10. D.C. circuits

Exam analysis

Most tested topics

10.1 Practical circuits

Group	Official symbols and discriminating cues
sources	cell: one long/short parallel-line pair; battery of cells: repeated pairs; power supply: two terminals; a.c. power supply: two terminals with ~
connections	junction of conductors: connected wires with a filled dot; switch: two contacts with a movable open/closed link; earth: vertical connection to three decreasing horizontal bars

Group	Official symbols and discriminating cues
resistive/light	lamp: circle with cross; fixed resistor: rectangle; variable resistor: rectangle with diagonal arrow; heater: segmented rectangle
sensors/control	thermistor: resistor with diagonal temperature mark; LDR: resistor with arrows pointing towards it; potentiometer: three-terminal resistor with sliding contact
semiconductor/storage	diode: one-direction diode mark with barrier; LED: diode plus two arrows pointing outwards; capacitor: two parallel plates

Group	Official symbols and discriminating cues
sound	electric bell and buzzer use opposite dome/bowl cues; microphone has a receiving diaphragm; loudspeaker has an outward cone
machines	motor: circle marked M; generator: box marked G
meters/display	ammeter: circle A; voltmeter: circle V; galvanometer: circle with pointer; oscilloscope: circle with trace

Use the exact symbol, orientation and number of terminals required. The long cell line is the positive terminal. Arrows pointing towards an LDR represent incoming light; arrows pointing away from a diode identify an LED. A filled dot means joined conductors; crossing lines without a junction mark are not automatically connected.

Component names are not substitutes for symbols in a requested circuit diagram. Reversing a cell, diode or LED, changing a sliding contact, or adding/removing a junction changes the electrical meaning.

To draw a circuit: list the required components and measurements; identify which components share one current path and which form branches; place standard symbols with correct polarity/orientation; connect straight wires and mark only true junctions; then label values and check a complete path between supply terminals.

Measurement	Connection	Why
current through a component	ammeter in series in that branch	the same branch current passes through the meter
p.d. across a component	voltmeter in parallel across its two terminals	the meter compares those two potentials

To interpret a diagram, ignore where components are drawn on the page and trace electrical nodes instead. Points joined by ideal wire belong to the same node. Components connected between the same two nodes are parallel; components on an unbranched path are series. Then check switch state, source polarity and diode direction before predicting current.

For a nichrome-wire investigation, draw the wire as a resistor in a closed loop with the cell and ammeter in series, and connect a voltmeter across only the wire. This measures the wire's I and V without including the ammeter's p.d. in the voltmeter reading.

A voltmeter in series can almost break the circuit because of its high resistance; an ammeter across a supply can cause a dangerously large current because of its low resistance. A line crossing is not a junction unless the diagram marks a connection.

The electromotive force ε of a source is the energy transferred by the source per unit charge in driving charge around the complete circuit. It converts chemical, mechanical or another stored form into electrical energy.

$ε = W/Q and W = εQ$

ε is measured in volts, with 1 V = 1 J C⁻¹. W is the total source energy supplied to the complete circuit, including energy transferred in the external circuit and any internal resistance.

A 9.0 V battery drives 250 C around its circuit. The energy converted by the battery is W = εQ = 9.0 × 250 = 2.25 × 10³ J. If internal resistance is present, not all of this reaches the external load.

E.m.f. is not a force in newtons. It characterises the source's energy conversion per coulomb and can remain constant even while loaded terminal p.d. changes.

Feature	E.m.f. ε of a source	Potential difference V across a component
meaning	energy supplied by source per unit charge	energy transferred from electrical form per unit charge
energy conversion	chemical/mechanical/etc. → electrical	electrical → thermal/mechanical/light/etc.
location	across the source's energy-raising process	between the two terminals of the named component
unit	V = J C⁻¹	V = J C⁻¹

For each coulomb around a complete steady circuit, energy supplied by the source equals the sum of energy transferred per coulomb in external components and inside the source. Thus ε equals the sum of all p.d. drops around the loop.

A 1.5 V cell supplies 1.5 J C⁻¹. If 0.20 J C⁻¹ is transferred internally, the terminal p.d. available to the external circuit is 1.3 V; the two numbers describe different energy destinations.

Equal numerical values do not make e.m.f. and p.d. the same process. With zero current, internal transfer is negligible and terminal p.d. approaches ε; under load, internal loss can make terminal p.d. smaller.

$ε = V + Ir so V = ε − Ir$

A real source is modelled as e.m.f. ε in series with internal resistance r. When current I is delivered, Ir is the p.d. lost inside the source and I²r is the internal power dissipation; the terminal p.d. V is what remains for the external circuit.

Decreasing external resistance increases total current. The internal drop Ir then increases, so terminal p.d. falls even if ε is unchanged. Increasing external resistance does the reverse. With an ideal high-resistance voltmeter and open circuit, I ≈ 0 and V ≈ ε.

For ε = 6.0 V, r = 0.50 Ω and I = 2.0 A, V = 6.0 − (2.0)(0.50) = 5.0 V. Internal power is I²r = 2.0 W, while power delivered at the terminals is VI = 10 W.

A graph of terminal V against delivered I is a straight line V = ε − rI: the V-axis intercept is ε and the gradient is −r. A lower ε moves the intercept down; a lower r makes the line less steep.

Internal resistance is part of the source model, not an added external resistor. Current changes terminal p.d. through internal energy transfer; it does not necessarily change the source e.m.f.

10.2 Kirchhoff’s laws

$ΣI_in = ΣI_out or, with signed currents, ΣI = 0$

At a junction in a steady circuit, the total current entering equals the total current leaving. Since I = ΔQ/Δt, any difference sustained for time Δt would leave net charge ΔQ = (ΣI_in − ΣI_out)Δt accumulating at the junction. A steady node does not accumulate charge, so the totals are equal.

Draw an arrow for every branch and choose one sign convention before writing the equation. You may sum entering and leaving currents separately, or assign one direction positive and use an algebraic sum. A negative solved value reverses the assumed arrow.

At a node, 2.0 A and 0.50 A enter while 1.5 A and I leave. First law gives 2.0 + 0.50 = 1.5 + I, so I = 1.0 A leaving. In 5.0 s that branch carries Q = It = 5.0 C.

Current is not used up at a junction. Charge can divide between branches, but the rate of charge arrival and departure must balance in steady operation.

$Σ(emf) = Σ(p.d. drops) or, algebraically around a closed loop, ΣV = 0$

After one complete circuit loop, a charge returns to its starting point with the same energy per unit charge. Energy gained per coulomb in sources therefore equals energy transferred per coulomb in circuit components, including internal resistance when present.

Traversal	Potential change
through a source from − to +	+ε rise
through a source from + to −	−ε fall
through a resistor in the direction of conventional current	−IR drop
through a resistor opposite to conventional current	+IR rise

Traversing a loop in the current direction through a 12 V source and resistive drops of 4 V and 8 V gives +12 − 4 − 8 = 0. For each coulomb, 12 J is supplied and 12 J is transferred; no net energy is gained after the closed loop.

The law does not say every component has the same p.d. Signs are set by the chosen traversal and polarity; reversing the loop reverses every term but leaves the physical equation equivalent.

Let resistors R₁, R₂, …, Rₙ be in one unbranched series path. Let I be the current and V the total p.d. across the combination. The equivalent resistance R_T is defined by V = IR_T.

With no junction between the resistors, Kirchhoff's first law gives the same current I through every resistor.

$V = V₁ + V₂ + ··· + Vₙ$

$IR_T = IR₁ + IR₂ + ··· + IRₙ = I(R₁ + R₂ + ··· + Rₙ)$

$For I ≠ 0: R_T = R₁ + R₂ + ··· + Rₙ$

The derivation uses common current and additive p.d.s; it does not assume equal p.d.s. It applies to components in one series path, not to a network containing a branch.

$R_T = R₁ + R₂ + ··· and I = V/R_T$

Confirm there is one unbranched current path, add the resistances, use the total applied p.d. to find the common current, then use V_i = IR_i for each resistor. Check that the individual p.d.s sum to the supply p.d.

A 12 V supply is connected to 2.0 Ω, 3.0 Ω and 5.0 Ω in series. R_T = 10.0 Ω and I = 12/10.0 = 1.2 A. The drops are 2.4 V, 3.6 V and 6.0 V; they sum to 12.0 V.

$For series resistors: V₁/V₂ = R₁/R₂$

Series p.d.s are equal only when the resistances are equal. Adding another positive series resistance increases R_T and, for a fixed supply p.d., decreases the circuit current.

Let R₁, R₂, …, Rₙ connect between the same two nodes. Let V be the p.d. across the combination, I the total current and R_T the equivalent resistance, so I = V/R_T.

Kirchhoff's second law applied to loops through different branches shows that every branch has the same p.d. V.

$I = I₁ + I₂ + ··· + Iₙ$

$V/R_T = V/R₁ + V/R₂ + ··· + V/Rₙ$

$For V ≠ 0: 1/R_T = 1/R₁ + 1/R₂ + ··· + 1/Rₙ$

Parallel branches share p.d., not necessarily current. The reciprocal formula follows because branch currents add; directly adding the resistance values would describe a series path instead.

$1/R_T = Σ(1/R_i) and, for exactly two branches, R_T = R₁R₂/(R₁ + R₂)$

Confirm the resistors connect between the same two nodes. Add conductances 1/R_i, then invert the complete sum. The product-over-sum shortcut is valid for exactly two resistors; for three or more, use the general reciprocal relation or reduce in stages.

For 6.0 Ω, 3.0 Ω and 2.0 Ω in parallel, 1/R_T = 1/6 + 1/3 + 1/2 = 1.0 Ω⁻¹, so R_T = 1.0 Ω. Across 6.0 V, the branch currents are 1.0 A, 2.0 A and 3.0 A, summing to 6.0 A = V/R_T.

Check	Required result
magnitude	R_T is less than the smallest positive branch resistance
identical n resistors R	R_T = R/n
current	V/R_T equals the sum of V/R_i

Do not report the reciprocal sum as R_T: after calculating 1/R_T, invert it. Adding another conducting parallel branch lowers the equivalent resistance because it provides another route for current.

Name every branch current and choose arrows; write independent junction equations; choose independent closed loops and assign voltage signs consistently; replace resistor p.d.s by IR; solve the simultaneous equations; then check every junction and loop. Include internal resistance as a series resistor inside its source loop when relevant.

Example network: a 12 V ideal source and 2.0 Ω series resistor feed node A. From A, currents I₄ and I₆ return to the source through separate 4.0 Ω and 6.0 Ω branches. Let total current I reach A through the 2.0 Ω resistor.

$junction A: I = I₄ + I₆ 4 Ω loop: 12 = 2I + 4I₄ 6 Ω loop: 12 = 2I + 6I₆$

The two branch equations give I₄ = (12 − 2I)/4 and I₆ = (12 − 2I)/6. Substitute into I = I₄ + I₆: I = (12 − 2I)(1/4 + 1/6), so I = 30/11 = 2.73 A. Then I₄ = 18/11 = 1.64 A and I₆ = 12/11 = 1.09 A.

Check	Substitution	Result
junction	18/11 + 12/11	30/11 A = I
4 Ω loop	2(30/11) + 4(18/11)	12 V
6 Ω loop	2(30/11) + 6(12/11)	12 V

Do not write a single current through every branch. If a solved current is negative, retain the magnitude and reverse its assumed arrow; the algebra has identified the physical direction rather than failed.

10.3 Potential dividers

Two resistors R₁ and R₂ in series carry the same current. If the output is measured across R₂ and the supply is V_s, then the output is the fraction of total series resistance contributed by R₂.

$I = V_s/(R₁ + R₂), so V_out = IR₂ = V_s R₂/(R₁ + R₂)$

With R₁ = 2.0 kΩ and R₂ = 3.0 kΩ across 10 V, V_out = 10 × 3/(2 + 3) = 6.0 V across R₂. Across R₁ it is 4.0 V, and the two outputs sum to the supply.

A device of resistance R_L connected across R₂ is a parallel load. Replace R₂ by R_b = R₂R_L/(R₂ + R_L), then use V_out = V_sR_b/(R₁ + R_b). For R_L = 3.0 kΩ above, R_b = 1.5 kΩ and V_out falls from 6.0 V to 4.29 V.

A divider gives half the supply only when the two effective series resistances are equal. Always identify the two output terminals and include any load that changes the effective resistance.

A steady current in a uniform wire of constant cross-sectional area produces resistance R ∝ l and therefore p.d. V = IR ∝ l. The potential gradient k = V_wire/L is constant, so a balance length l represents p.d. E = kl.

Connect the test p.d. with polarity opposing the wire's p.d., slide the contact and locate the point where the galvanometer reads zero. Keep the same wire current, wire and temperature while comparing the two balance lengths.

$E₁ = kl₁, E₂ = kl₂, hence E₁/E₂ = l₁/l₂$

A 1.20 V reference balances at 60.0 cm. An unknown balances at 45.0 cm without changing the wire current. E_x/1.20 = 45.0/60.0, so E_x = 0.900 V.

Balance length is not itself a voltage. The ratio method fails if the potential gradient changes between readings, and the available wire p.d. must be large enough for the unknown to reach a balance point.

$At null: I_g = 0 ⇔ V_g = 0 ⇔ the galvanometer terminals are at equal potential$

A sensitive centre-zero galvanometer is a detector, not the measuring scale. Away from balance, its deflection direction shows which terminal is at higher potential; moving the contact until the direction changes and then narrowing the interval locates zero.

At balance no current is drawn through the galvanometer branch or from the test source. The test source therefore has no internal voltage loss in that branch, so a potentiometer can compare its e.m.f. without the loading caused by an ordinary finite-resistance voltmeter.

On a potentiometer wire, a contact left of balance gives one deflection and a contact right of balance gives the opposite deflection. The zero point between them is the length whose wire p.d. exactly opposes the test p.d.

A null reading does not mean every voltage or current in the apparatus is zero. The driver current still flows in the potentiometer wire; only the detector branch has zero current and equal endpoint potential.

Sensor	Increasing stimulus	Resistance response
NTC thermistor	temperature increases	R_NTC decreases
LDR	light intensity increases	R_LDR decreases

Output measured across	When sensor resistance decreases	Why
sensor	V_out decreases	V_out = V_sR_sensor/(R_fixed + R_sensor)
fixed resistor	V_out increases	the fixed resistor receives a larger fraction of V_s

Explain any response in four links: stimulus change → sensor resistance change → selected divider fraction change → output p.d. change. Reversing the sensor and fixed-resistor positions reverses the output trend for the same stimulus.

An NTC thermistor is the output resistor below a 20 kΩ fixed resistor across 60 V. At 20 kΩ, V_out = 30 V. Cooling raises the NTC resistance by 50% to 30 kΩ, so V_out = 60 × 30/(20 + 30) = 36 V. Heating would lower its resistance and lower this sensor-output voltage.

The simple ratio assumes negligible output loading and an approximately fixed terminal supply p.d. A finite load changes the selected effective resistance; appreciable source internal resistance can also make terminal p.d. change as total current changes.

10. D.C. circuits

Exam analysis

Most tested topics

10.1 Practical circuits

Recognise the complete Cambridge circuit-symbol set

Draw and read circuits from their topology

E.m.f. is source energy supplied per coulomb around the complete circuit

E.m.f. supplies energy; p.d. transfers it from electrical form

Internal resistance makes terminal p.d. depend on current

10.2 Kirchhoff’s laws

Kirchhoff's first law is charge conservation at a junction

Kirchhoff's second law is energy conservation around a loop

Derive the series-resistance formula from Kirchhoff's laws

Use series resistance and recover each voltage drop

Derive the parallel-resistance formula from both Kirchhoff laws

Use parallel resistance with reciprocal and physical checks

Solve a branched circuit with junction and loop equations

10.3 Potential dividers

A potential divider selects a fraction of the supply voltage

A potentiometer compares p.d.s by their null balance lengths

A galvanometer locates equality without loading the test source

Sensor placement sets the direction of a divider's voltage response