10. D.C. circuits

Circuit diagrams use agreed symbols for cells, power supplies, switches, resistors, lamps, diodes, meters and sensors so a circuit can be read without a picture of the hardware.

Keep wires joined only where junctions are marked and show meter connections according to their measuring role.

A voltmeter symbol is drawn across a component, while an ammeter symbol belongs in series with the branch current.

A symbol is not just decoration: reversing a diode or omitting a junction changes the circuit being described.

A circuit diagram represents the topology of components, supply polarity and junctions using standard symbols and straight labelled connections.

Trace a complete path from one supply terminal to the other, place ammeters in series and voltmeters in parallel, and label relevant values.

To measure a lamp’s current and voltage, put the ammeter in the lamp branch and the voltmeter across the lamp.

A voltmeter placed in series can greatly alter or effectively break a circuit; an ammeter across a supply can cause a dangerous short.

The e.m.f. ε of a source is the energy supplied by the source per coulomb as charge moves through it: ε=W/Q.

E.m.f. describes the source’s energy rise, whereas terminal potential difference can be lower when internal resistance causes a drop.

A 12 V ideal source supplies 12 J per coulomb; a real source may show less terminal voltage while delivering current.

E.m.f. is not a force in newtons and is not always identical to the voltage measured across a working battery.

E.m.f. is the source energy rise per coulomb; potential difference is the energy transferred per coulomb across a component.

In a complete loop, source energy is shared among component transfers and internal losses. Use the direction of energy transfer to distinguish rises from drops.

A cell with ε=1.5 V supplies 1.5 J per coulomb; a lamp may receive less terminal p.d. when the cell is delivering current.

E.m.f. is not automatically the same as every measured voltage in a working circuit.

A real source can be modelled with internal resistance r. When current I flows, terminal voltage V=ε−Ir, so some source energy is dissipated internally.

The lost voltage Ir grows with current. At open circuit I=0, terminal voltage approaches ε.

For ε=6.0 V, r=0.50 Ω and I=2.0 A, terminal voltage is 5.0 V.

Internal resistance is not a separate external component that must be drawn outside the source; it represents loss inside the source model.

At a circuit junction, total current entering equals total current leaving: ΣI_in=ΣI_out.

Choose current directions before solving; a negative answer means the true direction is opposite to the assumed arrow.

If 2.0 A and 0.5 A enter a node and one branch carries 1.5 A out, the remaining branch carries 1.0 A out.

Current is not “used up” at a junction; charge conservation determines the branch relation.

Around any closed circuit loop, the algebraic sum of e.m.f.s and potential changes is zero: ΣV=0.

Choose a loop direction and keep rises and drops signed consistently. Include internal resistance where the loop contains a real source.

A 12 V source feeding 4 V and 8 V drops satisfies 12−4−8=0.

A loop equation is not a claim that every component has the same voltage; it balances signed energy transfers per charge.

The combined resistance of a network is the single resistance that gives the same total current for the same applied voltage.

Reduce simple series or parallel sections step by step, preserving which elements share current or p.d. before applying the next rule.

Two 6 Ω resistors in parallel have equivalent resistance 3 Ω, then adding a 2 Ω resistor in series gives 5 Ω total.

Equivalent resistance is not the arithmetic sum for every network; parallel branches always give a value below the smallest branch resistance.

For series resistors, the same current flows through each and total resistance is the sum of individual resistances.

Use V_total=IR_total and note that p.d. divides in proportion to resistance when current is common.

4 Ω and 6 Ω in series give 10 Ω; at 2 A the voltage drops are 8 V and 12 V.

Series components do not share equal voltage unless their resistances happen to be equal.

Combine series and parallel sections where possible, then apply junction current conservation and loop energy conservation to the remaining circuit.

Label branch currents and polarities before writing equations; a negative solved current only reverses the assumed direction.

Reduce two parallel branches first, then use the source loop equation to find total current and branch voltages.

Equivalent resistance simplifies terminal behaviour but does not erase the branch currents or internal voltage distribution.

For parallel resistors, each branch has the same potential difference and currents add, giving 1/R_total=Σ(1/R_i).

Use reciprocal values carefully and check that the equivalent resistance is below the smallest branch resistance.

Two 6 Ω resistors in parallel give 1/R=1/6+1/6, so R_total=3 Ω.

Parallel branches do not carry equal current unless their resistances are equal; they share voltage, not necessarily current.

Assign unknown currents, use ΣI=0 at junctions and ΣV=0 around loops, then solve the simultaneous equations.

Choose current directions and loop orientations consistently; include source internal resistance and resistor drops where relevant.

A two-loop circuit can be solved by one junction equation plus one loop equation per independent loop.

A negative current is not a failed circuit—it means the chosen arrow was opposite to the physical current.

For series resistors R₁ and R₂ across supply V_s, the output across R₂ is V_out=V_s R₂/(R₁+R₂).

The output changes when one resistance changes; specify which resistor is measured across and account for loading by a connected device.

With equal resistors across 10 V, the midpoint is 5 V; replacing the lower resistor by a sensor shifts the output.

The output is not always half the supply; that occurs only for equal effective resistances and negligible load.

A potentiometer uses a uniform resistance wire so p.d. is proportional to length; a balance point compares an unknown p.d. with a known reference.

At balance, the detector carries no current, so the comparison is not disturbed by the test source. Keep wire current and uniformity conditions clear.

If a 1.20 V reference balances at 60 cm and an unknown balances at 45 cm on the same wire, the unknown is 0.90 V.

The balance length is not itself a voltage; proportionality requires the same wire current and uniform resistance per unit length.

A galvanometer responds to small current and shows the null condition when its pointer is zero, indicating no potential difference across it.

Use the detector to locate balance rather than to carry substantial circuit current. Reverse connections if needed to identify which side is at higher potential.

Sliding a contact along a potentiometer wire until the galvanometer reads zero identifies the comparison length.

A zero reading does not mean no voltages exist elsewhere; it means the galvanometer’s own terminals have equal potential.

A sensor resistor in a potential divider changes the output voltage when temperature or light intensity changes its resistance.

First state the sensor trend, then identify which resistor the output is measured across; the same resistance change can raise or lower output depending on placement.

An NTC thermistor lowers its resistance when heated; an LDR lowers resistance in brighter light. Both can trigger a voltage threshold in a divider.

The sensor trend alone does not determine output direction—circuit arrangement and loading matter.