Ohm's law, electrical power relationships, resistors in series and parallel, and the potential divider rule.

What this key area is asking

The SQA wants you to apply Ohm's law and the electrical power relationships, combine resistors in series and in parallel, and use the potential divider rule to find how a supply voltage is shared between resistors.

Ohm's law and power

A component is ohmic if $V$ is directly proportional to $I$ at constant temperature, giving a straight-line $V$ against $I$ graph through the origin whose gradient is the resistance. A filament lamp is non-ohmic: its resistance rises as it heats, so its $V$ against $I$ graph curves. The three power forms are all derived by substituting $V = IR$ into $P = IV$, so choose whichever form matches the quantities you are given.

Resistors in series and parallel

In a series circuit the current is the same everywhere, which is why an ammeter reads the same wherever it is placed in the loop. In a parallel circuit each branch sees the full supply voltage, so adding more branches gives the current more paths and lowers the total resistance.

The potential divider

A potential divider is just two resistors in series across a supply; the voltage across each is in proportion to its resistance. If $R_2$ is a thermistor or a light-dependent resistor, then $V_2$ changes with temperature or light, which is the basis of sensor circuits that trigger an output (for example, switching on a light when it gets dark).

Examples in context

A phone charger uses fixed and variable resistances to set output levels, and the heating in any resistor follows $P = I^2 R$, which is why undersized wires get hot. Christmas lights in series share the mains voltage so each small bulb sees only a fraction of it, but if one fails open the whole string goes out because the single current path is broken. Household sockets are wired in parallel so each appliance receives the full mains voltage and can be switched independently. Thermostats and dimmer-style sensor circuits use a potential divider with a thermistor or LDR so the output voltage tracks an environmental quantity.

Try this

Q1. Calculate the power dissipated in a $20\ \Omega$ resistor carrying a current of $0.50\ \text{A}$. [2 marks]

• Cue. $P = I^2 R = 0.50^2 \times 20 = 5.0$ W.

Q2. Two $10\ \Omega$ resistors are connected in parallel. State the total resistance. [1 mark]

• Cue. $\frac{1}{R_T} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10}$, so $R_T = 5.0\ \Omega$.

Q3. A $9.0\ \text{V}$ supply is across a $300\ \Omega$ and a $600\ \Omega$ resistor in series. Calculate the voltage across the $600\ \Omega$ resistor. [2 marks]

• Cue. $V = \frac{600}{300 + 600} \times 9.0 = \frac{600}{900} \times 9.0 = 6.0$ V.