# Electrical flow in a simple parallel circuit

I'm having trouble understanding something in one of my text books:

Let’s have a look at the implications of each circuit configuration. Figure 3.13 shows the Conventional representation of a parallel circuit. If you assume that the resistance of the wires can be neglected, then the voltage drop across each bulb is equal to the e.m.f. of the source, the dynamo. The Current flowing from the source is divided between each bulb depending on its resistance (remember I = E/R), Removing one of the bulbs would not affect the voltage drop across the other bulb and would therefore not affect the Current in it, although the overall Current from the dynamo would drop as the demand has been reduced.

Specifically, the line about removing one of the bulbs not affecting the voltage drop. My understanding is that the sum of voltage drops in a circuit must be equal to the output of the emf, but the way I'm reading the text suggests that if you have three bulbs in a parallel circuit and you remove one of them, the voltage drop remains the same - this is what I don't understand, though. If you a lamp is removed, then that's one less lamp consuming emf. Does it not get redistributed to the other two lamps?

Likewise, the text mentions the overall current from the dynamo dropping because demand has been reduced.

My initial understanding was that current from an emf is only affected by the emf - not the demands of components further along the circuit, as in this case. What have I misunderstood?

## 1 Answer

To (hopefully) answer both your questions simultaneously, think of the concept this way: let's say I have an electrical circuit which consists of a battery (your EMF) connected by wires to some unknown electrical setup within a black box. Nothing appears to be melting or catching on fire within sight (which would imply the likely existence of a short circuit or voltage conflict somewhere on the apparatus, as might occur if we connected a battery's positive terminal to its negative terminal using wires with basically no resistance); beyond that, however, we don't know anything else about the circuit and, worse, you don't seem to have any instruments in sight (i.e. oscilloscopes) for making useful measurements of the circuit's electrical properties. What, if anything, do we know about this circuit?

The answer is this: regardless of whatever setup of resistors, capacitors, inductors, and God-knows-what-else is inside that black box, the wire at the positive terminal of the battery must have a voltage greater than that of the negative end of the battery by the size of the EMF (i.e. if it's a 9V battery, the wire attached to the positive terminal will have a voltage 9V higher than the wire attached to the negative terminal).

That's it! No matter the configuration of the circuit in between the two terminals of the battery, that initial voltage difference is guaranteed to hold true (again, excepting the bizarre cases of short circuits/voltage conflicts, and as you may learn if you go on to learn some more advanced E&M, the case of induced EMFS from fluctuating magnetic fields...but that doesn't sound like something you need to worry about for right now!).

So to return to your original questions, I think you have your assumptions backwards in both questions. If you use a battery to power three lamps in parallel and remove one, the voltage across each lamp will not drop; both lamps are still connected to the same wires which are connected to the same terminals of the same battery (which we assume to provide a constant EMF). The current, on the other hand, absolutely can change if you alter the configuration of circuit components powered by one battery. If you need a real-world example, think about how plugging too many appliances into one outlet can blow out the circuit: too much current is getting drawn by too many devices in parallel!

If all else fails and this (massive) explanation still doesn't make sense, consult the case of resistors in series vs. resistors in parallel...I think you may be confusing the latter with the former. Hope this all helps!

• Very nice explanation :-) – David Z Jan 21 '12 at 3:27
• Hi Steve - thanks for your reply! What I'm still having trouble with is the idea that, if our battery is 9V and there are three lamps in circuit, we can say that the voltage drop for each is ~3V, but if we remove one of them, the voltage drop does not increase to 4.5V, but as the text implies, the voltage drop remains the same. This leaves me wondering where the 'missing' 3V goes and why it doesn't get redistributed amongst the other two lamps?! – James Jan 21 '12 at 11:45
• @James: The error in your comment is that the voltage drop for each is 3V only if they were in a series circuit. This is a parallel circuit, which is not the case where that rule applies. Think about this: Given some reference 0V, you can assign a specific voltage to every node (i.e. wire/junction) in the circuit. Voltage drops are the differences between nodes. In your example, there are only two nodes, so the drop across all of the components must be 9V (or -9V if you take the difference in the other direction). This is equivalent to Kirchhoff's voltage law — I suggest trying that out. – Kevin Reid Jan 21 '12 at 16:01