07 · Mixed

Problems that test understanding rather than recall

Calculation, design, diagnosis and model-limit problems with solution paths.

60–120 minMixed

In this lesson

Choose the method without being told which theorem step to apply.
Translate loading specifications into impedance requirements.
Diagnose source–load and measurement discrepancies from symptoms.
Make nodal analysis, superposition and source transformation converge on the same Thévenin/Norton one-port.
Recognize when a fixed linear equivalent is no longer sufficient.

Starting question

Can you still use Thévenin correctly when the exercise no longer tells you which method to use?

This chapter is deliberately problem-driven. The target is not faster substitution; it is model selection, independent verification, dimensional checks and knowing when a linear one-port model is no longer enough. DC-05 has already established independent/dependent sources, superposition and source transformation; here those tools must agree with the Thévenin/Norton port model rather than live as separate recipes.

choose a modelsolvecheck independentlydiagnose the limit

07.1A reliable problem-solving sequence

Define the port. Mark the two terminals, voltage reference and positive current direction.
Ask what must be preserved. For a linear one-port, the external load only needs the terminal V–I law.
Find the intercept and slope. Usually V_TH from the open-circuit network and R_TH from source deactivation or a test source.
Reconnect the load only after the equivalent is known.
Check by another route. Open/short, direct nodal solution, limiting case or measurement consistency.

07.2Problem A — verify the same equivalent two ways

A 15 V ideal source feeds 4.7 kΩ in series with a node that is connected to ground through 10 kΩ. Find the Thévenin equivalent seen from that node to ground.

V_OC = 15 · 10/(4.7 + 10) = 10.204 V

R_TH = 4.7 kΩ ∥ 10 kΩ = 3.197 kΩ

Now verify with the short-circuit current:

I_SC = 15 V / 4.7 kΩ = 3.191 mA

V_OC/I_SC = 10.204 V / 3.191 mA = 3.197 kΩ

Retrieval check — what should happen if the 10 kΩ resistor is made extremely large?

The open-circuit voltage tends toward 15 V and RTH tends toward 4.7 kΩ. The network approaches a 15 V source with a 4.7 kΩ series resistance.

07.3Problem B — turn an accuracy requirement into an instrument specification

A sensor can be modeled by V_TH and R_TH = 47 kΩ. You want the measuring instrument to introduce less than 0.25% DC loading error. What minimum input resistance is required?

ε = R_TH/(R_TH + R_in)

R_in ≥ (1 − ε)/ε · R_TH

R_in ≥ 0.9975/0.0025 · 47 kΩ = 18.75 MΩ

A 10 MΩ DMM is therefore not sufficient for the stated 0.25% requirement.

At 100 kHz, suppose the input also has 50 pF to ground. Its capacitive reactance magnitude is

|X_C| = 1/(2π · 100 kHz · 50 pF) = 31.8 kΩ

The input is no longer well represented by a 10 MΩ resistor. This is exactly the point where the DC resistance model must give way to Z_in(ω).

07.4Problem C — infer source resistance from a voltage droop

A source measures 5.00 V open circuit and 4.55 V when connected to 10 kΩ. Assume a linear resistive one-port.

Figure 7.1 — A measured loading discrepancy as a circuit model

The 5.00 V open-circuit reading fixes VTH. The 4.55 V loaded reading then reveals RTH.

verified topology

Voltage arrow head = positive reference; IL is defined from source toward load.

4.55/5.00 = 10 kΩ/(R_TH + 10 kΩ)

R_TH = 10 kΩ · (5.00/4.55 − 1) = 989 Ω

If the same 10 kΩ load must produce less than 1% droop, chapter 05 gives

R_TH ≤ 0.01/0.99 · 10 kΩ = 101 Ω

07.5Problem D — recognize when one fixed Thévenin model is insufficient

Suppose a measured one-port has a visibly curved V–I characteristic over the operating range. Can one constant V_TH and one constant R_TH reproduce the whole curve?

V = V_TH − R_TH I

The right-hand side is an affine line. A curved characteristic therefore cannot be represented globally by one fixed pair. A useful diagnostic is the local slope:

R_local = −dV/dI

If that slope varies substantially with operating point, the fixed linear model has reached its limit. Later courses will introduce nonlinear devices and local linearization; no diode or transistor model is required here.

07.6Problem E — one multi-source network, three routes to the same one-port

A 12 V source reaches node A through 3 kΩ, a 6 V source reaches the same node through 6 kΩ, and 2 kΩ connects A to ground. The output port is A–ground. This is the synthesis network from DC-05.

Route 1 — direct nodal analysis. With the port open:

(V−12 V)/(3 kΩ)+(V−6 V)/(6 kΩ)+V/(2 kΩ)=0

V_OC = 5 V

Route 2 — superposition. The 12 V source contributes 4 V at the port and the 6 V source contributes 1 V:

V_OC = 4 V + 1 V = 5 V

Route 3 — source transformation. Transform the two source branches:

12 V / 3 kΩ → 4 mA ∥ 3 kΩ

6 V / 6 kΩ → 1 mA ∥ 6 kΩ

I_N = 5 mA

R_N = 3 kΩ ∥ 6 kΩ ∥ 2 kΩ = 1 kΩ

Therefore the whole original network is equivalent at the port to 5 mA in parallel with 1 kΩ, or 5 V in series with 1 kΩ:

V_TH = 5 V

R_TH = 1 kΩ

07.7Independent challenge — recover VTH and RTH from two loaded measurements

A linear source gives 4.000 V across 1.00 kΩ and 5.333 V across 4.00 kΩ. Recover the Thévenin parameters without ever opening or shorting the source.

Hint 1

Write VL = VTH·RL/(RTH+RL) once for each load.

Hint 2

For RL=1 kΩ: 4(RTH+1k)=VTH·1k. For RL=4 kΩ: 5.333(RTH+4k)=VTH·4k.

Solution

R_TH = 0.500 kΩ

V_TH = 6.00 V

Check: 6·1/(0.5+1)=4.00 V and 6·4/(0.5+4)=5.333 V.

This technique is useful when an open circuit or short circuit is undesirable but controlled loads are available.

07.8Before you leave the problem set

Can you explain which circuit state produced each equation?
Can you verify the equivalent by a second method?
Did you check units and limiting cases?
Did you accidentally assume a nonlinear device model that has not been introduced?
Could a real measurement be unsafe even if the mathematical short-circuit quantity is well defined?

Continue to chapter 08 — references and extensions