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Spec mapping: OCR H556 Module 4.3 — Electrical circuits (resistors in series Rseries=∑Ri and in parallel 1/Rparallel=∑1/Ri; current and voltage divisions; ammeter vs voltmeter placement; mixed networks reducible by sequential series-parallel collapse). Refer to the official OCR H556 specification document for exact wording.
Every realistic electronic circuit is a combination of series and parallel resistor segments. To analyse them cleanly you need three things: (i) compact formulas for combining resistors, (ii) intuition for how current and voltage distribute across a configuration, and (iii) a systematic "reduce and re-expand" workflow for mixed networks. This lesson covers all three from Kirchhoff-1 and Ohm's law, gives worked examples of series, parallel and mixed circuits, and connects forward to potential dividers (lesson 12) and Kirchhoff-2 (lesson 11).
By the end you should be able to (a) compute total resistance and main current for any reducible network without hesitation; (b) identify when a network is not reducible (lesson 11 territory); and (c) get current/pd distributions correct first time without algebraic slips.
Two or more components are in series if they form a single path with no junctions between them — current must flow through each in turn. The defining property: the same current flows through every series component.
flowchart LR
PLUS["+"] --> R1["R_1"]
R1 --> R2["R_2"]
R2 --> R3["R_3"]
R3 --> MINUS["-"]
By Kirchhoff-1 at any junction within a series chain, the current is the same everywhere:
I=I1=I2=I3=…
Once you have the main current I, every component carries that same I and the pd across each follows from V=IR.
The pds across the components add up to the source pd (Kirchhoff-2, lesson 11):
Vtotal=V1+V2+V3+…
Apply V=IR to each component with the same I: V1=IR1, etc. Sum: Vtotal=I(R1+R2+R3). But also Vtotal=IRtotal, so:
Rseries=R1+R2+R3+…
Series resistances add directly. No reciprocals. A series chain of identical resistors R has total resistance nR.
R1=100 Ω, R2=220 Ω, R3=470 Ω in series across 9.0 V (negligible internal resistance).
Check: V1+V2+V3=9.01≈9.0 V ✓. The largest resistor drops the largest pd; the pd ratio matches the resistance ratio. This is the heart of the potential divider (lesson 12).
Two or more components are in parallel if they share the same pair of nodes. Defining property: the same pd exists across every parallel component.
flowchart LR
PLUS["+"] --> J1((Top node))
J1 --> R1["R_1"]
J1 --> R2["R_2"]
J1 --> R3["R_3"]
R1 --> J2((Bottom node))
R2 --> J2
R3 --> J2
J2 --> MINUS["-"]
V=V1=V2=V3=… — same across every branch.
Kirchhoff-1 at the top node: Itotal=I1+I2+I3+…
Apply I=V/R to each branch with the same V: Itotal=V(1/R1+1/R2+1/R3). But Itotal=V/Rparallel, so:
Rparallel1=R11+R21+R31+…
Reciprocals add. Parallel total is always less than the smallest individual resistor — adding a parallel branch always reduces total resistance.
Rparallel=R1+R2R1R2(two resistors only)
For three or more, use the reciprocal formula.
Same three resistors 100 Ω, 220 Ω, 470 Ω in parallel across 9.0 V.
Rparallel1=1001+2201+4701=0.01667 Ω−1⇒Rparallel=60.0 Ω
Three observations: (i) Rparallel=60 Ω<100 Ω. (ii) Smallest R carries largest I. (iii) Parallel draws 13× more current than the same resistors in series.
| Configuration | Same quantity | Distributed | Rule |
|---|---|---|---|
| Series | I | V | Vi∝Ri |
| Parallel | V | I | Ii∝1/Ri |
"Series: same I, voltage splits. Parallel: same V, current splits."
A 12 V battery feeds R1=4.0 Ω in series with the parallel pair R2=6.0 Ω, R3=3.0 Ω.
R1=2 Ω + R2=3 Ω in series; that pair in parallel with R3=6 Ω; the whole in series with R4=4 Ω.
A galvanometer of internal resistance rg=50 Ω deflects full-scale at Ifs=1.0 mA. To convert it into an ammeter reading 0 – 1.0 A full-scale, a shunt resistor Rs is placed in parallel with the galvanometer to bypass most of the current. Find the shunt resistance.
At full-scale: total current Itotal=1.0 A; galvanometer current Ig=1.0 mA. Shunt current: Is=Itotal−Ig=0.999 A.
The pd across the parallel combination is V=Igrg=0.001×50=0.050 V. This same pd appears across the shunt:
Rs=V/Is=0.050/0.999=0.0500 Ω
The shunt carries 99.9% of the current; the galvanometer carries 0.1%. To extend the range further, decrease Rs. This is exactly how a multimeter's "current range" rotary switch works — different shunt resistors are switched in for different full-scale ranges.
The same galvanometer (full-scale 1.0 mA, rg=50 Ω) is converted into a voltmeter reading 0 – 10 V by placing a multiplier resistor Rm in series with the galvanometer.
At full-scale: I=1.0 mA through the series chain (rg+Rm); applied pd V=10 V. So:
Rm=V/I−rg=10/0.001−50=10000−50=9950 Ω
A digital multimeter's voltmeter ranges work by switching different precision multiplier resistors into series. The dual roles — shunt for ammeter, multiplier for voltmeter — reflect the fundamental insight: ammeters need low effective resistance (parallel shunt drops the effective R to a tiny value), voltmeters need high effective resistance (series multiplier raises the effective R to a huge value).
A delta network of three 30 Ω resistors (one between each pair of three nodes A, B, C) is sometimes more convenient to analyse if converted to an equivalent Y network (also called a "star"). The star has three resistors meeting at a central fourth node, each connecting to one of A, B, C.
For three identical delta resistors RΔ=30 Ω, the equivalent star resistors are each:
RY=3RΔ=10 Ω
The transformation preserves the impedance seen between any pair of external nodes. Some otherwise-irreducible networks (delta-Y bridges, certain three-phase circuits) become reducible after a star-delta transformation. Beyond A-Level in detail, but the existence of this trick is useful general knowledge.
For any network: ∑Pi=VsupplyItotal.
Example 3 powers: P1=I2R1=16 W; P2=I22R2=2.67 W; P3=I32R3=5.33 W. Sum =24 W =VI ✓.
If powers do not sum, you have an arithmetic error somewhere in the analysis. Performing this cross-check on every multi-step problem catches algebraic slips that would otherwise be missed — and OCR mark schemes generally accept the conservation check as evidence of physical reasoning even if numerical work is rounded.
A useful intuition: in a series chain the current is fixed, so P=I2R grows linearly with R. The largest resistor dissipates the most power. In a parallel combination the voltage is fixed, so P=V2/R falls inversely with R. The smallest resistor dissipates the most power.
This explains many practical observations: in a series chain of bulbs (Christmas-tree lights, traditional design) the dimmer bulbs (higher R, lower-wattage) actually shine brighter — counter-intuitive but follows directly from P=I2R. In a parallel arrangement (modern domestic wiring), each bulb sees the full mains voltage and dissipates its rated power independently of the others.
flowchart LR
PS["Battery"] --> SW["Switch"]
SW --> AM["Ammeter (series, R → 0)"]
AM --> COMP["Component"]
COMP --> PS
COMP -.-> VM["Voltmeter (parallel, R → ∞)"]
Real meters: ammeter 0.01 – 0.1 Ω; digital voltmeter 107 – 109 Ω. Mixing the two roles up is a wiring disaster.
When a non-ideal meter is connected, it perturbs the circuit it is measuring — a "loading error". For a high-resistance circuit, the voltmeter's finite input resistance draws a small but non-zero current, reducing the measured pd below the true open-circuit value (lesson 12 covers this for potential dividers). For a low-resistance circuit, the ammeter's finite resistance adds to the series impedance, reducing the measured current below the true value (lesson 9 mentions this in the context of measuring internal resistance).
Rule of thumb: the meter perturbs the measurement by less than ∼1% if its input impedance (voltmeter) or output impedance (ammeter) differs from the local circuit impedance by at least a factor of 100. Modern digital multimeters easily satisfy this for everyday A-Level work but precision laboratory measurements (microvolt-level signals, nanoampere currents) require careful meter selection.
A Wheatstone bridge (four resistors in a diamond with a galvanometer across the diagonal) cannot be reduced by series-parallel rules. Use Kirchhoff-2 (lesson 11). Tell-tale sign: cannot label every resistor as "in series with X" or "in parallel with Y" without ambiguity.
Consider a square network with one resistor on each side (four sides — R1, R2, R3, R4 — plus a fifth resistor R5 connecting two diagonally opposite corners). Trying to label by series/parallel rules: R5 is in neither series nor parallel with any single side; it sits across a pair of nodes whose voltage relationship to the source nodes depends on the combined network. No purely local series-parallel reduction works.
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