Battery emf & Internal Resistance — Battery emf and internal resistance models - Physics 2

SECTION 1

Historical Context & Motivation

The study of electromotive force (emf) and internal resistance grew directly from early attempts to harness electrochemistry for practical electrical work. In the late eighteenth century, Alessandro Volta constructed the first true battery—the voltaic pile—and immediately noticed something troubling: the device could not sustain a steady current indefinitely, and the voltage available at the terminals dropped as the current drawn from the pile increased. This observation hinted that something inside the battery itself was impeding current flow, a phenomenon that would eventually be formalized as the internal resistance of an electrochemical cell.

1800 • Volta's Pile

Alessandro Volta demonstrates the first electrochemical battery, stacking alternating zinc and copper discs separated by brine-soaked cloth. He notes that terminal voltage diminishes under sustained current draw, providing the earliest empirical evidence of internal losses.

1827 • Ohm's Law Published

Georg Simon Ohm publishes Die galvanische Kette, mathematisch bearbeitet, establishing the proportional relationship between voltage and current. Ohm explicitly includes an internal resistance term in his circuit equation, laying the mathematical foundation for battery models.

1834 • Faraday's Electrochemistry

Michael Faraday formulates his laws of electrolysis, linking the quantity of charge passed through a cell to the chemical reactions at the electrodes. His work reveals the electrochemical origin of emf and clarifies the distinction between the energy source inside the battery and the resistive losses within it.

1871 • Kirchhoff's Circuit Laws Extended

Gustav Kirchhoff's voltage and current laws, first articulated in the 1840s, become systematically applied to circuits containing batteries with internal resistance. This enables rigorous analysis of multi-loop networks and maximum power transfer.

1897 • Thévenin's Theorem Generalized

Léon Charles Thévenin's theorem, originally published in 1883, gains widespread adoption as engineers recognize that any linear circuit with sources can be replaced by a single ideal voltage source in series with a single resistance—precisely the structure of the battery internal-resistance model.

The central question that motivates this topic is deceptively simple: why does a battery labeled 9 V deliver something less than 9 V to a circuit, and how does that shortfall depend on the current being drawn? Answering this question requires distinguishing between the ideal emf of the battery—its open-circuit voltage determined by electrochemistry—and the terminal voltage that appears across its external terminals when current flows through the internal resistance.

SECTION 2

Core Principles & Definitions

A real battery is best understood as the combination of two conceptually distinct elements: an ideal source of emf and a parasitic internal resistance. The emf (ε) represents the maximum potential difference the battery can provide—the work done per unit charge by the electrochemical reactions converting chemical energy into electrical energy. When no current flows (the open-circuit condition), the terminal voltage equals ε exactly. The internal resistance (r) accounts for all dissipative mechanisms inside the battery, including the finite conductivity of the electrolyte, the resistance of electrode materials, and contact impedances at electrode–electrolyte interfaces. Together, these two elements fully describe the linear battery model used throughout introductory circuit analysis.

Electromotive Force (emf, ε)

The open-circuit voltage produced by the battery's internal chemistry. Despite its name, emf is not a force but a potential difference measured in volts. It equals the energy supplied per coulomb of charge that traverses the battery from the negative to the positive terminal internally.

Internal Resistance (r)

A lumped resistance modeling all dissipative processes inside the cell. It causes an Ir voltage drop within the battery whenever current flows, reducing the terminal voltage below ε. Typical values range from milliohms (lead-acid) to several ohms (coin cells).

Terminal Voltage (V_term)

The actual voltage measured across the battery's external terminals under load. Given by V_term = ε − Ir when the battery is discharging. This is the voltage available to drive current through the external circuit.

Open-Circuit vs. Closed-Circuit

When the circuit is open (I = 0), V_term = ε. When a load is connected and current flows, V_term drops linearly with current. This distinction is fundamental to every circuit measurement involving batteries.

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KEY TAKEAWAY

Think of a battery as a water pump with a narrow pipe inside it. The pump (emf) tries to push water (charge) at a certain pressure (voltage), but the narrow internal pipe (internal resistance) restricts the flow. The faster you try to push water through (higher current), the more pressure you lose inside the pump itself—so less pressure (terminal voltage) remains for the external plumbing (external circuit). At zero flow, you measure the pump's full rated pressure.

SECTION 3

Visual Explanation — The Battery Model Circuit

The dashed purple box encloses the battery's internal model: an ideal emf source ε in series with the internal resistance r. The external load resistor R (green) completes the circuit. The terminal voltage V_term (yellow dashed line) is measured between terminals A and B, and it equals ε minus the voltage drop Ir across the internal resistance.

The circuit diagram above captures the essential physics of every real battery. Inside the dashed boundary, the ideal emf source ε converts chemical energy into electrical potential energy—it raises the potential of each coulomb of charge by exactly ε volts. Immediately in series with this ideal source sits the internal resistance r, which dissipates some of that energy as heat whenever current I flows through the battery. The net result is that the voltage available at the external terminals—V_term—is always less than ε when the battery is delivering current. This model is a linearization: it assumes r is constant regardless of current magnitude, which is an excellent approximation for moderate currents and forms the basis of virtually all introductory circuit analysis.

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Sign Convention Note

When a battery is being charged (current forced into its positive terminal by an external source), the terminal voltage is V_term = ε + Ir, because the internal resistance now adds to the voltage required to push current backward through the cell. This distinction is critical in charging-circuit analyses.

SECTION 4

Mathematical Framework

The mathematical description of the battery with internal resistance follows directly from Kirchhoff's voltage law (KVL) applied around a single loop. Consider a circuit consisting of a battery (emf ε, internal resistance r) connected to an external load resistance R. Traversing the loop in the direction of conventional current, the sum of all potential rises and drops must equal zero. The emf provides a rise of ε, the internal resistance produces a drop of Ir, and the external resistance produces a drop of IR. Setting the algebraic sum to zero yields the fundamental loop equation.

KIRCHHOFF LOOP EQUATION

ε − Ir − IR = 0

ε = electromotive force (V), I = current (A), r = internal resistance (Ω), R = external load resistance (Ω). This equation is the starting point for all single-loop battery-circuit problems.

CURRENT IN THE CIRCUIT

I = ε / (R + r)

Solving the loop equation for current. The total resistance in the denominator is the sum of external and internal resistances, consistent with their series arrangement. Short-circuit current (R = 0) is I_sc = ε / r.

TERMINAL VOLTAGE

V_term = ε − Ir = IR = εR / (R + r)

The terminal voltage decreases linearly with current. Equivalently, it equals the voltage across the external load (IR). The rightmost form, obtained by substituting I = ε/(R + r), shows how V_term depends on the ratio R/(R + r).

POWER DELIVERED TO LOAD

P_load = I²R = ε²R / (R + r)²

Power delivered to the external load is maximized when R = r (the maximum power transfer theorem). At that point, P_max = ε² / (4r), and half the total power is dissipated internally.

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Maximum Power Transfer

To prove that P_load is maximized at R = r, differentiate P_load = ε²R/(R + r)² with respect to R and set dP/dR = 0. Using the quotient rule: dP/dR = ε²(r − R)/(R + r)³ = 0, which gives R = r. The second derivative confirms this is a maximum. At maximum power transfer, the efficiency is only 50%—a tradeoff between power delivery and energy efficiency that pervades electrical engineering design.

SECTION 5

Terminal Voltage vs. Current — The V–I Characteristic

One of the most informative representations of a real battery is its V–I characteristic curve: a plot of terminal voltage against the current drawn from the battery. The equation V_term = ε − Ir is the equation of a straight line with y-intercept ε and slope −r. This linear relationship holds over a wide range of currents for most batteries, although deviations appear at very low currents (where activation losses in the electrochemistry matter) and at very high currents (where concentration gradients within the electrolyte become significant). The linear region, however, is the domain of the standard internal-resistance model taught in introductory physics.

The V–I characteristic of a real battery is a straight line with y-intercept ε (the emf, shown as the cyan dot at I = 0) and slope −r (indicated in violet). The yellow dashed line shows an ideal battery with zero internal resistance. The green dashed lines mark a typical operating point where both V_term and I can be read. The red point represents the short-circuit condition (R = 0), where I = ε/r and V_term = 0.

This graph reveals several experimentally accessible quantities. The y-intercept gives the emf ε directly—measure the terminal voltage with no load connected, and you have the emf. The magnitude of the slope equals the internal resistance r. In laboratory practice, one connects the battery to several different resistive loads, measures the resulting current and terminal voltage for each, and performs a linear regression to extract ε (intercept) and r (negative of slope). This technique is far more accurate than a single measurement and also reveals whether the linear model is appropriate for the battery under test.

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Experimental Tip

A voltmeter with sufficiently high input impedance (≥ 10 MΩ) draws negligible current. Connecting such a voltmeter across a battery's terminals with no other load effectively gives V_term ≈ ε. An ammeter in series and a variable resistor allow you to map out the entire V–I line. Always take data quickly to avoid heating effects that change r.

SECTION 6

Worked Example — Finding emf and Internal Resistance

A battery of unknown emf ε and internal resistance r is connected to an external resistor. When the external resistance is R₁ = 10.0 Ω, the current is measured to be I₁ = 0.50 A. When the external resistance is changed to R₂ = 4.0 Ω, the current increases to I₂ = 1.0 A. Determine ε and r, then find the terminal voltage under each condition.

Determining ε and r from Two Measurements

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Step 1 — Write the loop equation for each case

Applying ε = I(R + r) for each measurement: Case 1: ε = I₁(R₁ + r) = 0.50(10.0 + r) Case 2: ε = I₂(R₂ + r) = 1.0(4.0 + r)
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Step 2 — Set the two expressions for ε equal

Since ε is the same in both cases: 0.50(10.0 + r) = 1.0(4.0 + r). Expanding: 5.0 + 0.50r = 4.0 + 1.0r.
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Step 3 — Solve for the internal resistance r

Rearranging: 5.0 − 4.0 = 1.0r − 0.50r, so 1.0 = 0.50r.

r = 2.0 Ω
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Step 4 — Substitute back to find ε

Using Case 1: ε = 0.50 × (10.0 + 2.0) = 0.50 × 12.0.

ε = 6.0 V
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Step 5 — Calculate terminal voltages

Case 1: V_term = ε − I₁r = 6.0 − (0.50)(2.0) = 6.0 − 1.0 = 5.0 V. Equivalently, V_term = I₁R₁ = 0.50 × 10.0 = 5.0 V ✓ Case 2: V_term = 6.0 − (1.0)(2.0) = 4.0 V.

V₁ = 5.0 V, V₂ = 4.0 V
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Step 6 — Interpret the results

When the load resistance decreased from 10.0 Ω to 4.0 Ω, the current doubled, and the terminal voltage dropped from 5.0 V to 4.0 V. The "lost" voltage in each case (1.0 V and 2.0 V, respectively) is dissipated as heat in the internal resistance. Note that in Case 2, fully one-third of the emf is wasted internally—a situation common with aged or high-r batteries.

SECTION 7

Strengths & Limitations of the Linear Battery Model

The constant-emf, constant-internal-resistance model is remarkably useful, but it is important to understand where it succeeds and where it breaks down. A clear picture of its strengths and limitations prepares you for the more sophisticated models encountered in electrochemistry and power-electronics courses.

Aspect	Strengths	Limitations
Simplicity	Only two parameters (ε, r) needed. Easily combined with KVL/KCL for multi-loop analysis.	Oversimplifies real electrochemistry, which involves nonlinear activation and concentration losses.
Accuracy	Excellent for moderate, steady-state currents far from short-circuit or near-zero-current extremes.	Fails at very low currents (activation overpotential) and very high currents (mass-transport limited).
Temperature	Works well at constant temperature; r can be treated as fixed.	Internal resistance varies with temperature (decreases as T rises in many chemistries), so r is not truly constant.
State of Charge	Adequate for short-duration problems where the battery's charge level doesn't change appreciably.	Both ε and r change as the battery discharges. A nearly dead battery has a substantially lower ε and higher r.
AC / Transients	Provides a reasonable first-order DC approximation for circuit design.	Real batteries exhibit frequency-dependent impedance (capacitive and inductive components), requiring full EIS models.

Comparison of the linear battery model's applicability

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KEY TAKEAWAY

The linear model V_term = ε − Ir is analogous to Hooke's law (F = −kx) in mechanics: it captures the dominant behavior with a simple linear relationship, works beautifully within its regime of validity, and serves as the foundation upon which more complex models (nonlinear spring behavior, electrochemical impedance spectroscopy) are built. Just as you wouldn't use Hooke's law for a spring stretched to its breaking point, you shouldn't apply the linear battery model to extreme currents or deeply discharged cells.

SECTION 8

Connection to Advanced Battery Models

The introductory model of a battery as an ideal emf source in series with a single resistor is the first rung on a ladder of increasingly sophisticated representations used in electrochemistry and electrical engineering. Understanding how these models relate to one another provides valuable context for why the simple model works and when you need to move beyond it.

Feature	Introductory Model (ε, r)	Advanced Models
Parameters	Two: constant ε and constant r	Many: open-circuit voltage OCV(SOC), charge-transfer resistance, double-layer capacitance, Warburg impedance, SOC-dependent parameters
V–I Curve	Perfectly linear	Nonlinear: includes activation overpotential (logarithmic via Tafel equation) and concentration overpotential at high currents
Frequency Domain	Pure resistance (real impedance only)	Complex impedance Z(ω) with resistive, capacitive, and diffusive elements; characterized via Electrochemical Impedance Spectroscopy (EIS)
State of Charge	Not modeled; ε is fixed	OCV is a function of SOC; coulomb counting or Kalman filters track SOC dynamically
Typical Use	Introductory circuit analysis, quick engineering estimates	Battery management systems (BMS), electric vehicle design, grid-scale energy storage modeling

Introductory vs. advanced battery models

The most widely used advanced circuit model is the Randles equivalent circuit, which replaces the single internal resistor with a series combination of an ohmic resistance R_Ω (electrolyte resistance), a parallel combination of charge-transfer resistance R_ct and double-layer capacitance C_dl, and a Warburg impedance element representing diffusion. In the DC steady-state limit (ω → 0), the Randles circuit collapses to a single effective resistance—recovering the introductory model as a special case. This is precisely why the simple model works so well for steady-state problems: it is the low-frequency limit of the full electrochemical description.

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Looking Ahead

In more advanced courses, you will encounter Thévenin and Norton equivalent circuits for arbitrary networks. The battery with internal resistance is itself a Thévenin equivalent: V_Th = ε and R_Th = r. The Norton equivalent replaces this with a current source I_N = ε/r in parallel with r. Mastering the battery model now gives you a concrete physical example of these powerful circuit-analysis theorems.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student measures the voltage across a fresh AA battery with a high-impedance voltmeter and reads 1.58 V. She then connects the battery to a small flashlight bulb and measures the voltage again, finding 1.41 V. Explain, using the internal-resistance model, why the voltage reading decreased. Would connecting two identical bulbs in parallel cause the voltage to drop even further? Justify your answer.

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With the voltmeter alone, essentially no current flows, so the terminal voltage equals the emf: V_term ≈ ε = 1.58 V. When the bulb is connected, current I flows through the circuit, producing a voltage drop Ir across the internal resistance. The terminal voltage drops to V_term = ε − Ir = 1.41 V, so Ir = 0.17 V. Connecting two bulbs in parallel halves the effective external resistance, approximately doubling the current. The internal voltage drop would then increase to roughly 2 × 0.17 = 0.34 V, giving V_term ≈ 1.58 − 0.34 = 1.24 V—a further reduction. The key insight is that higher current always means a larger Ir drop and a lower terminal voltage.

Question 2: BASIC CALCULATION

A battery has an emf of ε = 12.0 V and an internal resistance of r = 0.50 Ω. It is connected to an external load resistance of R = 5.5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, and (c) the power dissipated internally.

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(a) Using I = ε/(R + r) = 12.0/(5.5 + 0.50) = 12.0/6.0 = 2.0 A. (b) The terminal voltage is V_term = ε − Ir = 12.0 − (2.0)(0.50) = 12.0 − 1.0 = 11.0 V. Alternatively, V_term = IR = (2.0)(5.5) = 11.0 V, confirming consistency. (c) The power dissipated internally is P_int = I²r = (2.0)²(0.50) = 2.0 W. This represents 2.0/24.0 ≈ 8.3% of the total power delivered by the emf source, with the remaining 22.0 W going to the load.

Question 3: INTERMEDIATE

Two identical batteries, each with emf ε = 9.0 V and internal resistance r = 1.0 Ω, are connected in series with an external resistor R = 8.0 Ω. Find the current in the circuit and the terminal voltage across each battery. Then repeat the calculation if the two batteries are instead connected in parallel (both positive terminals joined, both negative terminals joined) across the same 8.0 Ω load.

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Series connection: The total emf is 2ε = 18.0 V and the total internal resistance is 2r = 2.0 Ω. The current is I = 18.0/(8.0 + 2.0) = 1.8 A. Each battery's terminal voltage is ε − Ir = 9.0 − (1.8)(1.0) = 7.2 V. Parallel connection: Two identical batteries in parallel have an equivalent emf of ε = 9.0 V and an equivalent internal resistance of r/2 = 0.50 Ω (parallel combination of two 1.0 Ω resistors). The current from the combination is I = 9.0/(8.0 + 0.50) = 9.0/8.5 ≈ 1.06 A. The terminal voltage is 9.0 − (1.06)(0.50) ≈ 8.47 V. Each battery supplies half the total current (≈ 0.53 A), so each battery's terminal voltage is 9.0 − (0.53)(1.0) ≈ 8.47 V—consistent. The parallel configuration provides a higher terminal voltage because each battery carries less current.

Question 4: APPLIED

A car battery (ε = 12.6 V, r = 0.080 Ω) must deliver 150 A to a starter motor during engine cranking. (a) What is the terminal voltage during cranking? (b) The dashboard voltmeter, which reads terminal voltage, drops to 10.5 V after the battery has been in service for a year. Assuming ε has not changed significantly, what is the new internal resistance? (c) Discuss what physical changes inside the battery might cause this increase in r.

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(a) V_term = ε − Ir = 12.6 − (150)(0.080) = 12.6 − 12.0 = 0.6 V. Wait—this seems too low. Let's recheck: Ir = 150 × 0.080 = 12.0 V, giving V_term = 0.6 V. This is unrealistically low for a starter; in practice the starter motor's resistance limits current. More realistically, if V_term ≈ 10.5 V, then I = (ε − V_term)/r = 2.1/0.080 ≈ 26 A, a more typical cranking current for the given r. However, answering the problem as stated: V_term = 12.6 − 150(0.080) = 0.6 V during cranking. (b) With the aged battery, if V_term = 10.5 V at I = 150 A: r = (ε − V_term)/I = (12.6 − 10.5)/150 = 2.1/150 = 0.014 Ω. This is actually lower than before, which is contradictory. The problem likely intends for part (a) to use a more typical cranking current. Taking the problem at face value with 150 A: r_new = (12.6 − 10.5)/150 = 0.014 Ω. (c) Physical changes that increase r include sulfation of lead plates (PbSO₄ buildup reducing active surface area), corrosion of terminal connections, electrolyte stratification, and loss of electrolyte concentration—all of which increase the effective resistance to ionic and electronic current flow within the cell.

Question 5: CRITICAL THINKING

Prove that the power delivered to an external load R by a battery with emf ε and internal resistance r is maximized when R = r. Then show that at this optimal point, the efficiency η = P_load/P_total is exactly 50%. Discuss why a power utility would never operate at the maximum power transfer condition and what condition they optimize for instead.

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The power delivered to the load is P_load = I²R = [ε/(R + r)]²R = ε²R/(R + r)². To find the maximum, differentiate with respect to R using the quotient rule: dP/dR = ε²[(R + r)² − R × 2(R + r)]/(R + r)⁴ = ε²(R + r − 2R)/(R + r)³ = ε²(r − R)/(R + r)³. Setting dP/dR = 0 gives r − R = 0, hence R = r. The second derivative is negative at this point, confirming a maximum. At R = r, the current is I = ε/(2r), the power to the load is P_load = I²r = ε²/(4r), the power dissipated internally is P_int = I²r = ε²/(4r), and the total power is ε²/(2r). The efficiency is η = [ε²/(4r)] / [ε²/(2r)] = 1/2 = 50%. A power utility's source impedance is very low (milliohms), and the load impedance is many orders of magnitude higher. They operate far from R = r, in the regime R ≫ r, where efficiency η = R/(R + r) ≈ 1. They optimize for maximum efficiency (minimal energy waste), not maximum power transfer. The 50% waste at R = r would be economically and thermally catastrophic at megawatt scales.

SUMMARY

Lesson Summary

Summary

A real battery is modeled as an ideal emf source ε in series with an internal resistance r. The emf represents the open-circuit voltage generated by the battery's electrochemistry, while the internal resistance accounts for all dissipative processes within the cell. The terminal voltage under load is V_term = ε − Ir (discharging), which decreases linearly with current. The circuit current is I = ε/(R + r), where R is the external load resistance. Both ε and r can be extracted experimentally from the V–I characteristic: the y-intercept gives ε and the magnitude of the slope gives r.

The maximum power transfer theorem states that the load receives maximum power when R = r, at which point the efficiency is only 50%. This linear model is a Thévenin equivalent and serves as the DC, steady-state limit of more advanced electrochemical impedance models. It is accurate for moderate currents and constant temperature but breaks down at extreme currents, varying state of charge, or when AC/transient behavior matters. Mastering this foundational model equips you to analyze any circuit containing real voltage sources and to appreciate the approximations inherent in the idealized circuits often encountered in textbooks.