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Equivalent resistance, derived rather than memorised

Equivalent resistance is the single resistor that a network of resistors could be replaced by without the rest of the circuit noticing. For two resistors in series and for two in parallel:

$$R_{\text{eq}} = R_1 + R_2 \qquad\text{and}\qquad \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}$$

Both extend to any number of resistors. Both are usually handed over as rules to memorise, and memorised rules are the ones that get applied to the wrong arrangement. Neither is an axiom; each follows in two lines from charge conservation and energy conservation, and the derivation is worth doing once because it tells you which quantity is shared and which one adds.

Where the two rules come from

Series means the resistors sit end to end on a single path, with no junction between them. Charge has nowhere else to go, so the same current $I$ passes through every one of them. That is the shared quantity. The potential drops add, because the energy a charge loses crossing the chain is the sum of what it loses crossing each piece:

$$V = V_1 + V_2 = IR_1 + IR_2 = I(R_1 + R_2)$$

Compare that with $V = IR_{\text{eq}}$ and the series rule is done. Resistance in series adds because the drops add and the current is common.

Parallel means both ends of every resistor are tied to the same pair of nodes. The work done on a charge moving between two points cannot depend on which route it takes, so every branch has the same potential difference $V$ across it. That is the shared quantity this time. The currents add, because charge arriving at the junction has to leave it:

$$I = I_1 + I_2 = \frac{V}{R_1} + \frac{V}{R_2} = V\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$

Compare with $I = V/R_{\text{eq}}$ and the parallel rule is done. Each rule uses both conservation laws, one to say what is common and one to say what adds; the two arrangements just swap which is which.

A ladder, from the inside out

Take a 12 V battery. From its positive terminal a resistor $R_1 = 2\ \Omega$ leads to a node; from that node two branches run in parallel to a second node, which returns to the battery. The first branch is $R_2 = 4\ \Omega$ in series with a parallel pair, $R_3 = 6\ \Omega$ and $R_4 = 3\ \Omega$. The second branch is a single $R_5 = 12\ \Omega$.

Reduce it from the innermost combination outward. The pair $R_3$ and $R_4$ share a node at each end, so they are parallel:

$$R_{34} = \left(\frac{1}{6} + \frac{1}{3}\right)^{-1} = \left(\frac{1}{6} + \frac{2}{6}\right)^{-1} = \left(\frac{1}{2}\right)^{-1} = 2\ \Omega$$

That 2 $\Omega$ now sits end to end with $R_2$, carrying the same current, so it is series:

$$R_{\text{branch}} = 4 + 2 = 6\ \Omega$$

That 6 $\Omega$ branch and the 12 $\Omega$ branch span the same two nodes, so they are parallel:

$$R_{\text{par}} = \left(\frac{1}{6} + \frac{1}{12}\right)^{-1} = \left(\frac{3}{12}\right)^{-1} = 4\ \Omega$$

And that sits in series with $R_1$:

$$R_{\text{eq}} = 2 + 4 = 6\ \Omega$$

Six ohms across 12 V draws $I = 12/6 = 2$ A.

Check it by walking the current back through the network. $R_1$ carries the whole 2 A, dropping $2 \times 2 = 4$ V, which leaves 8 V across the parallel section. The 12 $\Omega$ branch therefore carries $8/12 = 0.667$ A and the 6 $\Omega$ branch carries $8/6 = 1.333$ A. Those sum to 2.000 A, which they must, since the two branches are the only routes back to the battery. Inside the 6 $\Omega$ branch, $R_2$ drops $1.333 \times 4 = 5.333$ V, leaving 2.667 V across the $R_3$ and $R_4$ pair, so $R_3$ carries $2.667/6 = 0.444$ A and $R_4$ carries $2.667/3 = 0.889$ A. Those sum to 1.333 A again.

Check it a third way, on power, which is independent of all the branch bookkeeping. The battery delivers $P = VI = 12 \times 2 = 24$ W. Add up $I^2R$ over the five resistors: $8.000 + 7.111 + 1.185 + 2.370 + 5.333 = 24.00$ W. The energy the battery supplies is exactly the energy the resistors dissipate, so nothing was double-counted and no branch was dropped.

Two checks worth running every time

A parallel equivalent is always smaller than the smallest branch. Adding another path can only make it easier for charge to get across, never harder, and the algebra says the same thing: $1/R_{\text{eq}}$ is a sum of positive terms and so exceeds $1/R_{\text{min}}$, which makes $R_{\text{eq}} < R_{\text{min}}$. Above it was 2 $\Omega$ from a 6 and a 3, then 4 $\Omega$ from a 6 and a 12. Both pass.

There is a floor as well. Every branch resistance is at least $R_{\text{min}}$, so each reciprocal is at most $1/R_{\text{min}}$ and the sum of $n$ of them is at most $n/R_{\text{min}}$. A parallel combination of $n$ branches therefore lands between $R_{\text{min}}/n$ and $R_{\text{min}}$, and $n$ identical resistors of resistance $R$ give exactly $R/n$, the bottom of that window.

Series is the mirror image and even easier: the sum of positive numbers exceeds each one, so a series equivalent is always larger than the largest resistor in the chain.

The two ways this goes wrong

The first is adding parallel resistors as if they were in series. In the ladder above that turns the 6 and the 12 into 18 $\Omega$, giving $R_{\text{eq}} = 2 + 18 = 20\ \Omega$ and a supply current of $12/20 = 0.6$ A instead of 2 A. The answer is off by a factor of more than three, and in the wrong direction: adding a second path made the circuit draw less current, which is the opposite of what a second path does. The sanity check catches it before the arithmetic does, because 18 $\Omega$ is not smaller than 6 $\Omega$.

The second is computing $1/R_1 + 1/R_2$ correctly and then reporting that as the answer. Here $1/6 + 1/12 = 0.25$, and 0.25 $\Omega$ looks like a number, so it gets written down. It is a conductance in siemens, not a resistance, and the window test kills it instantly: two branches with a smallest member of 6 $\Omega$ must give between 3 and 6 $\Omega$. The reciprocal at the end is not decoration. Worked problems on the equivalent resistance skill nest the combinations deliberately so that the reciprocal comes up more than once per circuit, which is exactly where it gets forgotten.

Not every network yields to this. A bridge, in which a resistor connects the midpoints of two branches, has no pair of resistors that share both ends and no pair that carries a common current, so there is nothing to collapse. Those need Kirchhoff’s rules and a system of equations. Series and parallel reduction is the fast path, and part of using it well is recognising quickly when it does not apply. The simulations browser has circuits you can change a resistance in and watch the currents redistribute, which is the quickest way to see that the 12 $\Omega$ branch really does take half the current of the 6 $\Omega$ one.

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