Components, Ohm's law & resistor networks

Components

Current, voltage and resistance

• Current (I) is the rate of flow of electric charge, measured in amps (A). Electrons flow from − to +, but circuit diagrams use conventional current, from + to −. The arrow on a diode or transistor symbol points the conventional way.
• Voltage (V), or potential difference, is the energy given to each coulomb of charge, measured in volts (V). A 1.5 V cell gives 1.5 J to every coulomb.
• Resistance (R) opposes the current, measured in ohms (Ω). Metals are good conductors; plastic, rubber and glass are insulators.
• Direct current (d.c.) flows one way only and comes from batteries and low-voltage supplies (usually 3–12 V). Alternating current (a.c.) keeps reversing direction: UK mains is 230 V a.c.
• An ammeter goes in series (the current must flow through it). A voltmeter goes in parallel, across the component.

Switches, diodes and LEDs

• Push-to-make (PTM) switches are normally open (NO): current flows only while pressed. Push-to-break (PTB) switches are normally closed (NC).
• A microswitch has three terminals: common (C), NO and NC. With a roller lever it works as a limit switch that detects position, e.g. a barrier reaching the top or a lift door closing.
• SPST (single pole, single throw) is a simple on/off switch. A DPDT (double pole, double throw) switch can reverse a motor by swapping its connections.
• A diode lets current flow one way only, from anode to cathode. A protection diode across a relay coil protects the transistor when the coil switches off.
• An LED gives light when current flows through it the right way (the cathode is the short leg, next to the flat on the case). It needs a series resistor to limit the current: R = (V_s − V_LED) ÷ I. E.g. a 2 V LED that needs 20 mA on a 9 V supply: R = (9 − 2) ÷ 0.02 = 350 Ω, so choose the next value up (390 Ω).

Ohm's law and power

• V = I R (volts = amps × ohms).
• P = V I, also P = I² R and P = V² ÷ R (watts).
• Convert kΩ to Ω (4.7 kΩ = 4700 Ω) and mA to A (18 mA = 0.018 A).

Series and parallel resistors

• Series: R_T = R₁ + R₂ + … The same current flows through each.
• Parallel: 1/R_T = 1/R₁ + 1/R₂ + …, or for two resistors R_T = (R₁ × R₂) ÷ (R₁ + R₂). The total is always less than the smallest resistor. Each branch has the same voltage.
• For mixed networks, work out the parallel part first, then add the series parts.

Effect questions: if a resistance increases in a series circuit, the current falls and the voltage across the other resistors falls.

Kirchhoff's laws and combined circuits

• Kirchhoff's first law (parallel): the current splits at a junction, and the branch currents add up to the total: I_T = I₁ + I₂ + …
• Kirchhoff's second law (series): the voltages around a series loop add up to the supply: V_s = V₁ + V₂ + …
• Each branch of a parallel circuit gets the full supply voltage.

Worked example: a 24 Ω resistor in series with 10 Ω and 48 Ω in parallel, on 12 V.

1. Parallel part: (10 × 48) ÷ (10 + 48) = 8.28 Ω. Total R_T = 24 + 8.28 = 32.3 Ω.
2. Circuit current: I = 12 ÷ 32.3 = 0.372 A.
3. Voltage across the 24 Ω: 0.372 × 24 = 8.92 V.
4. Voltage across the parallel pair (second law): 12 − 8.92 = 3.08 V.
5. Current in the 10 Ω: 3.08 ÷ 10 = 0.308 A; in the 48 Ω: 3.08 ÷ 48 = 0.064 A. Check (first law): 0.308 + 0.064 = 0.372 A.

Design challenge: bike light circuit

The brief. A Dundee cycling club wants a rear light with two LEDs in parallel, each with a 220 Ω resistor, run from a 6 V battery pack.

1. Draw the circuit using correct symbols.
2. Calculate the total resistance of the two resistors in parallel.
3. Calculate the total current drawn from the battery.
4. Explain why the LEDs are wired in parallel rather than series.

Circuit worked examples

• Series: 10 Ω and 17 Ω on 9 V: R_T = 27 Ω, I = 9 ÷ 27 = 0.33 A, V₁ = 0.33 × 10 = 3.3 V, V₂ = 5.7 V. Check: 3.3 + 5.7 = 9 V.
• Two in parallel: product over sum. 12 Ω and 8 Ω: R_T = (12 × 8) ÷ (12 + 8) = 4.8 Ω. On 12 V: I = 2.5 A, splitting into 1 A and 1.5 A. Check: the branch currents add up to the total.
• Convert prefixes first: 30 mA = 0.03 A, 1.5 kΩ = 1500 Ω.
• Practise building circuits in Yenka to check your answers.

Why simulate circuits?

Explain one advantage with cause and effect: components can't be damaged in a simulation (cause), so testing costs less (effect); components can be swapped instantly, so designs are changed and retested quickly; faults are highlighted, so they are easier to correct; it's safer, as faults don't cause real hazards. Software such as Yenka is used in class and in the assignment.

Key terms

Quick check