Why is it $\Delta V$ (as opposed to $ V$) in these problems?

Question

Maybe this is a vague (and also really basic) question but I'm hoping some of you guys understand it. If not I will provide more details. Just trying not to make this a specific homework question.

So in physics we've been learning the basics of circuit stuff. We are using equations like $\triangle V=IR$ and $P=I\triangle V$. These are the equations that the book gives. However, my teacher wrote down $V=IR$ and . I will admit I haven't paid attention in a while, but this is kind of tripping me up since it's really basic.

Because of this, I'm getting confused in these problems. So why is it $\triangle V$? It makes no sense to me, because in this problem I'm supposed to use the terminal voltage of a battery as $\triangle V$.

I hope this isn't a dumb/confusing question.

Answer 1

The key concept is potential difference: that is, the amount of current that flows depends on the different in voltage between the two terminals of the resistor. If I have a battery with voltage $V$ in series with two resistors of 10 and 20 Ohm respectively, then we know the current through the two resistors has to be the same (because they are in series). It follows then that the voltage across one resistor must be different than the voltage across the other resistor.

But if the only voltage we know about is $V$, how can that be??

The answer is - it's not the voltage of the battery that matters to the individual resistors: only the voltage across its terminals. The voltage across the 10 Ohm resistor will end up being $\frac13 V$, while the voltage across the other is $\frac23 V$. So for one resistor, $\Delta V = \frac13 V$ and for the other, it will be $\Delta V = \frac23 V$.

Often people won't write the $\Delta$ because it's assumed that everyone remembers how electricity works. But sometimes you get a question like this, and you are reminded that no, not everyone knows - yet.

Let me know if that clears it up.

Answer 2

Common usage can sometimes be a curse and cause problems in the understanding of a topic.
In this case the term voltage being used as a synonymy for both potential and potential difference.+

A simple circuit illustrates this point.

Analysis of such a circuit will yield the following results.

The potential at node $C$ is defined with the inverted T symbol as being $0 \rm V$.
The potential at node $A, V_A = 4 \rm V$ and the potential at node $B, V_B = 3 \rm V$ with a current $I = 1 \rm A$ flowing through the circuit.

Potential difference across the voltage source $V_{ \rm AC} = 4 \rm V$, potential difference across the $1 \Omega$ resistor $V_{ \rm AB} = 1 \rm V$ and potential difference across the $3 \Omega$ resistor $V_{ \rm AB} = 3 \rm V$ assuming that $V_{\rm AC}$ means the potential of node $A$

You will note that the labelling in the circuit diagram above has $V_{\rm AB}$ rather then $\Delta V$ or $\Delta V_{\rm AB}$ but all are equivalent and can be put into an equation which links "voltage" to current and resistance.

Furthermore $V_{\rm AB} = V_{\rm A} - V_{\rm B} = \Delta V_{1 \Omega} = \Delta V = V$ all are used to mean the same thing.

Probably the best way forward is to avoid the term voltage and use the term potential difference and apply the definition of resistance $V=IR$ to circuit problems with appropriate subscripts to $V$ (or $\Delta V$) if the circuit is non trivial. So the cyclic nature of the subscripts when applying Kirchhoff's second law to the circuit can be used as check. $V_{\rm AC} - V_{\mathbf {AB}} - V_{\mathbf {BC}} = 0$

The use of $\Delta V$ for potential difference seems to be more common in electrostatics than it is in current electricity where $V$ tends to be used.

I must confess that I have only use $\Delta V$ in current electricity in the context of incremental resistance $\frac {\Delta V}{\Delta I}$ for non linear circuit elements.