A complete circuit: emf and internal resistance

A battery is not a perfect “pump”: it has an internal resistance r. Change the emf ε, r and the load R, open the switch, cause a short circuit — and watch where the volts and the watts go. The wires are coloured by potential, and the white dot shows the journey of one charge around the whole circuit.

wire colour = potential: blue = 0 V, red = ε ●moving charges (bigger current — faster) the white dot is the same charge on the diagram and on the profile ✱sparks — heat (in R it is useful, in r it is a loss)

Potential profile: the journey of one charge

The white dot is one charge. The source “lifts” it by ε (like a pump), then it slides down two slopes: first inside the source (across r), then through the external circuit (across R). The steeper the slope, the larger the resistance. After a full lap the charge is back at the same level.

The idea in 30 seconds

Emf is a pump for charge

A battery does not create charges — they are already in the wires. It works like a pump in a water pipe: it lifts the charge to a “height” ε (a volt is height measured in energy per unit of charge). After that the charge slides on its own around the circuit and gives away energy on the way — it lights a bulb and warms the wires.

A battery has resistance of its own

Inside any source there is chemistry, electrolyte, plates — and they obstruct the current too. This is the internal resistance r. So part of the “height” ε is lost already inside: ε = U_R + U_r. What is left at the terminals is U_R = ε − I·r — that is what the voltmeter sees and what the load receives.

The load changes everything

The larger R is, the smaller the current and the loss inside — the terminal voltage approaches ε (with an open circuit it equals ε). The smaller R is, the bigger the current and the more the voltage “sags”. At R = 0 — a short circuit: the current is ε/r and all the heat is released inside the source.

Try it yourself

Click the switch on the diagram — the circuit opens and the voltmeter shows the full emf ε. Close the switch: the reading becomes smaller. Where did the difference go? Look at the orange part of the “Where the emf goes” bar.
Move R from 50 Ω down to zero. The current grows and the terminal voltage falls. This is the “sag” of a battery under load — and the dot on the U(R) graph runs down.
Run the “Short circuit” experiment (R = 0): the voltmeter shows 0 V, the current is at its maximum and the battery turns red. Why is connecting the terminals with a wire a bad idea?
Find the R at which the power in the load is the largest (the green curve on the bottom-right graph). Check: it is R = r. And what is the efficiency? Only 50 %!
Raise r to 8 Ω — you get a “flat” battery: the current is small and less than half of the emf is left at the terminals. Now choose the “Lab” mode and find ε and r of a hidden battery.

Where you meet it

A car battery: its r is a few hundredths of an ohm, so it gives the starter hundreds of amperes. But you must not connect its terminals with a spanner — the metal would glow red-hot.
Why a battery “goes flat”: over time its r grows and the torch gets less and less voltage. A voltmeter with no load still reads almost the same 1.5 V — but under load it “collapses”.
Fuses and circuit breakers in a switchboard: they break the circuit when the current is too large — before a short circuit can melt everything.
Radio engineering: a transmitter and an antenna are matched (R ≈ r) to deliver the maximum power — even though the efficiency is then only 50 %.

Four graphs of the circuit

The yellow dot is the current state. Move the sliders and follow it.

U(I): the line of the source and the line of the load. Their crossing is the operating point. The green rectangle is the useful power, the red one is the loss inside

I(R): the current falls as R grows; at R = 0 only r limits it (dashed — the short-circuit current)

U(R): the terminal voltage rises toward ε but never reaches it; the orange curve is the loss inside the source

The universal picture (the axis is R/r): the power in the load is the largest at R = r, but the efficiency there is only 50 %

Mode

Quick experiments

ε emf of the source

r internal resistance

R load resistance

What runs along the wire

Playback speed

The numbers come from exact formulas and do not depend on the frame rate or the “playback speed”. The speed of the running dots is only schematic: the bigger the current, the faster they go, but for huge currents it is capped so that the dots do not blur into a stripe. The real drift speed of electrons in a wire is a fraction of a millimetre per second. The current direction is conventional: from the “+” of the source through the external circuit to the “−”.

KEY FORMULAS

Ohm’s law for a complete circuit

$$I = \frac{\varepsilon}{R + r}$$

The current is set by the emf and the sum of the external and internal resistances

▶ Detailed Explanation

r is the resistance of the source itself (a battery, an accumulator, a generator): part of the electrical energy turns into heat inside it before the current even reaches the load. That is why the denominator holds not R alone but R + r — the resistance of the whole loop. Numbers: ε = 12 V, r = 1 Ω, R = 5 Ω → I = 12/(5 + 1) = 2 A. Forget the internal resistance and you get 12/5 = 2.4 A — 20 % too high, exactly the mistake that costs a mark on an exam. The limiting cases: R = 0 (a short circuit) gives the largest possible current ε/r; R → ∞ (an open circuit) gives I → 0.

Terminal voltage and the split of the emf

$$\varepsilon = U_R + U_r,\qquad U_R = IR = \varepsilon - Ir$$

The emf splits into the terminal voltage (what the load gets) and the loss inside the source

▶ Detailed Explanation

A voltmeter across the terminals reads not the emf but U_R = ε − I·r: on its way through the source itself the charge loses U_r = I·r. The bigger the current, the stronger the “sag”. Numbers: ε = 12 V, I = 2 A, r = 1 Ω → U_r = 2 V, U_R = 12 − 2 = 10 V. Check through the load: I·R = 2·5 = 10 V. It matches. Only with an open circuit is I = 0, so U_r = 0 and U_R = ε: that is exactly how the emf is measured with a voltmeter. The “Where the emf goes” bar in the simulator is always split into these two parts.

The loop rule

$$\varepsilon - Ir - IR = 0$$

After a full lap the charge returns to the same potential: the rise in the source equals the sum of the drops

▶ Detailed Explanation

This is Ohm’s law for a complete circuit written as a “balance of heights”. The source lifts the charge by ε; across the internal resistance the potential falls by I·r, across the load by I·R. After a full lap the charge is at the same level it started from, so the rises and the drops are equal. Numbers: +12 − 2 − 10 = 0. On the “potential profile” this is a broken line that closes up: the white dot comes back to the same point. The steepness of a slope on the profile is the resistance: the 5 Ω load is “steeper” than the 1 Ω internal one. With the switch open there is no current and the whole ε falls across the gap of the switch.

Short circuit and open circuit

$$R = 0:\ I_{\text{sc}} = \frac{\varepsilon}{r},\ U_R = 0;\qquad R \to \infty:\ I = 0,\ U_R = \varepsilon$$

The two extremes: the current is limited by r alone, or there is no current at all

▶ Detailed Explanation

At R = 0 the current has no load and only the internal resistance limits it. For ε = 12 V and r = 1 Ω that is 12 A, and the whole power I²r = 144 W is released inside the source — it overheats quickly. A car battery has r ≈ 0.01 Ω and its short-circuit current reaches hundreds of amperes. With an open circuit (R → ∞ or an open switch) there is no current, nothing is lost inside the source, and the voltmeter across the terminals reads exactly ε. Measurements are based on this: ε is read by a voltmeter on an open battery; r = (ε − U)/I comes from one measurement with the circuit closed. This is exactly what you need to do in the “Lab” mode.

The current–voltage line of a source

$$U = \varepsilon - I\,r$$

The terminal voltage falls linearly as the current grows; the slope of the line is r

▶ Detailed Explanation

On the U(I) graph the source is a straight line starting at the point (0; ε) and reaching the current axis at I_sc = ε/r. The load is another straight line, U = I·R, through the origin. The operating point of the circuit is their crossing: both equations hold there at once. When you change R you turn the load line like a pointer, and the operating point slides along the source line. The area of the green rectangle U·I is the useful power in the load; the red rectangle above it (height I·r, width I) is the loss inside. Together they make the rectangle ε·I — the total power.

The power balance

$$P = \varepsilon I = I^2 R + I^2 r$$

The whole power of the source splits into the useful part (in the load) and the loss (heating inside the source)

▶ Detailed Explanation

Multiply the loop rule ε = I·R + I·r by the current I and you get the power balance: what the source delivers (εI) equals the heat in the load (I²R) plus the heat inside the source (I²r). Numbers: ε = 12 V, I = 2 A → P = 24 W. The load gets I²R = 4·5 = 20 W, and I²r = 4·1 = 4 W is lost inside the source: 20 + 4 = 24. These two numbers are shown in the “Where the power goes” bar. In a short circuit the useful power is zero and the entire power εI = ε²/r heats the source itself.

Efficiency of the source

$$\eta = \frac{P_R}{P} = \frac{R}{R + r} = \frac{U_R}{\varepsilon}$$

The share of the power that reaches the load; it grows with R

▶ Detailed Explanation

The efficiency is the ratio of the useful power to the total power. Because the current is the same, it reduces to the ratio of the resistances R/(R + r) or of the voltages U_R/ε. Numbers: R = 5 Ω, r = 1 Ω → η = 5/6 ≈ 83 %. To get 90 % you need R = 9r = 9 Ω. At R = r the efficiency is exactly 50 %, and at R ≪ r almost all of the energy heats the source itself. Note: the efficiency does not depend on the emf. If you double ε, the current and all the powers grow, but the share that reaches the load stays the same.

The maximum power in the load

$$P_R = \frac{\varepsilon^2 R}{(R + r)^2}\ \le\ \frac{\varepsilon^2}{4r},\qquad P_{\max}\ \Leftrightarrow\ R = r$$

The power in the load is the largest at R = r, but the efficiency is then only 50 %

▶ Detailed Explanation

The power in the load P_R = I²R first grows with R (the current hardly falls while R is larger), then decreases (the current falls faster). The maximum is exactly at R = r. It can also be shown without a derivative: (R + r)² ≥ 4Rr, hence P_R ≤ ε²/(4r), with equality at R = r. Numbers: ε = 12 V, r = 1 Ω. R = 1 Ω: I = 6 A, P_R = 36 W = ε²/(4r) — the maximum. R = 5 Ω: P_R = 20 W. R = 0.5 Ω: I = 8 A, P_R = 32 W. Both below and above r the power is smaller. On the universal graph (the axis is R/r) the power curve and the efficiency curve cross at R = 3r, where both equal 75 %. Matching R ≈ r pays off when power matters (a radio transmitter) and does not when saving energy matters.

Ohm’s law for a complete circuit: I = ε/(R + r). A source is not a perfect “voltage tap”: it has a resistance r of its own. So the current is limited not only by the load R but by the battery itself, and the real current is always smaller than ε/R. The most common misconception is to confuse the emf with the terminal voltage. The emf ε is the energy the source gives to a unit of charge (measured in volts). A voltmeter across the terminals of a working source reads U_R = ε − I·r = I·R. Part of the emf, U_r = I·r, is “eaten” inside the source, and always ε = U_R + U_r. Only with an open circuit (I = 0) does the voltmeter read exactly ε — that is how the emf is measured. The load controls everything. The larger R is, the smaller the current, the smaller the loss inside and the higher the terminal voltage; the smaller R is, the bigger the current and the more the voltage “sags”. At R = 0 (a short circuit) the current reaches I_sc = ε/r, the terminals show 0 V, and the whole power εI turns into heat inside the source. That is why a short circuit is dangerous and circuits are protected by fuses. The energy of the circuit. The total power of the source P = εI splits into the useful power P_R = I²R and the loss P_r = I²r. The efficiency of the source η = P_R/P = R/(R + r) grows with R, but the power in the load P_R = ε²R/(R + r)² has a maximum at R = r: P_max = ε²/(4r), and the efficiency there is only 50 %. So to avoid wasting energy one takes R ≫ r, and to get the maximum power (for example in radio engineering) R ≈ r. How to use the simulator. Move the three sliders or click the switch right on the diagram. The colour of a wire shows the potential (blue — 0 V, red — ε), and the “potential profile” shows how one charge is lifted by ε in the source and slides down two slopes — across r and across R. The “Predict” mode checks whether you understand how the current, voltage, efficiency and power change; in the “Lab” you find the hidden ε and r from the readings of the voltmeter and the ammeter — just as it is done in a school laboratory.