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Physics Lesson 15.3.4 - Ohm's Law for the Whole Circuit

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Welcome to our Physics lesson on Ohm's Law for the Whole Circuit, this is the fourth lesson of our suite of physics lessons covering the topic of Electric Potential Difference (Voltage). Ohm's Law, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Ohm's Law for the Whole Circuit

The battery has its own internal resistance, which we denote as r. We can draw this conclusion when we touch a battery immediately after it has been operating. We can feel the hotness of battery, which indicates the conversion of electricity into heat inside the battery.

In addition, from the previous tutorial, we know that a conductor has its own internal resistance, which depends from the type of material, length and thickness of conductor. This causes a drop in the potential difference at the resistor. Look at the figure below.

Physics Tutorials: This image provides visual information for the physics tutorial Electric Potential Difference (Voltage). Ohm's Law

We can write for the relationship between potential differences and the electromotive force in the above circuit:

emf = ΔVbattery + ΔVAB + ΔVBC + ΔVCD

If we write the total resistance of the circuit by Rtot, we obtain (applying Ohm's Law):

emf=I ∙ r + I ∙ RAB + I ∙ R + I ∙ RCD

Given that RABCD = Rwire, we obtain

emf = I ∙ r + I ∙ R + I ∙ Rw

or

emf = I ∙ (r + R + Rw)

The above equation gives Ohm's Law for the whole circuit. We can write it in a shorter way as

emf = I ∙ Rtot

Example 4

A 20 m long copper wire of 4 mm2 thickness (ρ = 1.69 × 10-8 Ω ∙ m) is used to build an electrical circuit that contains a 24 V battery and a resistor. The battery has a 0.6 Ω internal resistance. What is the resistance of the resistor if the current flowing through the circuit is equal to 12A? Use the figure shown in the theory as a reference.

Solution 4

Clues:

L = 20 m
A = 4 mm2 = 4 × 10-6 m2
emf = 24 V
r = 0.6 Ω
I = 12 A
R = ?

First, we find the resistance of wire. We have

Rw = ρ ∙ L/A
= (1.69 × 10-8 Ω ∙ m) ∙ (20m)/4 × 10-6 m2 )
= 8.45 × 10-2 Ω
= 0.0845 Ω

Applying Ohm's Law for the whole circuit, we obtain

emf = I ∙ (r + R + Rw)
r + R + Rw = Rtot = emf/I
= 24 V/12 A
= 2 Ω

Thus,

R = Rtot - r - Rw
= 2 Ω - 0.6 Ω - 0.0845 Ω
= 1.3155 Ω

You have reached the end of Physics lesson 15.3.4 Ohm's Law for the Whole Circuit. There are 4 lessons in this physics tutorial covering Electric Potential Difference (Voltage). Ohm's Law, you can access all the lessons from this tutorial below.

More Electric Potential Difference (Voltage). Ohm's Law Lessons and Learning Resources

Electrodynamics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
15.3Electric Potential Difference (Voltage). Ohm's Law
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
15.3.1Recap on Electric Potential and Potential Difference of a Charged Object
15.3.2Potential Difference Produced by a Battery. Electromotive Force
15.3.3Ohm's Law for a Part of the Circuit
15.3.4Ohm's Law for the Whole Circuit

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