A level physics

Gustav Robert Kirchhoff (1824 – 1887) was a pioneer in the study of the radiation given off by hot objects and was the first person to use the term ‘black body radiation’. He also made groundbreaking contributions to what was then the ‘new’ science of spectroscopy.

High school students first encounter his name when studying electric circuits. Kirchhoff developed laws which describe the behaviour of electric circuits and, rightly, these laws still bear his name.

Newton needed three laws to explain the whole of motion; Kirchhoff needed only two to explain the behaviour of all circuits.

Kirchhoff’s First Law (aka KCL or Kirchhoff’s Current Law)

This law is a consequence of the Principle of Conservation of Electric Charge.

The algebraic sum of all the currents flowing through all the wires in a network that meet at a point is zero
Oxford Dictionary of Physics (2015)

‘Algebraic sum’ means that we must take account of whether the electric currents are positive or negative; or, in other words, their direction.

This can be stated more simply as: the sum of electric currents flowing into a junction is equal to the sum of the electric currents flowing out of the junction.

Kirchhoff’s Second Law (aka KVL or Kirchhoff’s Voltage Law)

This law is a consequence of the Principle of Conservation of Energy.

The algebraic sum of the e.m.f.s within any closed circuit is equal to the sum of the products of the currents and the resistances in the various portions of the circuit.
Oxford Dictionary of Physics (2015)

This can be more directly understood as saying that in any closed loop of the circuit, the sum of the energies gained by the charge carriers as they pass through parts of the loop with a positive potential difference is equal to the sum of the energies lost by the charge carriers as they pass through parts of the circuit with electrical resistance.

In other words, the sum of the positive potential differences (e.m.f.s) is equal to the sum of the negative potential differences around any closed loop of a circuit.

Applying Kirchhoff’s Laws to a circuit problem

*Find the current flowing through each resistor in this circuit*

Applying Kirchhoff’s First Law

But wait — is the current I₂ flowing in the right way? Surely it should be coming out of the positive terminal of the 7 V battery, no?

Perhaps. The direction of I₂ is simply my semi-educated guess at its direction given the relative strength of the 27 V cell and the 7 V cell.

But the happy truth of using Kirchhoff’s First Law is that . . . it doesn’t matter. Even if we have guessed the direction wrong, after we have gone through the process all that will happen is that we will get the correct numerical value for I₂ but our mistake will be revealed by the fact that it will have a negative value.

If we look at the junction highlighted in yellow, we can see that the algebriac sum of currents is: I₁ – I₂ – I₃ = 0.

We can rewrite this as: I₁ = I₂ + I₃.

And that’s about as far as we can get using just Kirchhoff’s First Law. We have three unknowns so somehow we need to obtain two more independent expressions of their relationships to solve this circuitous conundrum.

Luckily, we still have Kirchhoff Second Law to bring into play…

Applying Kirchhoff’s Second Law (Part 1 of 2)

Before starting, I find it immensely helpful to indicate which ends of the components have a positive potential and which have a negative potential. This process is part of a general maxim that I try to apply to all areas of physics problem solving: why think hard when your diagram can do the thinking for you? (See here for a similar process applied to dynamics problems.)

*An example of ‘Why think hard when the diagram will do it for you?’*

Going around loop L1 in the direction shown by the arrow:

The emf 27 V will be positive as the potential increases (as we are moving from – to +).
I₃R₃ will be negative as the potential decreases (as we are moving from + to -).
I₁R₁ will be negative as the potential decreases (as we are moving from + to -).

This gives us:

Applying Kirchhoff’s Second Law (Part 2 of 2)

*Yet another example of ‘Let the diagram do the hard thinking for you!’*

Going around Loop L2 we find that:

The emf 7 V will be positive as the potential increases (as we are moving from – to +).
I₂R₂ will be positive as the potential increases (as we are moving from – to +).
I₃R₃ will be negative as the potential decreases (as we are moving from + to -).

This gives us:

The rest is history algebra

Going through a rather involved process of using simultaneous equations for solving for I₁, I₂ and I₃ . . .

(NB There’s probably a quicker way than the way I chose, but I got there in the end and that’s the important thing. AND I guessed the direction of I₂ correctly. *Pats himself on the back*).

Conclusion

Kirchhoff’s Laws: I hope you give them a spin, whether or not you decide to use the procedure outlined above 🙂

Nuclear binding energy and binding energy per nucleon are difficult concepts for A-level physics students to grasp. I have found the ‘pool table analogy’ that follows helpful for students to wrap their heads around these concepts.

Background

Since mass and energy are not independent entities, their separate conservation principles are properly a single one — the principle of conservation of mass-energy. Mass can be created or destroyed , but when this happens, an equivalent amount of energy simultaneously vanishes or comes into being, and vice versa. Mass and energy are different aspects of the same thing.
Beiser 1987: 29

E = mc²

There, I’ve said it. This is the first time I have directly referred to this equation since starting this blog in 2013. I suppose I have been more concerned with the ‘andallthat‘-ness side of things rather than E=mc². Well, no more! E=mc² will form the very centre of this post. (And about time too!)

The E is for ‘rest energy’: that is to say, the energy an object of mass m has simply by virtue of being. It is half the energy that would be liberated if it met its antimatter doppelganger and particles and antiparticles annihilated each other. A scientist in a popular novel sternly advised a person witnessing an annihilation event to ‘Shield your eyes!’ because of the flash of electromagnetic radiation that would be produced.

Well, you could if you wanted to, but it wouldn’t do much good since the radiation would be in the form of gamma rays which are to human eyes what the sound waves from a silent dog whistle are to human ears: beyond the frequency range that we can detect.

The main problem is likely to be the amount of energy released since the conversion factor is c²: that is to say, the velocity of light squared. For perspective, it is estimated that the atomic bomb detonated over Hiroshima achieved its devastation by directly converting only 0.0007 kg of matter into energy. (That would be 0.002% of the 38.5 kg of enriched uranium in the bomb.)

Matter contains a lot of energy locked away as ‘rest energy’. But these processes which liberate rest energy are mercifully rare, aren’t they?

No, they’re not. As Arthur Beiser put it in his classic Concepts of Modern Physics:

In fact, processes in which rest energy is liberated are very familiar. It is simply that we do not usually think of them in such terms. In every chemical reaction that evolves energy, a certain amount of matter disappears, but the lost mass is so small a fraction of the total mass of the reacting substances that it is imperceptible. Hence the ‘law’ of conservation of mass in chemistry.
Beiser 1987: 29

Building a helium atom

The constituents of a helium nucleus have a greater mass when separated than they do when they’re joined together.

Here, I’ll prove it to you:

The change in mass due to the loss of energy as the constituents come together is appreciable as a significant fraction of its original mass. Although 0.0293/4.0319*100% = 0.7% may not seem like a lot, it’s enough of a difference to keep the Sun shining.

The loss of energy is called the binding energy and for a helium atom it corresponds to a release of 27 MeV (mega electron volts) or 4.4 x 10^-12 joules. Since there are four nucleons (particles that make up a nucleus) then the binding energy per nucleon (which is a guide to the stability of the new nucleus) is some 7 MeV.

But why must systems lose energy in order to become more stable?

The Pool Table Analogy for binding energy

Imagine four balls on a pool table as shown.

The balls have the freedom to move anywhere on the table in their ‘unbound’ configuration.

However, what if they were knocked into the corner pocket?

To enter the ‘bound’ configuration they must lose energy: in the case of the pool balls we are talking about gravitational potential energy, a matter of some 0.30 J per ball or a total energy loss of 4 x 0.30 = 1.2 joules.

The binding energy of a pool table ‘helium nucleus’ is thus some 1.2 joules while the ‘binding energy per nucleon’ is 0.30 J. In other words, we would have to supply 1.2 J of energy to the ‘helium nucleus’ to break the forces binding the particles together so they can move freely apart from each other.

Just as a real helium nucleus, the pool table system becomes more stable when some of its constituents lose energy and less stable when they gain energy.

Reference

Beiser, A. (1987). Concepts of modern physics. McGraw-Hill Companies.

e=mc2andallthat

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Teaching Circuits using Kirchoff’s Laws