Terminal voltage equals E minus I r: understanding internal resistance in real sources.

Outline (brief skeleton)

• Hook: why voltage isn’t just a fixed number in real batteries

• Define the players: E (emf), I (current), r (internal resistance), V (terminal voltage)

• The core idea: as current flows, internal resistance drops some voltage, so V = E − I r

• Quick derivation in plain terms, plus a simple example

• Why the other choices are wrong (quick check for understanding)

• Real-world implications: batteries, power banks, and circuits you actually build

• A mental model you can carry to exams and labs

• Practice-ready takeaway and a tiny problem

• Friendly wrap-up

Understanding voltage with internal resistance: the real-life version of V = E − I r

Let me ask you a simple question. When you connect a flashlight to a battery, does the light always glow with the same brightness, no matter how long you keep it on? In a classroom idealization, you might treat the battery as a perfect source that never loses voltage. But in the real world, a battery is messier. It has something called internal resistance. That little resistance inside the battery—the r in the formula—acts like a tiny friction inside the conduit that current uses to flow.

What are E, I, r, and V anyway? Here’s the thing in plain terms:

• E is the electromotive force, the “open-circuit” voltage the source would provide with no current flowing. Think of it as the battery’s ideal voltage.

• I is the current flowing through the circuit, the rate at which charges move.

• r is the internal resistance of the source. It’s not something you can see with the naked eye, but it’s the reason you feel a drop in voltage as current ramps up.

• V is the terminal voltage, the voltage you actually measure across the device you’re powering (your lamp, your phone, whatever is connected outside the source).

Now to the heart of the matter: why the terminal voltage is not simply E, but E minus a little extra drop. When current flows, it must push charges through that resistance inside the battery. The energy lost to internal resistance is I times r (that is, the current times the internal resistance). According to Kirchoff’s ideas about energy balance, the voltage drop across the internal resistance is I r. So the voltage left for the outside world—the terminal voltage—shrinks by that amount. Put another way, V = E − I r.

That tiny minus sign is not a quirk; it’s the natural consequence of current doing work against the internal friction inside the source. If r were zero, we’d have V = E all the time, which is the ideal battery picture. In the real world, r isn’t zero, so V slips as I climbs.

A quick, down-to-earth derivation (without heavy math)

• Start with E, the no-load voltage.

• When you draw current I, all of that current has to push through the internal resistance r.

• The voltage drop across that internal resistance is I r.

• The external terminals must supply the remainder, so V = E − I r.

Think of it like this: you have a water tank (the battery) pushing water through a pipe with a tiny throttle inside it (the internal resistance). The deeper you open the faucet (the larger the current I), the more pressure is lost inside the pipe. By the time the water leaves the outside faucet (the terminal), its pressure is a bit lower than the tank’s no-load pressure. That “bit lower” is your V = E − I r.

Why the other options don’t fit

• B. V = E + I r would imply the external voltage grows as you draw more current, which isn’t how real sources behave. More current, more internal loss, less terminal voltage.

• C. V = I r − E would be negative when E is larger than I r in typical small circuits, which isn’t the usual relationship we measure at the terminals.

• D. V = E / (I r) mixes up units and concepts; dividing voltage by a product of current and resistance isn’t the physics here.

In short: only V = E − I r captures the correct balance between the source’s emf, the current, and the energy lost inside the source.

A tangible feel with real-world examples

• Battery-powered gadgets: imagine your smartphone battery. When you’re idling, the current is small, so I r is small and V is close to E. When you run a heavy app or play a game, I shoots up, and that extra drop inside the battery makes the phone feel like it’s a little weaker, even though the nominal emf hasn’t changed.

• Power banks and chargers: many power banks advertise a voltage close to 5 V at the output, but under heavy load the terminal voltage can sag a bit due to internal resistance. This is exactly why high-drain devices might see a faster battery drain feel than under light use—more current means a bigger drop inside the bank.

• Real batteries in devices: car batteries, AA cells, solar cells—each has its own internal r. In high-current situations (cranking a car engine, for example), the drop I r can be substantial, which is why a battery’s performance is often described by both E and r.

A mental model that sticks

• If you plot V against I for a fixed battery, you get a straight line with slope −r and y-intercept E. The bigger the internal resistance, the steeper the slope; the bigger the drop as soon as you push current.

• If r = 0, the line would be flat at V = E. That’s the ideal world, which helps you see what the internal resistance is doing in the real one.

What this means for your circuit work

• Measuring V across a source isn’t just a single number. You’re seeing the effect of current you’re drawing. If you want to know the true emf E, you’d need to measure with essentially no current (an open circuit). In practice, you infer E from the open-circuit voltage or from circuit models.

• When you design a circuit, paying attention to r matters. If your device needs a stable voltage, you want a source with a small internal resistance, or you design the circuit to regulate the voltage after measuring it at the terminals.

• Think of r as a built-in limitation. It’s why high-output devices sometimes need better power sources or regulators.

Tiny checklist you can carry to labs and questions

• Identify E: what is the emf of the source, in volts?

• Note I: what current is flowing in the circuit?

• Look for r: what is the internal resistance of the source? If you’re unsure, you can model the source as a small resistor in series with an ideal voltage source.

• Use V = E − I r to find the terminal voltage.

• Check the edge cases: with no current (I ≈ 0), V ≈ E. With a large current and a noticeable r, V drops noticeably.

A quick practice thought experiment

Suppose you have a 9 V battery with an internal resistance of 0.2 Ω. If you draw a current of 1 A, what’s the terminal voltage?

V = E − I r = 9 − (1)(0.2) = 9 − 0.2 = 8.8 V

If you crank up the current to 5 A, V = 9 − (5)(0.2) = 9 − 1 = 8 V.

See how the terminal voltage slides downward as current climbs? That’s the lived reality inside every battery.

A few practical notes that prevent confusion

• Always mind the sign. The term I r is a drop, not a gain. That’s why it’s subtracted.

• If you’re comparing two sources, don’t just look at E. Look at E and r together. A source with a higher E but large r might give you a lower terminal voltage under load than a source with a smaller E and tiny r.

• When you read labels or datasheets for power sources, you’ll often see both the emf and an internal resistance or an internal resistance at a specific temperature. Temperature matters, because r can shift a bit with heat.

A gentle closer

Voltage isn’t a fixed badge on a battery. It’s a conversation between the source’s ideal force, the current you’re pulling, and the friction hidden inside the source itself. The clean relationship V = E − I r makes that conversation precise, actionable, and surprisingly intuitive once you picture the symmetry: a little drop inside, a bit of a drop outside, and a balance that shifts with every click of a switch or each heartbeat of a circuit.

If you keep that image in mind, you’ll find the concept becomes not just a formula to memorize, but a practical lens for analyzing real circuits. And when you see V, E, I, and r together, you’ll understand why the numbers you measure can feel both familiar and a touch dynamic—the battery talking back to the circuit, just a little.

Final takeaway

• The correct relationship is V = E − I r. It’s the straightforward way to account for the voltage that’s lost inside the source as current flows. Remember it next time you’re evaluating a circuit with any real power source, and you’ll be better equipped to predict behavior, diagnose quirks, and design smarter, steadier systems.

If you’d like, I can walk through a couple more example scenarios or set up a tiny, hands-on exercise you can try with a digital multimeter to see the effect of internal resistance in action.