2.7 Series and parallel circuits and I–V characteristics

How is current shared at a junction?

A junction is any point where three or more wires meet. Know the rule and the reason for it: "explain" questions ask for the reason.

Current is the rate of flow of charge, \(I = \Delta Q/\Delta t\). Charge cannot be created or destroyed, and in a steady circuit it does not pile up at a junction. So in every second, the charge arriving at a junction must equal the charge leaving it, and dividing both by the time gives exactly the statement about currents. In a series loop, with no junctions, the current is therefore the same at every point. A component transfers energy from the charge; it does not use up charge, so it does not use up current.

The smaller current flows through the branch with the larger resistance: both branches have the same p.d., so \(I = V/R\) gives more current to the smaller resistance.

How is potential difference shared around a circuit?

P.d. is energy transferred per unit charge, \(V = W/Q\). A coulomb gains energy in the source, transfers it in the components, and returns to its start with the energy it began with. Energy is conserved, so the energy transferred in the components must equal the energy supplied, and dividing by the charge turns energies into p.d.s. (With negligible internal resistance, e.m.f. = terminal p.d.; see page 2.9.)

• Series: the p.d.s across the components add up to the supply p.d. Because the current is the same in each component, \(V = IR\) means the p.d. across each resistor is proportional to its resistance.
• Parallel: every branch connects the same two points, so every branch has the same p.d. A coulomb passes through only one branch and transfers the same energy whichever branch it takes.

A graph of potential against position round the circuit makes the energy picture visible. Connecting wires have negligible resistance, so the potential along a wire is constant. Each resistor causes a drop, and the drops together use up the whole of the source p.d.

How do you derive the equations for resistors in series and in parallel?

Both results come from combining the two conservation laws with \(V = IR\). Learn the derivations as three short steps each; the final equations are in the formulae list, the derivations are not.

Resistors in series

1. The same current \(I\) passes through each resistor (conservation of charge).
2. The p.d.s add: \(V = V_1 + V_2 + V_3\) (conservation of energy).
3. Write each p.d. as \(IR\): \(IR = IR_1 + IR_2 + IR_3\), then divide by \(I\).

The total is always larger than the largest single resistor: every coulomb has to pass through all of them.

Resistors in parallel

1. The same p.d. \(V\) is across each resistor (conservation of energy: each branch joins the same two points).
2. The currents add: \(I = I_1 + I_2 + I_3\) (conservation of charge at the junction).
3. Write each current as \(V/R\): \(\dfrac{V}{R} = \dfrac{V}{R_1} + \dfrac{V}{R_2} + \dfrac{V}{R_3}\), then divide by \(V\).

The total is always smaller than the smallest single resistor. Each extra branch is an extra path for charge, so for the same p.d. more current flows, and \(R = V/I\) falls.

How do you analyse a combined series–parallel network?

Simplify inwards to a single resistance, find the supply current, then work back outwards. Use the same four steps every time:

1. Replace each parallel group with its single equivalent resistance.
2. Add the series resistances to get the total; find the supply current, \(I = V/R_{\text{total}}\).
3. Work back out: the p.d. across each series part is \(I\) times its resistance, and these p.d.s add up to the supply p.d.
4. Each branch of a parallel group has the group's p.d.; branch current = that p.d. ÷ branch resistance. The branch currents add up to the current into the group.

Changing any one resistor changes every current and p.d. in the network, because the total resistance, and therefore the supply current, changes.

A follow-up thought shows how the whole network responds. If the 30 Ω resistor were removed, the total would rise to \(10 + 20 = 30\ \Omega\), the supply current would fall to \(6.0/30 = 0.20\) A, and the p.d. across the 20 Ω resistor would rise to \(0.20 \times 20 = 4.0\) V. Less p.d. is "lost" across the 10 Ω resistor because less current flows through it.

What do the I–V graphs of common components look like?

An I–V characteristic is a graph of the current \(I\) through a component against the p.d. \(V\) across it. It includes negative values, obtained by reversing the connections, because some components behave differently in the two directions. A component is ohmic if \(I \propto V\): its graph is a straight line through the origin. For a metal this needs constant temperature (Ohm's law, page 2.6).

On I–V axes a steeper line from the origin means a smaller resistance. Why the lamp and thermistor change resistance with temperature is explained on page 2.10.

How do you read resistance from an I–V graph and solve non-ohmic circuits?

Resistance is defined as \(R = V/I\) at a point. It does not mean the gradient, and it does not mean \(\Delta V/\Delta I\) between two points. The two methods agree only for a straight line through the origin.

With a lamp in a circuit you cannot share the p.d. in a ratio, because the lamp's resistance depends on the current through it. Write down both conditions and use the graph.

The same answer can be found by drawing. Rearranged, the resistor condition is \(I = (8.0 - V_{\text{lamp}})/16\), a straight line on the lamp's I–V axes. Where it crosses the lamp curve, both conditions hold at once.

How do you measure an I–V characteristic?

The ammeter goes in series with the component; the voltmeter goes in parallel with the component only.

• Varying the p.d.: a variable resistor used as a potential divider (above) lets the p.d. start from zero. A variable resistor simply in series with the component is easier to set up, but the p.d. cannot fall to zero, so the region near the origin is missed.
• Readings: record I for a range of V, in equal steps across the whole range; take extra readings where the graph curves most.
• Negative values: reverse the connections to the component (or the supply) and repeat.
• Diode: add a protective resistor in series so the forward current cannot become large; use a milliammeter; take small steps between about 0.4 V and 0.8 V where the current rises steeply.
• Lamp and thermistor: wait for each reading to settle, because the component needs time to reach a steady temperature. The lamp gets hot, so do not touch it.
• Graph: plot I against V and draw a smooth curve (or straight line) of best fit through the points.

Exam tips for 2.7

Common misconceptions

How this topic connects

2.6 Current, p.d. and resistance

\(I = \Delta Q/\Delta t\), \(V = W/Q\) and \(R = V/I\) are the definitions that turn conservation of charge and energy into the rules on this page; power in each resistor follows from \(P = VI\).

2.8 Resistivity and the potential divider

A potential divider is a series circuit sharing p.d. in the ratio of resistances: section 2 of this page applied to sensors with thermistors and LDRs.

2.9 E.m.f. and internal resistance

Kirchhoff's second law with the internal resistance included gives \(V = \varepsilon - Ir\); the source's internal resistance is simply one more resistor in series.

2.10 Resistance, temperature and light

Explains the shapes on this page: lattice vibrations raise a metal filament's resistance, while extra conduction electrons lower a thermistor's.

Key equations

Check your understanding

Specification checklist

• (67) I can explain, using conservation of charge, why the total current into a junction equals the total current out, and use this to find unknown currents.
• (67) I can explain, using conservation of energy, why the p.d.s around a closed loop add up to the source p.d., and why parallel branches have the same p.d.
• (68) I can derive \(R = R_1 + R_2 + R_3\) for resistors in series and \(1/R = 1/R_1 + 1/R_2 + 1/R_3\) for resistors in parallel.
• (68) I can calculate the total resistance, currents and p.d.s in a network of resistors in series and in parallel.
• (70) I can sketch and recognise the I–V graphs of an ohmic conductor, a filament lamp, an NTC thermistor and a semiconductor diode.
• (70) I can find a component's resistance from its I–V graph using V/I at a point, and interpret what the shape of the graph shows.
• (70) I can describe how to obtain an I–V characteristic using a variable resistor or potential divider, an ammeter and a voltmeter.