Combination of resistances

Series: same current, voltages add. When resistors are joined end-to-end so charge has only one path to follow, the same current $I$ passes through every resistor in turn. Each resistor develops its own share of the total potential difference, and since $V = V_1+V_2+V_3$ with $V_1=IR_1$ etc., dividing through by the common $I$ gives the series formula:

where: $R_s$ is the single equivalent resistance and $R_1, R_2, R_3, \dots$ are the individual resistors in the chain. Because every term adds positively, $R_s$ is always greater than the largest resistor in the group — stacking resistors end-to-end can only make the path harder for charge to travel.

Parallel: same voltage, currents add. When several resistors are connected between the same two points, each one feels the full potential difference $V$ of the source, but the total current $I$ from the source splits among them, with more current taking the lower-resistance branch. Since $I = I_1+I_2+I_3$ and $I_1 = V/R_1$ etc., dividing through by the common $V$ gives:

where: $R_p$ is the equivalent resistance of the parallel group. Adding a branch always opens a new path for current, so $R_p$ is always smaller than even the smallest individual resistor.

Mixed circuits. Most real diagrams combine both — say, two resistors in parallel, with that block wired in series to a third. The rule is always the same: spot a clearly-grouped series or parallel sub-set, collapse it to one equivalent resistor, redraw the (now simpler) circuit, and repeat until a single resistance remains.

Two resistors of $6\ \Omega$ and $3\ \Omega$ are connected in parallel, and this parallel block is connected in series with a $4\ \Omega$ resistor across a $12\ \text{V}$ battery of negligible internal resistance. Find the total current drawn from the battery and the current through each parallel branch.

Step 1. Reduce the parallel pair first, since it is the clearly-grouped sub-circuit. Using the reciprocal formula, $\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{6}+\dfrac{2}{6} = \dfrac{3}{6} = \dfrac{1}{2}$, so $R_p = 2\ \Omega$ (the product-over-sum check gives the same value: $6\times3/(6+3) = 2\ \Omega$).

Step 2. Redraw the simplified circuit. The $2\ \Omega$ parallel block is now in series with the $4\ \Omega$ resistor, so $R_{\text{total}} = 2 + 4 = 6\ \Omega$.

Step 3. Apply Ohm's law to the whole circuit to get the total current: $I = V/R_{\text{total}} = 12/6 = 2\ \text{A}$. This full $2\ \text{A}$ flows through the $4\ \Omega$ resistor, since it is the single path before the current reaches the parallel block.

Step 4. Split the current across the parallel branches. The parallel block has $V_p = I \times R_p = 2 \times 2 = 4\ \text{V}$ across it (check: $12 - I \times 4 = 12 - 8 = 4\ \text{V}$, consistent). So the branch currents are $I_6 = 4/6 = 0.67\ \text{A}$ and $I_3 = 4/3 = 1.33\ \text{A}$, and indeed $0.67 + 1.33 = 2\ \text{A}$, matching Step 3.

Answer: total current $= 2\ \text{A}$, splitting into $0.67\ \text{A}$ through the $6\ \Omega$ branch and $1.33\ \text{A}$ through the $3\ \Omega$ branch — more current through the smaller resistance, as expected.

• Adding parallel resistances directly like series ones (writing $R_p = R_1+R_2$) — the parallel rule needs reciprocals throughout, and forgetting to invert the final sum to recover $R_p$ is the single most common slip.
• Believing $R_p$ can exceed the smallest branch resistance — physically, every extra parallel path can only make current flow more easily, so $R_p$ must always be smaller than every individual resistor in the group.
• Mixing up which quantity is shared — in series it is current that is common while voltage divides; in parallel it is voltage that is common while current divides.
• Trying to solve a mixed circuit in one jump instead of reducing it stage by stage — always collapse the clearly-grouped series or parallel sub-part first, redraw, then continue.

Ask the tutor: Want to practise spotting which parts of a tricky mixed-circuit diagram are in series versus parallel, or try a few more product-over-sum shortcuts?