage changes too, so a genuine optimum can exist.

Enumerate the six possibilities with \(sm = 12\):

The best arrangement is \(6 \times 2\), and it is precisely the one satisfying

Why matching applies here when Problem 11 said it should not. The difference is that \(V_{TH}\) is not held constant: rearranging trades voltage against resistance, so raising \(R_{TH}\) also raises \(V_{TH}\). Writing \(m = 12/s\) and hence \(R_{TH} = s^2/12\):

which does have an interior maximum, at \(s = 6\). The competing effects are back.

Note the symmetry. The \(3 \times 4\) and \(12 \times 1\) arrangements both give 7.68 W, with \(R_{TH} = 0.75\) and \(12\ \Omega\) — and \(0.75 \times 12 = 9 = R_L^2\), exactly Problem 5's reciprocal pairing.

What it says. Given a linear source with fixed \(V_{TH}\) and fixed \(R_{TH} > 0\), and a load free to take any positive value, the load receives most power when \(R_L = R_{TH}\), and that power is \(V_{TH}^2/4R_{TH} = V_{oc}I_{sc}/4\).

The four assumptions, each with a problem showing what happens without it:

The three misapplications worth naming explicitly:

Its place in Part 1. Sets 9 to 12 established that a linear two-terminal network is a straight line in the \((V,I)\) plane. This set asks where on that line to operate, and answers: at the midpoint, \((V_{oc}/2,\ I_{sc}/2)\). That is the whole of it — the rest is knowing when the question is the right one to ask.

What carries forward. In AC circuits the condition becomes \(Z_L = Z_{TH}^{*}\) — the complex conjugate — because the reactance must be cancelled as well as the resistance matched (Set 26). The derivation is the same two lines with \(|Z|\) in place of \(R\), and every caution in this set survives the translation unchanged.