Apply superposition theorem: Question bank — Hardware deep dive

Refresh the recipe in the parent first: Apply superposition theorem.

True or false — justify

Superposition applies to any circuit with more than one source.

False — it needs the circuit to be linear; a diode or a squaring relation like power breaks it even with many sources.

A killed ideal voltage source becomes an open circuit.

False — zero volts means both terminals at the same potential, which is a wire (short), not a gap.

A killed ideal current source becomes a short circuit.

False — zero amps means no path is forced through it, which is an open circuit; shorting it would let uncontrolled current flow.

Dependent sources are switched off along with the independent ones.

False — dependent sources track real circuit variables, so they must stay active in every sub-analysis; killing them deletes actual physics.

The individual source contributions are always added as positive magnitudes.

False — you add them algebraically; a contribution can be negative if it drives the target voltage/current the opposite way.

Total power dissipated in a resistor equals the sum of the powers from each source acting alone.

False — power is

i^{2} R

, and

(i_{1} + i_{2})^{2} \neq = i_{1}^{2} + i_{2}^{2}

; you superpose current or voltage first, then square once. See Why power does not superpose.

If a resistor's total current is zero, superposition tells us no source contributed to it.

False — the source contributions can be equal and opposite, cancelling to zero total while each is individually nonzero.

Superposition changes the answer you'd get from nodal or mesh analysis.

False — it must agree exactly; linearity guarantees the split-and-add answer equals the direct solve (parent Example 2 shows the check).

Adding a second, identical source doubles every voltage and current in a linear circuit.

True — this is homogeneity/additivity: scaling or duplicating the input scales the whole linear response (Linearity and homogeneity).

Spot the error

"To kill the 10 V source I disconnected it, leaving an open gap."

Error: opening it changes the circuit's topology. A dead voltage source is 0 V = a short; the wire must stay so current can still flow through that branch.

"I superposed the two currents through

R

, then added their two

i^{2} R

powers."

Error: you may add the currents, but not the powers separately. Compute total current first, then

P = i_{total}^{2} R

once.

"There's a current-controlled voltage source, so in each sub-circuit I turned it off with the others."

Error: it's dependent, not independent. It stays live and re-computes its value from the controlling variable in every sub-circuit.

"Contribution from source A is

+ 3

V and from B is

2

V, so the node is

5

V."

Error: you never fixed a reference polarity. If B's contribution actually opposes A's chosen positive direction it is

- 2

V, giving

1

"The circuit has a diode, but I linearised nothing and just superposed the two batteries."

Error: a diode is nonlinear; superposition is invalid unless the diode is replaced by a valid linear model within its operating region.

"I killed all sources at once, solved the empty circuit, and got zero — so superposition gives zero."

Error: you kill sources one-at-a-time, leaving exactly one active per sub-circuit; killing them all just solves nothing.

Why questions

Why is a dead voltage source a short but a dead current source an open?

A voltage source defines the voltage (0 V ⇒ same potential ⇒ wire); a current source defines the current (0 A ⇒ no forced path ⇒ gap).

Why does superposition follow from Kirchhoff's laws?

KCL and KVL are linear equations (sums equal zero), so the system is

A x = b

; splitting

b

into per-source pieces and adding solutions is guaranteed by linear algebra (Kirchhoff's Voltage and Current Laws).

Why must dependent sources stay on?

Their value is a function of a circuit variable; freezing them to zero would erase a real constraint the true circuit enforces at all times.

Why can we not superpose power even though we can superpose voltage?

Voltage combines through a linear map, but power involves a square (nonlinear); squaring a sum introduces cross-terms that the individual squares miss.

Why does superposition break for a transistor amplifier at large signals?

The device is nonlinear over that range; superposition needs proportional-and-additive relations, which only hold for a linearised small-signal model.

Why do we keep a consistent reference direction across all sub-circuits?

So each contribution carries a meaningful sign; without a shared reference the algebraic sum mixes incompatible polarities.

Why is superposition often more work than nodal analysis, yet still taught?

It builds physical insight into how each source contributes and underlies Thevenin and Norton equivalents and small-signal analysis — value beyond just the number.

Edge cases

A circuit with exactly one independent source — is superposition meaningful?

Trivially yes: the "sum" has one term, so the theorem reduces to solving the circuit once; nothing new but not wrong.

A circuit with only dependent sources and no independent source — what does superposition give?

Every response is zero; with no independent excitation there is nothing for the dependent sources to feed on.

Two sources whose contributions to a node exactly cancel — is that a violation?

No — that's superposition working correctly; equal-and-opposite contributions sum to zero net, a legitimate result.

A resistor of

0 Ω

(a wire) sits where you kill a voltage source — any conflict?

No conflict: the dead source is already a wire, so it merges with the existing wire; the topology stays linear.

An ideal current source in series with a resistor gets its source killed — what happens to that branch?

It becomes an open circuit, so no current flows through that resistor regardless of its value in that sub-circuit.

The target is the voltage across a current source that's active — can superposition still find it?

Yes — the source fixes its current, but its terminal voltage is a circuit unknown you solve for, summing contributions like any other voltage.

Superposing across a purely capacitive/inductive AC circuit — allowed?

Yes, as long as elements stay linear; the branch relations

i = C d v / d t

and

v = L d i / d t

are linear, so superposition holds (per-frequency in phasor form).

Recall Quick self-scoring

Missed the "kill a voltage source" traps? Re-read the parent's "V-Short, I-Open" mnemonic. Missed power questions? Study Why power does not superpose. Missed the dependent-source items? They are the single most common exam slip — dependent sources never switch off.

Connections

Apply superposition theorem
Linearity and homogeneity
Kirchhoff's Voltage and Current Laws
Voltage divider and current divider
Thevenin and Norton equivalents
Nodal and mesh analysis
Why power does not superpose