Use Thevenin equivalent circuits: Question bank — Hardware deep dive

Before we start, two words we lean on constantly:

True or false — justify

The equivalent is identical to the original circuit inside the box.

False — it only reproduces the terminal behaviour (

V

I

). Internal node voltages and internal power dissipation generally differ; only what the load "sees" is preserved.

V_{t h}

always equals the largest source voltage in the network.

False —

V_{t h}

is the open-circuit terminal voltage, which is what remains after internal resistive drops; it is usually smaller (e.g. 4 V from a 12 V source in the parent example).

A circuit made only of resistors and one battery has a Thevenin equivalent.

True — resistors and independent sources are linear elements, so the straight-line terminal law

V = V_{t h} - I R_{t h}

holds and the theorem applies.

A circuit containing a diode has a valid Thevenin equivalent.

False — a diode is non-linear, so the

V

–

I

relation is a curve, not a straight line; superposition fails and no single

(V_{t h}, R_{t h})

pair reproduces it.

If the network contains a dependent source, it still has a Thevenin equivalent.

True — dependent sources are linear (output proportional to a circuit variable), so the equivalent exists; you just cannot deactivate the dependent source when finding

R_{t h}

R_{t h}

can be negative.

True in principle — a network with dependent sources can present a negative terminal resistance (the line

V = V_{t h} - I R_{t h}

slopes upward). The test-source method will simply return a negative

V_{t es t} / I_{t es t}

Turning off a voltage source means removing it and leaving a gap.

False — an ideal voltage source at

0 V

is a short (a wire), not a gap. You remove current sources by leaving a gap; you short voltage sources.

The Thevenin resistance depends on which load you attach.

False —

R_{t h}

is a property of the network as seen from its terminals with sources deactivated; it is fixed before any load is chosen.

Spot the error

"To find

R_{t h}

with a dependent source, deactivate every source including the dependent one."

Wrong — a dependent source tracks a circuit variable, so zeroing it changes the network's behaviour. Leave dependent sources active and inject a test source:

R_{t h} = V_{t es t} / I_{t es t}

R_{1}

and

R_{2}

are in series in the original divider, so

R_{t h} = R_{1} + R_{2}

Wrong — you must first deactivate the source. Shorting the battery ties one end of

R_{1}

to ground, so from the terminals

R_{1}

and

R_{2}

run in parallel:

R_{t h} = R_{1} ∥ R_{2}

V_{t h}

is measured with the load connected, across the load."

Wrong —

V_{t h}

is the open-circuit voltage, so the load must be removed. With the load on, the terminal voltage is

V_{t h} - I R_{t h}

, not

V_{t h}

"Since

R_{t h} = V_{t h} / I_{sc}

, I can find

R_{t h}

without ever removing the source."

The relation is correct, but note

I_{sc}

is measured with sources active and terminals shorted — this is the alternative to deactivation, not a way to skip understanding it. Both routes give the same

R_{t h}

"A current source in the network means

R_{t h}

can't be computed by deactivation."

Wrong — an independent current source is deactivated by opening it (leaving a gap), and deactivation works fine. Only dependent sources block the deactivation method.

"The short-circuit current equals

V_{t h} / R_{L}

Wrong — short circuit means the terminals are joined directly (

R_{L} = 0

V = 0

), so

I_{sc} = V_{t h} / R_{t h}

. The load is not part of

I_{sc}

Why questions

Why must the network be linear for Thevenin to work?

Linearity guarantees a straight-line

V

–

I

relation, and only a straight line is fully described by two numbers (intercept

V_{t h}

, slope

- R_{t h}

). Superposition, the proof's engine, is also legal only for linear elements.

Why is the slope of the terminal line

- R_{t h}

rather than

+ R_{t h}

As you pull more current

I

out of the terminals, the internal resistance drops more voltage, so terminal voltage falls:

V = V_{t h} - I R_{t h}

. The minus sign is the falling voltage as load draws current.

Why does maximum power transfer occur at

R_{L} = R_{t h}

and not at

R_{L} = 0

R_{L} = \infty

R_{L} = 0

the voltage across the load is zero; at

R_{L} = \infty

the current is zero — both give zero power. The peak of

P_{L} = (V_{t h} / (R_{t h} + R_{L}))^{2} R_{L}

sits in between, exactly where

R_{L} = R_{t h}

Why can the same

R_{t h}

be found three different ways (deactivation, test source,

V_{t h} / I_{sc}

All three probe the same fixed slope of the terminal line. Deactivation reads the resistance directly; the test source forces it;

V_{t h} / I_{sc}

reads it from two points on the line. Different measurements of one property must agree.

Why does shorting a voltage source (not opening it) correctly model "turned off"?

"Off" means the source contributes zero voltage. A wire holds zero volts across itself, so replacing the source with a wire enforces exactly the

0 V

condition while still letting current pass through that branch.

Edge cases

What is

R_{t h}

for a single ideal voltage source alone across the terminals?

Zero — deactivating the source shorts the terminals, so looking in you see a wire,

R_{t h} = 0

. This is a "stiff" source: terminal voltage never sags with load.

What is

R_{t h}

for a single ideal current source alone across the terminals?

Infinite — deactivating opens the source, leaving no path, so

R_{t h} = \infty

. The Norton form is natural here; the Thevenin voltage is undefined (infinite) for a bare ideal current source.

V_{t h} = 0

but the network still has resistors, is the equivalent trivial?

Not quite —

V_{t h} = 0

means no net driving voltage, but

R_{t h}

can be non-zero, so the box behaves as a pure resistor

R_{t h}

to any external source you attach.

Terminals are shorted from the start (

R_{L} = 0

): what does the terminal law give?

V = 0

, so

0 = V_{t h} - I_{sc} R_{t h}

, giving the short-circuit current

I_{sc} = V_{t h} / R_{t h}

. This is one endpoint of the terminal line.

Terminals are left open (

R_{L} = \infty

): what does the terminal law give?

I = 0

, so

V = V_{t h}

. This is the other endpoint of the terminal line — the definition of

V_{t h}

itself.

R_{t h} = 0

, what is the maximum power transferred to a load?

Unbounded in the ideal model — matching

R_{L} = R_{t h} = 0

gives

P_{ma x} = V_{t h}^{2} / (4 R_{t h}) \to \infty

, which is why real sources always carry some internal resistance to cap the current.

Can two different-looking circuits share the exact same Thevenin equivalent?

Yes — any two networks with the same open-circuit voltage and the same terminal resistance are indistinguishable from the terminals. Thevenin's theorem is precisely the statement that only

(V_{t h}, R_{t h})

matters externally.

Recall One-line self-test

Cover everything above. Can you state, from memory, the three ways to get $R_{t h}$ and exactly when each is legal? ::: Deactivation (no dependent sources), test source $V_{t es t} / I_{t es t}$ (dependent sources present), and $V_{t h} / I_{sc}$ (any linear network) — all read the same slope of the terminal line.

Connections

Norton Equivalent Circuits — the dual: swaps $V_{t h}$ intercept for the $I_{sc}$ short-circuit endpoint.
Superposition Theorem — the linearity tool these traps keep circling back to.
Maximum Power Transfer Theorem — several edge cases here live at its $R_{L} = R_{t h}$ peak.
Source Transformation — clarifies the "short a V-source / open an I-source" traps.
Voltage and Current Dividers — the divider trap in "Spot the error".
Nodal and Mesh Analysis — general routes to $V_{oc}$ and $I_{sc}$ underneath these questions.