\(\def \u#1{\,\mathrm{#1}}\) \(\def \abs#1{\left|#1\right|}\) \(\def \ast{*}\) \(\def \deg{^{\circ}}\) \(\def \tau{\uptau}\) \(\def \ten#1{\times 10^{#1}}\) \(\def \redcancel#1{{\color{red}\cancel{#1}}}\) \(\def \BLUE#1{{\color{blue} #1}}\) \(\def \RED#1{{\color{red} #1}}\) \(\def \PURPLE#1{{\color{purple} #1}}\) \(\def \th#1,#2{#1,\!#2}\) \(\def \lshift#1#2{\underset{\Leftarrow\atop{#2}}#1}}\) \(\def \rshift#1#2{\underset{\Rightarrow\atop{#2}}#1}}\) \(\def \dotspot{{\color{lightgray}{\circ}}}\) \(\def \ccw{\circlearrowleft}\) \(\def \cw{\circlearrowright}\)
Chapter 16: Circuits
9.

Resistance Reduction

Why should we care about equivalent resistance? One reason is this: if we swap out one object with an object with the same equivalent resistance, then nothing changes as far as the rest of the circuit is concerned (in our simplified view of circuits, anyway).

Wideexample

(no alternate text)

In (a), we have two resistors wired in parallel with the 6V battery. Because the drop across both resistors is 6V, there is $I={6\u{V}\over 4\Omega}=1.5\u{A}$ through the 4Ω resistor and $I={6\u{V}\over 2\Omega} = 3\u{A}$ through the 2Ω resistor, and the total current through both is $4.5\u{A}$.

Now we draw a dotted line around the two resistors so that there are two terminals sticking out of the box. The equivalent resistance of these two parallel resistors is given by $\frac1{R_{eq}} = \frac1{4\Omega} + \frac1{2\Omega} = {3\over4\Omega}$ so $R_{eq} = {4\over3}\Omega$.

If we replace the set of resistors with a single resistor having the same resistance ${4\over3}\u{\Omega}$, then the current through it is $I={\Delta V\over R} = {6\u{V}\over (4/3)\u{\Omega}} = 4.5\u{A}$ which is the same current as we had before. As far as the battery is concerned, the circuit hasn't changed.

Resistance Reduction

You can use this process, called resistance reduction, to simplify a circuit. You might do this because you only care about the current out of the battery, or you might use it to solve for the current through the battery, let's say, which will then make it easier to solve for the rest of the circuit.

Here's the process:

  1. Find two resistors which are either in series or in parallel with each other.
  2. Replace that resistor with their equivalent resistance.
  3. Continue until you have a single resistor, or until you can no longer replace resistors.

Wideexample

We start with the circuit in (a). It may surprise you that there are no resistors in parallel here! Remember that "parallel" means "having the same potential drop", and has nothing to do with geometry. Also, while you may be tempted to say "the 3Ω resistor is in parallel with the 5Ω AND 1Ω resistor", that's not how we use the word parallel here: none of the resistors have the same drops, and so none of them in parallel.

However, the 5Ω and 1Ω resistors are in series, so we can replace them with a single resistor having the resistance $R_{eq} = 5\u{\Omega} + 1\u{\Omega} = 6\u{\Omega}$.

(b) Now the new 6Ω resistor is in parallel with the 3Ω resistor, so we can replace those with $R_{eq}$ where $\frac1{R_{eq}} = \frac1{3\u{\Omega}} + \frac1{6\u{\Omega}} = \frac1{2\u{\Omega}}$, which means that $R_{eq} = 2\u{\Omega}$.

(c) The new $2\u{\Omega}$ resistor is in series with the $4\u{\Omega}$ resistor, so we can replace it with a $2+4=6\u{\Omega}$ resistor.

(d) We're now down to a single resistor, so the equivalent resistance of the original set of resistors is $6\u{\Omega}$. If the battery has an emf of $6\u{V}$, then the current coming out of the battery would be $I={6\u{V}\over 6\u{\Omega}} = 1\u{A}$, and the power provided by the battery is $P=I\Delta V = (1\u{A})(6\u{V}) = 6\u{W}$.