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AS Level · Physics 9702

D.C. Circuits

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  Cambridge International AS & A Level Physics (9702)

D.C. Circuits

The big idea: Charge is conserved at every junction and energy is conserved around every loop. Kirchhoff's two laws are those two conservation laws, and every circuit calculation follows from them.

Summary — What This Chapter Covers
  • E.m.f. = energy transferred per unit charge in driving charge round a complete circuit; p.d. = energy transferred per unit charge from electrical to other forms in a component.
  • Internal resistance makes the terminal p.d. fall as current rises: V = E − Ir.
  • Kirchhoff's first law (conservation of charge): current into a junction = current out.
  • Kirchhoff's second law (conservation of energy): around any loop, sum of e.m.f.s = sum of p.d.s.
  • Series: R = R1 + R2 + …; parallel: 1/R = 1/R1 + 1/R2 + ….
  • A potential divider splits a supply p.d. in the ratio of the resistances; a potentiometer compares p.d.s using a null method.
1. Practical Circuits
Practical circuits

E.m.f., p.d. and internal resistance

You must know the circuit symbols listed in the syllabus: cell, battery, power supply, a.c. supply, switch, lamp, fixed and variable resistor, thermistor, LDR, potentiometer, diode, LED, ammeter, voltmeter, galvanometer, oscilloscope, fuse, earth, junction, and others. Ammeters go in series (ideally zero resistance); voltmeters go in parallel (ideally infinite resistance).

The e.m.f. of a source is the energy transferred (from chemical or other forms to electrical) per unit charge in driving charge round a complete circuit. The p.d. across a component is the energy transferred per unit charge from electrical to other forms. Both are measured in volts.

Real sources have internal resistance r. Some of the e.m.f. is "lost" driving current through r, so the terminal p.d. V = E − Ir. The larger the current, the lower the terminal p.d.; only on open circuit (I = 0) does V = E.

Internal resistance E = I(R + r)     V = E − Ir = IR
Worked Example

A battery of e.m.f. 9.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor. Find the current, the terminal p.d. and the power dissipated in the resistor.

Practice Question

A cell has terminal p.d. 1.40 V when supplying 0.20 A and 1.30 V when supplying 0.40 A. Find its e.m.f. and internal resistance.

2. Kirchhoff's Laws
Kirchhoff's laws

Two conservation laws

Kirchhoff's first law: the sum of the currents entering a junction equals the sum leaving it. This follows from conservation of charge.

Kirchhoff's second law: in any closed loop, the sum of the e.m.f.s equals the sum of the p.d.s. This follows from conservation of energy.

Series (same current I): by the second law V = V1 + V2, so IR = IR1 + IR2 and R = R1 + R2. Parallel (same p.d. V): by the first law I = I1 + I2, so V/R = V/R1 + V/R2 and 1/R = 1/R1 + 1/R2. The combined resistance in parallel is always less than the smallest single resistor.

Combining resistors series: R = R1 + R2 + …     parallel: 1/R = 1/R1 + 1/R2 + …
Worked Example

A 6.0 Ω resistor is in series with a parallel pair of 3.0 Ω and 6.0 Ω, across a 12 V supply of negligible internal resistance. Find the current in each resistor.

Worked Example
Practice Question
3. Potential Dividers
Potential dividers
Memorise
Kirchhoff
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  • Sharing a p.d. and comparing p.d.s
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