Current into a junction equals current out of it
- At any junction the sum of the currents entering equals the sum of the currents leaving.
- So a 0.90 A0.90\ \text{A}0.90 A current arriving at a junction and splitting into branches carrying 0.60 A0.60\ \text{A}0.60 A and 0.30 A0.30\ \text{A}0.30 A balances exactly.
- The reason is that charge is conserved: electrons are neither created nor destroyed at a junction.
- Charge also cannot pile up there, so whatever arrives each second has to leave each second.
- That gives you the standard route to a missing branch current: subtract the branches you know from the total.
P.d.s across series components add up to the supply p.d.
- The total p.d. across a set of components in series equals the sum of the individual p.d.s across each of them.
- Each coulomb is given energy by the source and then delivers it in stages, one component at a time.
- Since energy is conserved, the energy delivered across all the stages equals the energy the source handed out at the start.
- So a 6.0 V6.0\ \text{V}6.0 V supply feeding two lamps in series that drop 4.0 V4.0\ \text{V}4.0 V and 2.0 V2.0\ \text{V}2.0 V balances exactly.
- The component with the largest resistance takes the largest share, because V=IRV = IRV=IR and the current through each of them is the same.
Every parallel branch has the same p.d. across it
- The p.d. across a parallel arrangement is the same as the p.d. across any one branch in it.
- Both ends of every branch are joined to the same two junctions, so every branch spans the same pair of points.
- Each coulomb passes through only one branch and delivers all of its energy there, whichever branch it happens to take.
- So a 3 Ω3\ \Omega3 Ω and a 6 Ω6\ \Omega6 Ω resistor in parallel both have the full supply p.d. across them, but they carry different currents.
- The smaller resistance takes the larger current, because I=VRI = \dfrac{V}{R}I=RV with the same VVV on both.
Common Mistake
- Do not share the supply p.d. between parallel branches; each branch gets the whole of it.
- Do not share the current between series components; the same current passes through all of them.
- The junction rule is about charge being conserved and the series p.d. rule is about energy being conserved, so quote the right one.
- The larger branch current goes through the smaller resistance, which is easy to write down back to front under pressure.
Exam technique
- Label every current and p.d. on the diagram before you calculate, because a labelled diagram makes the missing quantity obvious.
- For an "explain" mark on the junction rule, say charge is conserved and that charge does not build up at the junction.
- In a mixed circuit, deal with the parallel section first, then treat it as a single resistance in the series loop.
- Finish by checking that your branch currents add to the supply current and your series p.d.s add to the supply p.d.
Self review
- State the rule linking the currents entering and leaving a junction, and explain why it holds.
- A supply current of 1.2 A1.2\ \text{A}1.2 A splits into two branches, one carrying 0.5 A0.5\ \text{A}0.5 A. Find the other.
- How do the p.d.s across components in series relate to the supply p.d.?
- What is the p.d. across each branch of a parallel arrangement connected to a 9.0 V9.0\ \text{V}9.0 V supply?
- Two resistors in series across 10 V10\ \text{V}10 V drop 6 V6\ \text{V}6 V across the first. Which resistor is larger, and what is the p.d. across the second?
