01 · Foundational

Independent ideal sources & source models

An ideal voltage source fixes voltage and lets current be determined by the network; an ideal current source fixes current and lets voltage be determined by the network. Learn the terminal laws, power signs, limiting cases and where the idealization breaks.

120–150 minFoundational
In this lesson
  • State the terminal law of an independent ideal voltage source and identify which terminal variable remains network-determined.
  • State the terminal law of an independent ideal current source and identify which terminal variable remains network-determined.
  • Predict how load resistance changes current for an ideal voltage source and voltage for an ideal current source.
  • Use passive-sign power bookkeeping to determine whether an ideal source absorbs or delivers power.
  • Explain why ideal-source short/open limits can demand unbounded current or voltage.
  • Distinguish source deactivation as an analysis operation from physically switching off a real instrument.
  • Relate ideal-source models to practical voltage/current limits without prematurely replacing them by Thévenin/Norton models.
Starting question

What does a circuit source actually guarantee?

An ideal source is best understood as a terminal constraint. An independent ideal voltage source prescribes its voltage regardless of the current required by the rest of the ideal circuit. An independent ideal current source prescribes its current regardless of the voltage required by the rest of the ideal circuit. The word independent means the specified source value is not controlled by another circuit variable.

01.1An ideal voltage source fixes V, not I

Figure 1.1 — An ideal voltage source fixes the load voltage

The ideal source and load resistor share the same two nodes. The source therefore fixes VOUT at 10 V; the resistor current is then determined by the load.

verified topology
Ideal DC voltage source driving one resistor+−VS10 VR5 kΩVOUTGND
The one question is: which terminal variable is imposed directly by the ideal voltage source?

Let an ideal voltage source have its positive terminal at node A and negative terminal at node B. Its defining terminal law is simply

VA − VB = VS

VS is specified independently of the current through the source. The source model does not give an additional equation such as I=V/R because no finite internal resistance belongs to the ideal voltage-source element.

That does not mean the current is arbitrary after the whole circuit is assembled. Kirchhoff's laws and the connected elements determine the current needed to satisfy the fixed source voltage.

01.2The load determines how much current a voltage source must supply

Figure 1.2 — The voltage source does not prescribe the load current

The arrow defines positive resistor current from VOUT toward ground. With 10 V and 5 kΩ, Ohm's law gives 2 mA. A different load produces a different current while the ideal source still holds 10 V.

verified topology
Ideal DC voltage source driving one resistor+−VS10 VR5 kΩVOUTGND
No pedagogical voltage arrow is added here: this figure addresses only the load-determined current.

With VS=10 V and RL=5 kΩ, the resistor shares the same terminal voltage as the ideal source:

IL = VS/RL = 10 V/(5 kΩ) = 2 mA

If RL becomes 1 kΩ, the ideal source still imposes 10 V and the required current becomes 10 mA. If RL→∞, the branch approaches an open circuit and IL→0.

If RL→0 while VS remains nonzero,

IL=VS/RL → ∞

This is a limiting statement. A perfect short directly across a nonzero ideal voltage source would simultaneously demand V=0 and V=VS, which is an inconsistent set of ideal constraints. Real sources avoid this mathematical pathology through current limits, internal impedance, protection, heating and other nonideal physics.

01.3An ideal current source fixes I, not V

Figure 1.3 — An ideal current source fixes the current

The arrow inside the source is intrinsic to the component: it imposes 2 mA from ground toward VOUT. With one resistor return path, the same 2 mA must flow through the resistor.

verified topology
Ideal DC current source driving one resistorIS2 mAR5 kΩVOUTGND
The one question is: which terminal variable is imposed directly by the ideal current source?

For an ideal current source, the defining law is the current itself. If its arrow defines positive current from terminal n toward terminal p, then

I = IS

IS is prescribed independently of the voltage across the source. The surrounding network must develop whatever voltage is required to accommodate that current within the ideal model.

In Figure 1.3 the source injects 2 mA into VOUT. With only one resistor path back to ground, Kirchhoff's current law forces 2 mA through the resistor.

01.4The load determines how much voltage a current source must develop

Figure 1.4 — The load determines the current source voltage

The voltage reference runs from ground, the negative reference, toward positive VOUT. The imposed 2 mA through 5 kΩ requires 10 V here. Changing R changes the voltage demanded from the source.

verified topology
Ideal DC current source driving one resistorIS2 mAR5 kΩVOUTGND
The arrow inside the source remains intrinsic to the component; no additional pedagogical current arrow is added.

For IS=2 mA and RL=5 kΩ, Ohm's law on the load gives

VL = ISRL = (2 mA)(5 kΩ) = 10 V

If RL becomes 1 kΩ, the same ideal current needs only 2 V. If RL→0, VL→0 while the current remains 2 mA.

If RL→∞ while IS remains nonzero,

VL = ISRL → ∞

Again this is a limit. A perfect open circuit would demand branch current I=0 while a nonzero ideal current source simultaneously demands I=IS. Those constraints are inconsistent. A real current source instead reaches its finite compliance-voltage limit and can no longer maintain the commanded current.

01.5Terminal I–V relations make the duality explicit

Choose terminal voltage V on the horizontal axis and terminal current I on the vertical axis. Then an ideal independent voltage source satisfies

V = VS

for whatever current the external ideal network determines. Its I–V locus is therefore a vertical line at V=VS.

An ideal independent current source satisfies

I = IS

for whatever terminal voltage the external ideal network determines. Its locus is a horizontal line at I=IS.

These are constitutive/terminal laws, just as V=RI is the terminal law of an ideal resistor. The difference is which relationship each element imposes between its terminal variables.

01.6“Source” does not mean “always delivering power”

Power sign depends on the chosen voltage polarity and current reference. Under the passive-sign convention, current is positive when it enters the positive-voltage terminal and

P = VI

means positive P is absorbed power. If current instead leaves the positive terminal, the element delivers power and its absorbed power is negative.

For the 10 V voltage-source example, the 5 kΩ resistor absorbs

PR=(10 V)(2 mA)=20 mW

The source current leaves the source's positive terminal, so current entering that terminal is −2 mA:

PS=(10 V)(−2 mA)=−20 mW

The source delivers 20 mW. But if another part of a circuit forced positive current into the source's positive terminal, the same ideal voltage source would absorb power. Rechargeable-battery charging is a familiar real-world reminder that an element called a source need not deliver power at every instant, although an ideal voltage-source model is not itself a battery model.

01.7Ideal sources are intentionally extreme models

ModelExactly imposedAllowed by the ideal modelReal limitation
Ideal voltage sourceV=VSWhatever current the network demandsCurrent limit, output impedance, power/thermal limits, bandwidth
Ideal current sourceI=ISWhatever voltage the network demandsCompliance voltage, output impedance, power/thermal limits, bandwidth

The purpose of an ideal model is not to claim that infinite current or voltage exists in hardware. It isolates one dominant behavior so circuit equations stay transparent. Engineering judgment enters when we ask whether the required current, voltage, power and frequency range stay inside the real source's specifications.

Later, source resistance and terminal-equivalent models will quantify some nonideal behavior. Do not silently add those effects before the model says they are present.

01.8Setting an independent source to zero is an analysis operation

Linear-circuit analysis often asks us to set an independent source value to zero. Apply the element's own defining law literally.

For an independent ideal voltage source:

VS=0 V ⇒ V=0 V

Two terminals constrained to the same voltage are represented by an ideal short circuit.

For an independent ideal current source:

IS=0 A ⇒ I=0 A

A branch constrained to zero current is represented by an ideal open circuit.

Chapter 03 will use this operation carefully in superposition. Dependent sources, introduced in chapter 02, are not simply turned off during superposition because their values remain tied to circuit variables.

01.9A robust source-model reading workflow

  1. Identify the source type. Voltage or current? Independent or, later, dependent?
  2. Read the reference. For voltage, identify +/− terminals; for current, identify the arrow.
  3. Write the defining source equation first. V=VS or I=IS.
  4. Let the network determine the conjugate variable. Use Ohm's law and Kirchhoff's laws to find source current or source voltage.
  5. Check extreme demands. Does the ideal solution require implausibly large current, voltage or power?
  6. Compute signed power with an explicit convention. Do not assume a source must deliver.
  7. Only then compare with a real source specification. Current limit and compliance are model-validity questions, not corrections to be guessed.

01.10Practice for mastery

  1. An ideal 12 V source drives a 3 kΩ resistor. Find load current and resistor power.
  2. The same ideal 12 V source drives 300 Ω. What changes? Which quantity is still fixed by the source model?
  3. An ideal 1.5 mA source drives 4 kΩ. Find terminal voltage and resistor power.
  4. The same current source drives 40 kΩ. What voltage is required? Why might a real current source fail to realize the ideal answer?
  5. A 5 V ideal source carries 2 mA entering its positive terminal. Does the source absorb or deliver power?
  6. A 2 mA ideal current source points upward from ground into a node whose voltage is −3 V relative to ground. Using the upper node as the positive voltage reference, does the source absorb or deliver power?
  7. What replaces a 0 V independent ideal voltage source in a linear-analysis diagram? What replaces a 0 A independent ideal current source?
  8. Why is a nonzero ideal voltage source directly shorted, or a nonzero ideal current source directly opened, better described as inconsistent ideal constraints than as an ordinary finite circuit operating point?
Solutions

1. I=12 V/3 kΩ=4 mA. The resistor absorbs P=48 mW; the ideal source delivers 48 mW.

2. I=40 mA and resistor power is 480 mW. The source voltage remains exactly 12 V in the ideal model.

3. V=(1.5 mA)(4 kΩ)=6 V. The resistor absorbs 9 mW.

4. V=60 V. A real source may hit its finite compliance-voltage or power limit and then cease to maintain 1.5 mA.

5. Current enters the positive terminal, so P=(5 V)(2 mA)=+10 mW: the ideal source absorbs 10 mW.

6. The source arrow is upward, so current entering the positive upper terminal is −2 mA. P=V·Ientering+=(−3 V)(−2 mA)=+6 mW. The current source absorbs 6 mW.

7. A 0 V ideal voltage source is a short circuit; a 0 A ideal current source is an open circuit. These are model operations.

8. The direct short imposes V=0 while the nonzero voltage source imposes V=VS; the direct open imposes I=0 while the nonzero current source imposes I=IS. Both demand mutually incompatible ideal equations. Infinite current/voltage arises as a limiting statement when a finite load approaches zero/infinite resistance.

Chapter 02 will keep the source idea but remove independence: the source value will be controlled by another voltage or current elsewhere in the circuit. That is the language used to model amplification and coupling before introducing any specific transistor technology.