Using superposition, you find that source I₁ alone produces 4 V across resistor R, and source I₂ alone produces 3 V across R in the same direction. Both sources are now active. What is the power dissipated in R?
A25/R watts — add the individual power contributions (16/R + 9/R)
B49/R watts — first combine voltages (4 + 3 = 7 V), then compute power: 7²/R
C12.5/R watts — average the two power contributions
D7/R watts — superpose voltages and divide directly by R
Superposition applies to voltages and currents, not power. The total voltage is 7 V (correctly found by superposition). Power must then be computed from that combined voltage: P = 7²/R = 49/R. The tempting wrong answer (25/R) incorrectly superimposes the individual power contributions (16/R + 9/R), but P = V²/R is nonlinear — (4+3)² = 49, not 4² + 3² = 25. Superposing powers only works when the two contributions are orthogonal, which is not the case in a resistive circuit.
Question 2 Multiple Choice
To find the voltage contribution from source V₂ alone using superposition, how should the other voltage source V₁ be treated?
ARemove it from the circuit (open circuit) so it does not interfere
BReplace it with a short circuit (a wire) — an ideal voltage source set to zero volts becomes a short
CSet its value to a very small number approaching zero
DReplace it with an open circuit (a gap) — the same treatment as a current source
An ideal voltage source enforces a fixed voltage between two terminals. Setting it to zero means it enforces 0 V — which is the same as connecting the two terminals with a wire (short circuit). By contrast, turning off a current source means it forces zero current — which is equivalent to removing the connection (open circuit). These are opposite treatments and are a common source of error: voltage sources → short circuits, current sources → open circuits.
Question 3 True / False
The superposition principle can be used to calculate power contributions from each source separately and then add them to find total power in a linear circuit.
TTrue
FFalse
Answer: False
Superposition applies only to linear responses — voltages and currents. Power is nonlinear (P = V²/R = I²R), so power contributions from different sources do not simply add. The correct procedure is to use superposition to find the actual combined voltage or current, and then compute power from those combined quantities. Directly adding individual power contributions gives a wrong result whenever two sources produce nonzero individual contributions.
Question 4 True / False
In a linear circuit, doubling the magnitude of every independent source exactly doubles every node voltage and branch current in the circuit.
TTrue
FFalse
Answer: True
This is the homogeneity (scaling) property of linear systems. Because all element relationships (V = IR for resistors, etc.) are linear and the governing equations (KVL, KCL) are linear, the response scales proportionally with the inputs. This is a direct consequence of linearity and is why Thévenin/Norton equivalents work: the terminal behavior of any linear subcircuit is a linear function of its sources.
Question 5 Short Answer
Explain why superposition cannot be used to directly calculate total power dissipation in a circuit with multiple sources, even though it can be used to find total voltages and currents.
Think about your answer, then reveal below.
Model answer: Superposition requires linearity — the output must be a linear function of the inputs. Voltages and currents satisfy this because circuit equations (KVL, KCL, Ohm's law) are linear. Power is P = V²/R or I²R, which involves a square — a nonlinear operation. Because (V₁ + V₂)² ≠ V₁² + V₂² in general, the sum of individual power contributions is not equal to the power from the combined circuit. You must first find the combined voltage or current using superposition, then compute power from that combined value.
This is the most common error when students first apply superposition. The linearity of the theorem is not a formality — it has a hard boundary at any nonlinear quantity like power, stored energy (½CV², ½LI²), or any product of two circuit variables.