D.C. Circuits

Circuit symbols and circuit diagrams

A circuit diagram is a drawing of a real circuit in a shared shorthand. Every component has one agreed circuit symbol, so that anyone reading the diagram builds the same circuit in their head, and every wire is drawn as a straight line with right-angled corners. Where three or more wires meet, draw a solid dot: the dot is a junction, and junctions are where the current can split, so leaving one out changes the meaning of the diagram.

Two symbols carry a rule about how they are connected. An ammeter measures current, so the current must flow through it: it is connected in series, breaking the circuit at the point of interest. A voltmeter measures the potential difference between two points, so it is connected in parallel across the component. An ideal ammeter has zero resistance and an ideal voltmeter has infinite resistance, so that the meter does not change the very quantity it is measuring.

Read the symbol table below from left to right: first what the component is, then what it does in the circuit, then how it is connected up. The two sensing components are easy to confuse, so fix them now: a thermistor is drawn as a resistor with a diagonal line across it, and a light-dependent resistor is drawn as a resistor with two arrows pointing into it, standing for the light arriving at the surface.

Check the details that carry marks in a drawing question. The longer line of a cell symbol is its positive terminal, so the arrow of conventional current leaves that side; a diode conducts only when conventional current flows from the triangle towards the bar, and a light-emitting diode emits light when it does so; a switch is drawn open unless the question says the circuit is complete; and a fuse is drawn as a rectangle with a line through it, placed in the live wire of a mains circuit.

Electromotive force and potential difference

The electromotive force (e.m.f.) of a source is the energy transferred from other forms to electrical energy per unit charge in driving charge around a complete circuit. In a chemical cell the other form is chemical energy; in a generator it is mechanical energy. Because it is an energy per unit charge, the e.m.f. is measured in joules per coulomb, and one joule per coulomb is one volt.

The potential difference (p.d.) between two points is the energy transferred from electrical energy to other forms per unit charge as charge passes between those points. Across a lamp some of that energy becomes light and the rest heat; across a resistor it all becomes thermal energy. A p.d. is therefore always measured between two points, and you must say which two.

Both quantities are measured in volts, and that is why candidates mix them up. The distinction is the direction of the energy transfer. An e.m.f. puts energy into the circuit, at the source, while a p.d. takes energy out of the circuit, at the components. An e.m.f. and a p.d. are not two different things that happen to share a unit: they are the same energy per unit charge measured at opposite ends of the same transfer.

The energy transferred is given by W = QV, where Q is the charge that passes. A cell of e.m.f. 1.50 V transfers 1.50 J of chemical energy to electrical energy for every coulomb of charge that it drives round the circuit, and 2.0 C of charge passing through it therefore gains 2.0 x 1.50 = 3.0 J. The same 2.0 C passing through a 6.0 V lamp transfers 2.0 x 6.0 = 12 J out of the circuit.

• E.m.f.: energy transferred from other forms to electrical energy per unit charge, in the source, in driving charge around a complete circuit.
• P.d.: energy transferred from electrical energy to other forms per unit charge, between two points in the circuit.
• Both are measured in volts, and one volt is one joule per coulomb.
• Energy transferred W = QV, so a 1.50 V cell gives 1.50 J to every coulomb it drives round the circuit.

Internal resistance and the terminal potential difference

A real source of e.m.f. is made of materials, and those materials resist the flow of charge, so the source has its own internal resistance r. Charge passing through the source therefore transfers some energy inside the source itself as well as in the external circuit. That energy is not available outside, and the result is that the terminal potential difference - the p.d. across the terminals of the source while it is delivering current - is less than the e.m.f.

Apply Kirchhoff's second law to the single loop formed by the source and the external resistance R. The e.m.f. supplies energy, and the rest of the loop takes it out again, so E = I(R + r) = IR + Ir. The term Ir is called the lost volts, because it is the p.d. that is used up inside the source, and the terminal p.d. is therefore V = E - Ir. Increasing the current always increases the lost volts and lowers the terminal p.d.

A numerical check makes this concrete. A dry cell has E = 1.50 V and r = 0.40 ohm, and it is connected to a 2.6 ohm resistor. The total resistance is 2.6 + 0.40 = 3.0 ohm, so I = 1.50 / 3.0 = 0.50 A. The lost volts are Ir = 0.50 x 0.40 = 0.20 V, so the terminal p.d. is 1.50 - 0.20 = 1.30 V. Check it from the other side: the p.d. across the external resistor is IR = 0.50 x 2.6 = 1.30 V, which is exactly the terminal p.d., as it must be.

Plot the terminal p.d. against the current for a cell and you get a straight line. The intercept on the voltage axis is the e.m.f., because when no current flows there are no lost volts and the terminal p.d. equals the e.m.f., and the gradient is -r. A short circuit, with R = 0, gives the largest possible current E / r, and every volt of the e.m.f. is lost inside the source. This is why an old cell with a high internal resistance can still read nearly its full e.m.f. on a high-resistance voltmeter yet cannot light a lamp.

• E = I(R + r), so E = IR + Ir: the e.m.f. is shared between the external resistance and the internal resistance.
• Terminal p.d. V = E - Ir: it falls as the current rises, and equals the e.m.f. only when no current flows.
• Lost volts = Ir: the p.d. across the internal resistance, which is not available to the external circuit.
• Graph of terminal p.d. against current: intercept = E, gradient = -r.
• Short-circuit current = E / r, the current when the external resistance is zero.

Kirchhoff's first law: charge is conserved

Kirchhoff's first law states that the sum of the currents entering a junction is equal to the sum of the currents leaving it. Read it as a statement about the charge that flows past a point: the rate at which charge arrives at the junction equals the rate at which charge leaves it.

The law is a consequence of conservation of charge. Charge cannot be created or destroyed, and in a steady-state d.c. circuit it cannot pile up at a junction either, because if it did the junction would become charged and the current would change. So whatever charge flows in must flow out at the same rate, which is exactly what the law says. At a junction where 3.0 A enters and two branches leave, the branch currents might be 1.0 A and 2.0 A, and 1.0 + 2.0 = 3.0 A.

A series circuit is the simplest case. There are no junctions anywhere along the single path, so the current is the same at every point in the loop: every ammeter you could connect in series reads the same value, whatever the resistances in the loop are. This is the fact you use to justify writing one current I through two resistors in series when you derive the series rule for resistance.

One qualification matters. Charge can accumulate, in a capacitor, and the first law is then written for the total charge including what the capacitor stores. For the steady-state circuits in this topic, with cells, resistors and meters, the current into any junction equals the current out.

Kirchhoff's second law: energy is conserved

Kirchhoff's second law states that in any closed loop of a circuit, the sum of the e.m.f.s is equal to the sum of the potential differences, or equivalently that the algebraic sum of the potential changes around a closed loop is zero. It applies to every loop you can trace, however complicated the circuit is.

The law is a consequence of conservation of energy. Imagine a small positive charge being carried right round a loop and back to where it started. It ends at the same point, in the same electric field, so it has exactly the same potential energy as it had at the start. Every joule of energy the source gave it must therefore have been transferred in the components somewhere along the loop, and that is what the law records.

To apply it, pick a direction round the loop and go steadily round it, adding the potential changes with their signs. Traversing the source from its negative terminal to its positive terminal adds its e.m.f.; traversing a resistor in the direction of the current subtracts Ir; and going the other way through either of them reverses the sign. Round the loop the signed changes must add up to zero, whatever direction you chose, because choosing the other way round simply multiplies every sign in the equation by -1.

Test it on one loop. A cell of e.m.f. 6.0 V is connected to resistors of 4.0 ohm and 2.0 ohm in series. The total resistance is 6.0 ohm, so the current is 6.0 / 6.0 = 1.0 A, and the two p.d.s are 1.0 x 4.0 = 4.0 V and 1.0 x 2.0 = 2.0 V. Their sum is 4.0 + 2.0 = 6.0 V, which equals the e.m.f., so the loop balances.

Combined resistance in series and in parallel

Resistors in series. The same current I flows through both, because there is only one path between the junctions. Kirchhoff's second law says the e.m.f. of the source equals the sum of the p.d.s across the resistors, so V = V1 + V2. Substituting V1 = IR1, V2 = IR2 and V = IR gives IR = IR1 + IR2, and dividing by I leaves R = R1 + R2. In series, resistances simply add, and the combined resistance is always larger than any single one.

Resistors in parallel. Both resistors are connected between the same two junctions, so the p.d. V across each is the same. Kirchhoff's first law says the total current splits at the junction, so I = I1 + I2. Substituting I1 = V / R1, I2 = V / R2 and I = V / R gives V / R = V / R1 + V / R2, and dividing by V leaves 1 / R = 1 / R1 + 1 / R2. For two resistors this rearranges to the convenient form R = R1R2 / (R1 + R2), and for n equal resistors in parallel R = the resistance of one divided by n.

Numbers to keep in your head. Two resistors of 6.0 ohm and 3.0 ohm in series give 6.0 + 3.0 = 9.0 ohm; the same two in parallel give R = (6.0 x 3.0) / (6.0 + 3.0) = 18 / 9.0 = 2.0 ohm. Two equal 6.0 ohm resistors in parallel give 6.0 / 2 = 3.0 ohm. In every parallel case the combined resistance is smaller than the smallest resistor in the group, and that is the quickest check that you have not inverted your fraction wrongly.

The physical reason for the difference is worth stating in words. In parallel, the same p.d. now drives current through more than one path at once, so the total current drawn from the source is larger than either branch current alone. Since the combined resistance is the supply p.d. divided by that larger total current, it must be smaller than either branch resistance. Adding any resistor in parallel therefore always reduces the combined resistance, whereas adding one in series always increases it.

Solving circuit problems with Kirchhoff's laws

Work through a circuit methodically rather than in one leap. First mark the current in each branch with an arrow and a symbol, I1, I2, and so on, choosing directions at random - if you guess one wrong, the algebra will simply give you a negative answer and tell you the current really flows the other way. Then write Kirchhoff's first law at every junction, and Kirchhoff's second law round every independent loop. Finally solve the simultaneous equations, and check each answer by putting it back into one of the original equations.

A single-loop example with a parallel section. A cell of e.m.f. 9.0 V and internal resistance 1.0 ohm is connected to a 2.0 ohm resistor in series with a parallel pair of 6.0 ohm and 3.0 ohm. The parallel pair has a combined resistance of (6.0 x 3.0) / (6.0 + 3.0) = 2.0 ohm, so the external resistance is 2.0 + 2.0 = 4.0 ohm and the total resistance in the loop is 4.0 + 1.0 = 5.0 ohm. The current from the cell is therefore I = 9.0 / 5.0 = 1.8 A, and the terminal p.d. is 9.0 - 1.8 x 1.0 = 7.2 V.

Now follow the energy round the loop. The 2.0 ohm resistor takes 1.8 x 2.0 = 3.6 V, so the parallel pair has the remaining 7.2 - 3.6 = 3.6 V across it, and 3.6 + 3.6 + 1.8 = 9.0 V, which is the e.m.f. The current then splits at the junction: through the 6.0 ohm resistor it is 3.6 / 6.0 = 0.60 A, and through the 3.0 ohm resistor it is 3.6 / 3.0 = 1.20 A. Kirchhoff's first law confirms it, because 0.60 + 1.20 = 1.80 A recombines to leave the junction.

In a circuit with two sources, the same routine still works, but there are now two or more loops and you must use both laws together. Label the currents in the three branches, apply the junction rule to relate them, then write the loop rule once round each of the two smaller loops. Two independent equations are enough for two unknown currents. Check the signs carefully: the p.d. across a resistor is subtracted when you travel in the direction of its current, and a shared branch appears in both loop equations with its own sign in each.

Potential dividers, potentiometers and sensing resistors

A potential divider is two or more resistors in series across a supply, used to obtain a smaller, chosen potential difference. Because the same current flows through both resistors, the supply p.d. divides between them in proportion to their resistances: the larger resistance takes the larger share, and V_out / V_in = R2 / (R1 + R2) when the output is taken across R2. With 12 V across a 2.0 kohm and a 6.0 kohm resistor, the current is 12 / 8000 = 1.5 x 10-3 A and the p.d. across the 6.0 kohm resistor is 1.5 x 10-3 x 6000 = 9.0 V.

Replacing one resistor with a sensing resistor turns a physical quantity into a voltage. The resistance of an NTC thermistor falls as its temperature rises, so as the temperature goes up the total resistance falls, the current rises, and the p.d. across the fixed resistor in series with it rises while the p.d. across the thermistor falls. The resistance of a light-dependent resistor falls as the light intensity rises, so the same argument applies with light in place of temperature. Which way the output moves always depends on which component the output is taken across, so state that before you answer.

A potentiometer is a potential divider with a sliding contact. A steady current is passed through a uniform resistance wire, so the p.d. per unit length is constant, and tapping off a length L gives a p.d. proportional to L. Connect two cells in turn in series with a galvanometer and find the balance length for each: because the p.d. per unit length is the same throughout, E1 / E2 = L1 / L2, and the e.m.f.s are compared without ever knowing the current in the wire or the resistance per metre. Balance lengths of 62.0 cm and 48.0 cm for two cells therefore give a ratio of 62.0 / 48.0 = 1.29, so if one cell has an e.m.f. of 1.20 V the other has an e.m.f. of 1.20 x 1.29 = 1.55 V.

The galvanometer is used because it is a null detector: the contact is moved until the galvanometer reads zero, which shows that no current is flowing through it. At balance no current is drawn from the cell being tested, so there are no lost volts and the p.d. being matched is the e.m.f. of that cell, not its terminal p.d. This is what makes the potentiometer a way of measuring an e.m.f. rather than a p.d., and it is why it can be used to calibrate a voltmeter. The driver cell that pushes current through the wire must have an e.m.f. greater than either cell being compared, or the balance point lies beyond the end of the wire and no null can be found.