Ohm's law and the power relationships, and analysing series and parallel resistor networks for current, voltage, resistance and power.

What this key area is asking

The SQA wants you to apply Ohm's law and the power relationships, and to analyse series and parallel resistor networks to find currents, voltages, total resistance and power. These calculations underpin every input and output sub-system in the electronics area, so fluency here pays off throughout the course.

Ohm's law and power

Ohm's law links the three basic quantities; rearrange it as $I = V/R$ or $R = V/I$ as needed. The three power forms are all equivalent (substitute Ohm's law into $P = IV$ to get the others), so pick the one whose quantities you already have: use $P = I^2R$ when you know current and resistance, and $P = V^2/R$ when you know voltage and resistance. This avoids an extra step.

Series circuits

In a series circuit the components form a single loop, so there is only one path for the current.

• The current is the same through every component.
• The supply voltage is shared, so the voltages across the components add to the supply: $V_S = V_1 + V_2 + \dots$
• The total resistance is the sum: $R_T = R_1 + R_2 + \dots$

Because the current is common, the larger resistor drops the larger voltage and dissipates the larger power.

Parallel circuits

In a parallel circuit the components are connected across the same two points, giving several paths.

• The voltage is the same across every branch (equal to the supply, ignoring source resistance).
• The branch currents add to the total: $I_T = I_1 + I_2 + \dots$
• The total resistance combines by reciprocals: $\dfrac{1}{R_T} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$, and is always less than the smallest branch resistance.

For two resistors the reciprocal rule can be written $R_T = \dfrac{R_1 R_2}{R_1 + R_2}$, the product over the sum.

Putting it together

The standard method for any network is: combine resistors into a single equivalent total (parallel blocks first, then add series), use $I = V/R_T$ for the supply current, then work back through the circuit applying the series rule (same current) and parallel rule (same voltage) to find individual currents and voltages, and finally a power form for any dissipation asked for. Working from a clearly redrawn circuit prevents most errors.

Examples in context

These rules decide practical choices. House wiring uses parallel circuits so every appliance gets the full mains voltage and can be switched independently; if it were series, switching one off would break the loop for all. The total current drawn rises as more parallel appliances are switched on (currents add), which is why circuits are fused: too many high-power devices in parallel draw a total current that would overheat the wiring, and $P = I^2R$ in the cable resistance is exactly the heating the fuse guards against.

Try this

Q1. A 4 V supply drives 2 A through a resistor. Find its resistance. [2 marks]

• Cue. $R = V/I = 4/2 = 2\ \Omega$.

Q2. Two 10 ohm resistors are in series. State the total resistance. [1 mark]

• Cue. $R_T = 10 + 10 = 20\ \Omega$.

Q3. A resistor carries 0.5 A and has resistance 8 ohms. Find the power dissipated. [2 marks]

• Cue. $P = I^2R = 0.5^2 \times 8 = 2\ \text{W}$.