Electric Circuits Revision Notes for Grade 12 NSC Matric Physical Sciences

Evaluating Internal Resistance in Circuits

Understanding internal resistance

Internal resistance is a property that all real batteries possess. Unlike ideal batteries, real batteries have a small resistance inside them that opposes the flow of current. This internal resistance, denoted by r, causes the terminal voltage of a battery to be less than its electromotive force (emf) when current flows through the circuit.

The electromotive force (emf), represented by ε, is the maximum potential difference a battery can provide when no current flows. When current flows, some voltage is "lost" across the internal resistance, reducing the voltage available to the external circuit.

infoNote

Think of internal resistance as a "tax" that the battery charges on every unit of current flowing through it. The more current flows, the more voltage is "stolen" from the external circuit.

Key relationships and formulas

The fundamental relationship for circuits with internal resistance is:

$\varepsilon = V + Ir$

Where:

ε = emf of the battery (V)
V = terminal voltage across the battery (V)
I = current through the circuit (A)
r = internal resistance of the battery (Ω)

This equation tells us that the emf equals the terminal voltage plus the voltage drop across the internal resistance. When no current flows ( $I = 0$ ), the terminal voltage equals the emf.

The power dissipated by the internal resistance is calculated using:

$P = V_i I \text{ or } P = I^2r$

Where $V_i$ is the voltage drop across the internal resistance ( $V_i = Ir$ ).

chatImportant

Critical Concept: The emf is always greater than or equal to the terminal voltage. When current flows, some electrical energy is always "wasted" as heat in the internal resistance.

Approach to solving internal resistance problems

The approach to solving problems involving internal resistance builds on your existing knowledge of circuits. Think of the battery as an ideal emf source connected in series with a small resistor (the internal resistance). Then apply the same circuit analysis techniques you already know:

Identify the circuit configuration (series, parallel, or combination)
Calculate equivalent resistances where needed
Apply Ohm's law to find currents and voltages
Use the emf equation to relate terminal voltage, current, and internal resistance
Calculate power dissipation if required

infoNote

Key Strategy: Always treat the internal resistance as if it were just another resistor in series with the emf source. This mental model simplifies complex circuit analysis significantly.

Worked examples

lightbulbExample

Worked Example 1: Series circuit with internal resistance

Consider a circuit with a 23.0 V battery (internal resistance r = 0.1 Ω) connected to three resistors in series: $R_1 = 1.0$ Ω, $R_2 = 3.0$ Ω, and $R_3 = 7.5$ Ω. The current through the circuit is 2.0 A.

Step 1: Calculate the terminal voltage Since the resistors are in series, the current through each is the same (2.0 A).

Step 2: Find voltage drops across each resistor

$V_1 = IR_1 = (2.0)(1.0) = 2.0$ V
$V_2 = IR_2 = (2.0)(3.0) = 6.0$ V
$V_3 = IR_3 = (2.0)(7.5) = 15.0$ V

Step 3: Calculate the emf using the relationship The terminal voltage $V = V_1 + V_2 + V_3 = 2.0 + 6.0 + 15.0 = 23.0$ V Using $\varepsilon = V + Ir = 23.0 + (2.0)(0.1) = 23.2$ V

Step 4: Calculate power dissipated by internal resistance $P = I^2r = (2.0)^2(0.1) = 0.4$ W

lightbulbExample

Worked Example 2: Parallel circuit with internal resistance

A battery with terminal voltage 18 V is connected to two parallel resistors: $R_1 = 4.0$ Ω and $R_2 = 12$ Ω. The internal resistance is $r = 0.375$ Ω.

Step 1: Calculate the equivalent resistance of the parallel combination $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4.0} + \frac{1}{12} = \frac{1}{3.0}$ Therefore, $R = 3.0$ Ω

Step 2: Find the total current through the battery Using Ohm's law: $I = \frac{V}{R} = \frac{18}{3.0} = 6.0$ A

Step 3: Calculate current through each parallel resistor Since the voltage across each parallel resistor equals the terminal voltage:

$I_1 = \frac{V}{R_1} = \frac{18}{4.0} = 4.5$ A
$I_2 = \frac{V}{R_2} = \frac{18}{12} = 1.5$ A

Step 4: Determine the emf The voltage drop across the internal resistance: $V_r = Ir = (6.0)(0.375) = 2.25$ V Therefore: $\varepsilon = V + Ir = 18 + 2.25 = 20.25$ V

lightbulbExample

Worked Example 3: Complex series-parallel network

For complex circuits containing both series and parallel sections, break the problem down step by step:

Step 1: Identify parallel sections and calculate their equivalent resistances Step 2: Treat the circuit as a series combination of these equivalent resistances Step 3: Apply Ohm's law to find the total current Step 4: Work backwards to find individual currents and voltages Step 5: Apply the emf relationship

lightbulbExample

Worked Example 4: NSC exam question - Automotive circuit

This example shows a practical application with automotive tail lamps. The graph shows how the terminal voltage of the battery changes when additional load is connected, demonstrating the effect of internal resistance on circuit performance.

When only one lamp operates, the terminal voltage is 12 V. When both lamps operate (drawing more current), the terminal voltage drops to 9.6 V due to the increased voltage drop across the internal resistance.

Key insight: Higher current means greater voltage drop across internal resistance, reducing the voltage available to the external circuit.

Extension: Wheatstone bridge

The Wheatstone bridge is a special circuit configuration used to measure unknown resistance values precisely. It consists of four resistors arranged in a diamond pattern with a voltmeter connected between two opposite corners.

Principle: When the bridge is balanced (voltmeter reads zero), the ratio of resistances satisfies:

$R_x : R_3 = R_2 : R_1$

This can be rearranged to find the unknown resistance:

$R_x = \frac{R_2 \times R_3}{R_1}$

The bridge is balanced when no current flows through the voltmeter, which occurs when the potential difference between the two midpoints is zero.

infoNote

Bridge Balance Condition: The Wheatstone bridge is a powerful tool because it can measure resistance without needing to know the exact emf or internal resistance of the battery used.

chatImportant

Common Exam Tips:

Always check units in your calculations
Remember that emf is always greater than or equal to terminal voltage
Draw circuit diagrams to visualise complex problems
Work systematically from equivalent resistances to individual components
Check your answers by ensuring power conservation

bookmarkSummary

Key Points to Remember:

Internal resistance causes terminal voltage to be less than emf when current flows
The key relationship is $\varepsilon = V + Ir$
Treat internal resistance as a series resistor with the emf source
Higher current through internal resistance means greater voltage drop and power loss
The Wheatstone bridge provides a precise method for measuring unknown resistance

Electric Circuits (Grade 12 NSC Matric Physical Sciences): Revision Notes

Evaluating Internal Resistance in Circuits

Understanding internal resistance

Key relationships and formulas

Approach to solving internal resistance problems

Worked examples

Extension: Wheatstone bridge

Explore Grade 12 NSC Matric Physical Sciences Model Answers by Topics

Momentum and Impulse

Vertical Projectile Motion in One Dimension

Work, Energy, and Power

Doppler Effect