Difficulty: Intermediate Prerequisites: Ohm's Law, series and parallel resistors, Kirchhoff's rules basics, power dissipation formula.
Ideal batteries supply a fixed voltage no matter how much current you draw. Real batteries do not. Every real battery has internal resistance that causes its terminal voltage to drop under load. Understanding this is essential for any practical circuit work, from sizing batteries for a project to diagnosing why a car won't start on a cold morning. This topic builds on basic DC circuit analysis and shows you how to model a real source, then use equivalent resistance to solve the rest of the circuit.
A real battery is modelled as an ideal EMF in series with a small internal resistance r. The terminal voltage you actually measure is V_battery = EMF - I·r, which drops as you draw more current. Once you know the terminal voltage, the rest of the circuit solves with standard series/parallel reduction and Ohm's Law.
Electromotive force (EMF, ε)
The maximum potential difference a battery can provide when no current flows (open-circuit voltage). In simple terms, it is the battery's "ideal" voltage printed on the label.
Internal resistance (r)
The resistance inside the battery itself, caused by the chemistry and physical construction of the cell. Think of it as a small hidden resistor sitting inside the battery casing.
Terminal voltage (V_battery, V_terminal)
The voltage you actually measure across the battery's terminals when current is flowing. It equals EMF minus the voltage lost across the internal resistance: V_terminal = EMF - I·r.
Equivalent resistance (R_equiv)
A single resistance value that replaces an entire resistor network for the purpose of calculating total current from a source. In simple terms, it is what the battery "sees" when it looks out at the circuit.
Power dissipated (P)
The rate at which electrical energy is converted to heat in a resistor. Calculated as P = I²R, or equivalently P = V²/R.
Replace the physical battery with two elements in series: an ideal EMF source (voltage = EMF) and a resistor r (the internal resistance).
The terminal voltage is what appears across the external circuit: V_terminal = EMF - I_total · r.
When no current flows (open circuit), V_terminal = EMF. Under load, V_terminal is always less than EMF.
Identify which resistors are in series and which are in parallel, working outward from the load back towards the battery terminals.
Series: R_series = R_a + R_b + R_c + ...
Parallel: 1/R_parallel = 1/R_a + 1/R_b + ... , or for two resistors: R_parallel = (R_a · R_b) / (R_a + R_b).
The total equivalent resistance the battery sees is the combination of all external resistors.
If you know the EMF and the measured terminal voltage under load, you can find r.
r = R_equiv × (EMF - V_terminal) / EMF
Alternatively: r = (EMF - V_terminal) / I_total.
Once you have I_total through the main branch, the current splits at parallel junctions.
The branch with lower resistance carries more current.
For two parallel branches: I_branch = I_total × R_other / (R_branch + R_other).
Given values:
EMF (V) = 12 V, V_battery = 11.75 V
R₁ = R₃ = 57 Ω, R₄ = R₅ = 103 Ω, R₂ = 108 Ω
R₃, R₄, and R₅ are in series: R₃₄₅ = 57 + 103 + 103 = 263 Ω.
R₂ is in parallel with R₃₄₅: R_parallel = (108 × 263) / (108 + 263) = 76.56 Ω.
R₁ is in series with that parallel combination: R_equiv = 57 + 76.56 = 133.56 Ω.
I₁ = V_battery / R_equiv = 11.75 / 133.56 = 87.97 mA
Note the milliamp prefix. Watch your units.
r = R_equiv × (EMF - V_battery) / EMF = 133.56 × (12 - 11.75) / 12 = 2.8 Ω
V_battery = I₁R₁ + I₃R₃₄₅, so:
I₃ = (V_battery - I₁R₁) / R₃₄₅ = (11.75 - 0.08797 × 57) / 263 = 25.6 mA
The voltage across R₂ equals the voltage across R₃₄₅ (they are in parallel):
V₂ = I₃ × R₃₄₅ = 0.0256 × 263 = 6.73 V
P₂ = V₂² / R₂ = (6.73)² / 108 = 0.420 W
Shorting R₂ removes it from the circuit. The equivalent resistance of the external circuit decreases (the parallel combination is gone, leaving only R₁ + R₃₄₅ in one branch, but with a short across part of it). Lower external resistance means higher total current. Higher current means a larger voltage drop across the internal resistance r. Therefore the terminal voltage decreases.
Formula | Use |
|---|---|
V_terminal = EMF - I·r | Terminal voltage of a real battery |
r = (EMF - V_terminal) / I | Internal resistance from measured terminal voltage |
R_equiv = R₁ + (R₂ · R₃₄₅) / (R₂ + R₃₄₅) | Equivalent resistance for this circuit topology |
P = I²R = V²/R = IV | Power dissipated in a resistor |
I₃ = (V_terminal - I₁R₁) / R₃₄₅ | Branch current from known total current |
Internal resistance is why your phone charger delivers less voltage to the battery as charging current increases, and why car batteries struggle in cold weather (internal resistance rises with lower temperature, so the terminal voltage drops further under the high current demanded by the starter motor). Engineers design battery packs with cells in parallel partly to reduce the effective internal resistance of the pack.
"The EMF is the same as the voltage I measure with a multimeter." Not under load. A multimeter across a loaded battery reads the terminal voltage, which is EMF minus the internal voltage drop. You only measure EMF at open circuit (zero current).
"Internal resistance is constant." It varies with temperature, state of charge, and age of the battery. The constant-r model is a useful approximation, not a physical law.
"Shorting a parallel resistor increases the total resistance." It does the opposite. Shorting a resistor replaces it with a zero-ohm path, which decreases the equivalent resistance and increases the current drawn.
"Power dissipated in a resistor depends only on the current through it." Power depends on both current and resistance (P = I²R), or equivalently on voltage and resistance (P = V²/R). Make sure you use the voltage or current specific to that resistor, not the total circuit values.
⚠️ The internal resistance model (EMF in series with r) is a staple exam question. You may be given EMF and terminal voltage and asked to find r, or given r and asked for the terminal voltage under a specific load.
⚠️ Be careful with units. Currents in these problems often come out in milliamps. If you see a suspiciously small number, check whether you need to convert.
⚠️ "What happens when a resistor is shorted?" is a common conceptual question. Remember: shorting decreases resistance, increases current, increases internal voltage drop, and decreases terminal voltage.
⚠️ Power dissipation questions often require you to find the voltage across the specific resistor first, not just the total current. Use P = V²/R when you have the branch voltage, or P = I²R when you have the branch current.
True or false: A battery's terminal voltage is always less than its EMF. True when current is flowing (under load). At open circuit (I = 0), they are equal.
Fill in the blank: If the internal resistance of a battery is 3 Ω and the total current is 0.5 A, the voltage lost inside the battery is ___. 1.5 V.
True or false: Shorting a resistor that is in parallel with other components increases the total current drawn from the battery. True. The equivalent resistance decreases, so total current increases.
Fill in the blank: Power dissipated in a resistor can be calculated as P = I²R or as P = ___. V²/R (or equivalently IV).
Q: A battery has an EMF of 9 V and an internal resistance of 1.5 Ω. If the external circuit has an equivalent resistance of 18 Ω, what is the terminal voltage?
A: Total current I = 9 / (18 + 1.5) = 0.462 A. Terminal voltage = 9 - (0.462)(1.5) = 8.31 V.
Q: You measure 11.75 V across the terminals of a 12 V battery driving a 133.56 Ω load. What is the internal resistance?
A: r = R_equiv × (EMF - V_terminal) / EMF = 133.56 × 0.25 / 12 = 2.8 Ω.
Q: In a circuit where R₂ is in parallel with a series chain R₃₄₅, and R₂ is then shorted, what happens to the terminal voltage of a real battery powering the circuit?
A: The terminal voltage decreases. Shorting R₂ lowers the external equivalent resistance, so more current flows, and the increased current creates a larger voltage drop across the internal resistance.
Q: The current through a 263 Ω series chain is 25.6 mA. What voltage appears across that chain?
A: V = IR = 0.0256 × 263 = 6.73 V.
This material connects to Kirchhoff's rules (the same loop and junction equations apply once you include r in the loop). It also leads into Thevenin's theorem, where any linear circuit is reduced to a single voltage source in series with a single resistance, which is exactly the internal resistance model applied to an entire network rather than just a battery. The concept of power dissipation here reappears in AC circuits, where you additionally need to account for reactive components.
Internal resistance, real battery model, EMF vs terminal voltage, equivalent resistance, series parallel reduction, current divider, power dissipation, P = I²R, P = V²/R, voltage drop, battery under load, shorting a resistor, Kirchhoff's rules, PHYS 101, university physics, DC circuits