Simple circuits — OCR GCSE Physics

This equation links three circuit quantities. Potential difference, measured in volts (V), is the energy transferred per unit charge, so 1 V means 1 J of energy is transferred for each coulomb of charge passing. Multiplying charge in coulombs (C) by potential difference in volts (V) therefore gives energy transferred in joules (J): E = Q × V. For example, a charge of 4 C moving through a potential difference of 3 V transfers 4 × 3 = 12 J. Rearranged, Q = E ÷ V and V = E ÷ Q. In a multiple-choice question, check the units of each option and confirm which quantity is being asked for before substituting numbers.

Power is the rate of energy transfer, measured in watts (W), where 1 W equals 1 J transferred each second. In a circuit, power equals potential difference in volts (V) multiplied by current in amperes (A): P = V × I. For example, a component with a potential difference of 6 V across it and a current of 2 A through it transfers 6 × 2 = 12 W. Rearranged, V = P ÷ I and I = P ÷ V. In a multiple-choice question, identify which two quantities are given, check their units, then choose the option that multiplies them and gives the answer in watts.

This equation links the energy a device transfers to how quickly it works and for how long. Energy transferred in joules equals power in watts multiplied by time in seconds, so a 60 W lamp left on for 120 s transfers 60 × 120 = 7200 J. The same relationship works in the domestic units: energy in kilowatt-hours equals power in kilowatts multiplied by time in hours, so a 2 kW heater running for 3 h transfers 2 × 3 = 6 kW h. Keep each quantity in its matching unit pair: watts with seconds give joules, kilowatts with hours give kilowatt-hours. To convert, 1 kW h = 1000 W × 3600 s = 3 600 000 J. Rearranging gives power = energy ÷ time and time = energy ÷ power. In a multiple-choice question, check the units offered before substituting numbers, because mixing watts with hours or kilowatts with seconds is the usual trap.

A strong answer contrasts the two circuit layouts across three strands. For current: in a series circuit the current is the same at every point because there is one path, while in a parallel circuit the current splits between branches and the total current equals the sum of the branch currents. For potential difference: in series the supply p.d. is shared between components, while in parallel each branch has the same p.d. across it as the supply. For energy: in series each component transfers a share of the energy, while in parallel each branch transfers energy according to its own current and resistance. For measuring instruments: an ammeter goes in series with the component whose current is measured, and a voltmeter goes in parallel across the component whose p.d. is measured. Explain each difference with a reason, such as the single path in series forcing the same current everywhere.

A circuit diagram is a shared drawing code. Draw the long line of a cell as the positive terminal and the short thick line as the negative terminal, so the supply pushes conventional current from positive round the outside of the circuit to negative. Use the standard symbols: a cell is one long and one short line, a power supply is drawn as several cells or a labelled supply, a diode is a triangle with a bar and conducts one way only, an LDR is a resistor box with arrows pointing in, an NTC thermistor is a resistor box marked with a small t or a diagonal line, a filament lamp is a circle with a cross, an ammeter is a circle with A placed in series, a voltmeter is a circle with V placed in parallel across a component, a fixed resistor is a plain rectangle, a variable resistor is a rectangle with an arrow through it, and a switch is a break in the line with a hinged arm.

Current in a circuit is not set by the supply alone: it depends on the potential difference across the component and on the resistance of that component. Potential difference is defined as the energy transferred per unit charge, measured in volts (V), where one volt is one joule per coulomb. Current is measured in amperes (A) and resistance in ohms (Ω). For a fixed resistance, increasing the potential difference increases the current; for a fixed potential difference, increasing the resistance decreases the current. These relationships are summarised by V = I × R, so a 6 V supply across a 3 Ω resistor gives 2 A.

This criterion asks you to recall and use the relationship linking current I, resistance R and potential difference V, and to recognise that resistance is not always fixed. The relationship is V = I × R, rearranged as I = V ÷ R or R = V ÷ I. Apply it with consistent units: current in amperes, potential difference in volts, resistance in ohms. For example, a potential difference of 6 V across a 12 Ω resistor gives I = 6 ÷ 12 = 0.5 A. A fixed resistor keeps the same R whatever current flows, so doubling V doubles I. A filament lamp or thermistor behaves differently: its resistance changes as current changes, so the V–I graph curves rather than staying straight.

This criterion asks you to explain, not just state, why resistance can stay the same or change. For a fixed resistor at constant temperature, the ratio V ÷ I stays the same as current changes, so its V–I graph is a straight line through the origin and R is constant. In a filament lamp, a larger current makes the filament hotter, which increases its resistance, so the graph curves and R rises as current rises. A thermistor's resistance falls as its temperature rises. Explain the mechanism: heating changes how easily charge carriers move through the material, altering resistance. Link the shape of the graph to whether R is constant or changing.

A good answer treats each circuit as a designed test, not a list of parts. State the aim, then justify the layout: a series ammeter and a parallel voltmeter let you measure current through and potential difference across the component so resistance can be found from R = V ÷ I. Explain why a variable resistor or a range of fixed wires of different length or thickness changes current, giving several readings rather than one. For a filament lamp, describe plotting V against I and noting the curve that shows resistance rising as the lamp heats. For a diode, explain forward and reverse connection and why current flows mainly one way. For an NTC thermistor, describe heating it (for example in a water bath) and recording resistance falling as temperature rises. For an LDR, describe changing light intensity at a fixed distance and recording resistance falling as illumination increases.

A linear element gives a straight line through the origin on a potential difference–current graph, so its resistance stays constant as current changes; a fixed resistor at constant temperature behaves this way. A non-linear element gives a curve, so resistance changes with current. A filament lamp curves because the metal heats and its resistance rises. A diode conducts in forward bias and blocks in reverse, so its graph is not a straight line through the origin. An NTC thermistor and an LDR are non-linear with respect to temperature and light intensity. To judge linearity, plot the points, draw the line of best fit and check whether it is straight and passes through the origin. Read the gradient as resistance only where the line is straight.

Learners plot or read current–voltage graphs and use the shape of each curve to identify a component and explain how its resistance behaves. A metallic wire at constant temperature gives a straight line through the origin, so resistance stays constant. A filament lamp curves because heating raises resistance, so the current rises less steeply at higher voltages. A diode conducts in one direction only, giving almost no current in reverse and a steep rise once forward voltage is reached. An NTC thermistor drops in resistance as temperature rises, and an LDR drops in resistance as light intensity rises. The gradient of a current–voltage graph is not resistance itself; resistance is found from voltage divided by current at a chosen point.

Learners explain, without calculation, why combining resistors changes the net resistance. In series, the same current passes through each resistor in turn, so each adds an extra opposition to the flow; the net resistance equals the sum and is larger than either resistor. In parallel, the current splits between alternative branches, giving more paths for charge to flow; the net resistance is smaller than the smallest branch resistance. A useful check is that two equal resistors in parallel give half the resistance of one, while in series they give double. The explanation is qualitative: describe the effect of added opposition or extra paths rather than deriving a formula.

In a series circuit the current is the same at every point, the supply potential difference is shared between components, and resistances add: R_total = R₁ + R₂ + …. In a parallel circuit the potential difference across each branch equals the supply value, the branch currents add to give the total current, and the combined resistance is less than the smallest branch resistance. Use V = IR on each component or branch, then apply the series and parallel rules. Non-ohmic components change resistance with conditions: a filament lamp's resistance rises as it heats, a diode conducts in one direction only, an NTC thermistor's resistance falls as temperature rises, and an LDR's resistance falls as light intensity rises. Wire of greater length or smaller cross-section has more resistance.

A measurement circuit is designed around what is being measured and how the meter must be connected. An ammeter measures current and is placed in series with the component, so the same current flows through meter and component. A voltmeter measures potential difference and is placed in parallel across the component, so it experiences the full potential difference while drawing very little current. A test circuit may include a variable resistor to vary current or potential difference and so gather several readings, a switch to make and break the circuit safely, and a protective resistor to limit current. Choosing components suited to the range being tested, such as a thermistor for temperature or an LDR for light, lets the circuit act as a sensor. Explain each design choice by linking it to the quantity measured and the behaviour required.

Power transfer in a device is the rate at which energy is transferred to or from it. For a device carrying current I with potential difference V across it, the power is P = V × I. A 12 V lamp drawing 0.5 A transfers 12 × 0.5 = 6 W, so 6 J each second. Over a time t the energy transferred is E = P × t = V × I × t; the same lamp run for 30 s transfers 6 × 30 = 180 J. Rearranging gives I = P ÷ V and V = P ÷ I. A larger p.d. across the same current, or a larger current at the same p.d., means more power and faster energy transfer. In a resistor the energy becomes thermal energy; in a motor most becomes kinetic energy; in a lamp most becomes light and thermal energy.

This criterion asks you to select and apply the right equation from the set linking p.d. V, current I, charge Q, resistance R, power P, energy E and time t. The key relations are Q = I × t, V = I × R, P = V × I and E = P × t, with E = V × I × t combining them. For resistors in series the same current flows through each resistor, the p.d.s add, and the equivalent resistance is R = R₁ + R₂ + … . Two series resistors of 4 Ω and 6 Ω give 10 Ω; with a 5 V supply the current is 5 ÷ 10 = 0.5 A, and the p.d. across the 4 Ω resistor is 0.5 × 4 = 2 V. Choose the equation that matches the quantities given and the quantity required.