Questions: Electromotive Force (EMF) and Batteries

Question 1

A battery with EMF ε = 9 V and internal resistance r = 1 Ω is connected to an external resistor R = 8 Ω. What is the terminal voltage across the battery's terminals?

• A. 9 V — the terminal voltage always equals the EMF
• B. 8 V — the terminal voltage equals the voltage across the external resistor only
• C. 1 V — the terminal voltage equals the voltage drop across the internal resistance
• D. 8 V — calculated as V = ε − Ir = 9 − (1)(1) = 8 V

Answer: (D)

The current in the circuit is I = ε/(R + r) = 9/(8+1) = 1 A. The terminal voltage is V = ε − Ir = 9 − (1)(1) = 8 V. This equals the voltage across the external load, as expected: V = IR = (1)(8) = 8 V. Note that options B and D state the same number (8 V) but option B gives the wrong reasoning — the terminal voltage happens to equal the load voltage here, but the reason is V = ε − Ir, not a definition. Option A is the ideal-battery misconception; real batteries always have V < ε when current flows.

Question 2

A nearly discharged battery reads 12 V on a voltmeter when nothing is connected, but drops to 8 V when connected to a motor. What best explains this?

• A. The voltmeter drained the battery during measurement
• B. The motor reversed the current through the battery, reducing its voltage
• C. The battery's internal resistance increased as it discharged; the large current drawn by the motor creates a significant Ir voltage drop inside the battery
• D. The motor's back-EMF opposes the battery, reducing the effective voltage

Answer: (C)

The open-circuit voltage (12 V) approximates the EMF ε because no current flows and thus Ir ≈ 0. When the motor draws large current I, the internal resistance r (which increases as the battery discharges) causes a voltage drop Ir inside the battery, leaving V = ε − Ir = 8 V at the terminals. This is exactly why 'dead' batteries often read fine on a voltmeter but fail under load — their internal resistance has risen so much that any real current draw collapses the terminal voltage.

Question 3

Electromotive force is a force that pushes charges through the circuit.

• T. True
• F. False

Answer: False

Despite its name, EMF is not a force — it is energy per unit charge, measured in volts (joules per coulomb). It represents the work done by the source (chemical reactions, mechanical motion, etc.) per coulomb of charge moved through it. The naming is a historical accident; 'electromotive force' was coined before the modern distinction between force and energy was fully established. Treating EMF as a literal force leads to confusion when applying energy conservation (Kirchhoff's voltage law) to circuits.

Question 4

The terminal voltage of a real battery under load is always less than its EMF.

• T. True
• F. False

Answer: True

The terminal voltage is V = ε − Ir. Since I > 0 when current flows and r > 0 for any real battery, the product Ir > 0, so V < ε always. The terminal voltage equals the EMF only in the ideal limiting case r = 0 (no internal resistance) or I = 0 (open circuit). In real batteries, internal resistance is always positive, so current draw always causes a voltage sag at the terminals. This is why the measured voltage of a battery under load is the relevant quantity for circuit calculations, not the open-circuit EMF.

Question 5

Why does the terminal voltage of a battery decrease when it must supply a larger current? Explain using the internal resistance model.

Model answer: Every real battery has internal resistance r. When current I flows, this internal resistance causes a voltage drop Ir inside the battery itself. The terminal voltage — the voltage available to the external circuit — is what remains after that internal drop: V = ε − Ir. The larger the current drawn, the larger the Ir drop, and the lower the terminal voltage. The EMF ε is fixed by the battery's chemistry; it is the internal drop that eats into the usable voltage.

This model explains many practical observations: why headlights dim when you start a car (starter draws huge current, Ir increases, lights get less voltage); why a 'dead' battery reads fine on a voltmeter but fails under load (high internal resistance makes Ir large even for moderate I); and why short-circuiting a battery is dangerous (R → 0, I = ε/r can be enormous). The internal resistance model connects the abstract EMF concept to observable circuit behavior.