Cambridge AS & A Level9702

D.C. circuits

Physics 9702 Chapter Notes

What this chapter covers

D.C. circuits - Practical circuitsD.C. circuits - Kirchhoff’s lawsD.C. circuits - Potential dividers
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In this chapter

  1. 1Kirchhoff's First Law: The Current Law
  2. 2Kirchhoff's Second Law: The Voltage Law
  3. 3Resistors in Series
  4. 4Resistors in Parallel
  5. 5Ammeters and Voltmeters

1. Kirchhoff's First Law: The Current Law

Kirchhoff's First Law is based on the principle of conservation of charge. It states that the total current flowing into any junction (a point where wires meet) in a circuit must equal the total current flowing out of that junction. Imagine it like water flowing through pipes; you can't lose any water at a T-junction. Charge carriers (electrons) cannot be created or destroyed at a point, so what flows in must flow out. This law is essential for analysing how current splits in parallel circuits.

ΣI_in = ΣI_out

Key term

Junction: A point in an electrical circuit where three or more conductors meet.

Examiner insight

Examiners look for a clear statement of the law (sum of currents in = sum of currents out) and its correct application at a specific junction in the circuit diagram.

Fun fact

This simple rule is fundamental to designing complex microchips, where billions of tiny junctions direct the flow of current to perform calculations.

Worked example 12 marks

In the circuit junction shown, a current of 5.0 A flows in, and a current of 2.0 A flows out along one branch. Calculate the current, I, in the other branch and state its direction.

  1. 1

    Step 1: Apply Kirchhoff's First Law, which states that the sum of currents entering a junction equals the sum of currents leaving it.

  2. 2

    ΣI_in = ΣI_out

  3. 3

    Step 2: Identify the currents entering and leaving. The current entering is 5.0 A. The currents leaving are 2.0 A and the unknown current I.

  4. 4

    5.0 A = 2.0 A + I

  5. 5

    Step 3: Rearrange the equation to solve for I.

  6. 6

    I = 5.0 A - 2.0 A

  7. 7

    I = 3.0 A

  8. 8

    Step 4: State the direction. Since our calculated value for I is positive, our assumed direction (leaving the junction) is correct. The current is 3.0 A leaving the junction.

Recap

  • Kirchhoff's First Law deals with the conservation of charge.
  • The total current entering a junction is equal to the total current leaving it.
  • This law is used to determine how current splits in parallel branches.
  • A junction is any point where three or more wires connect.

Quick check

  1. Three wires meet at a junction. Currents of 1.5 A and 2.5 A flow into the junction. What is the magnitude and direction of the current in the third wire?2 marks

2. Kirchhoff's Second Law: The Voltage Law

Kirchhoff's Second Law is based on the principle of conservation of energy. It states that for any closed loop in a circuit, the sum of the electromotive forces (e.m.f.s) is equal to the sum of the potential differences (p.d.s). In simpler terms, the total energy supplied by the batteries in a loop must be equal to the total energy used by the components in that same loop. An e.m.f. is an 'energy source' (like a battery), providing energy per unit charge. A p.d. is an 'energy user' (like a resistor), dissipating energy per unit charge.

ΣE = ΣV

Σ(e.m.f.s) = Σ(p.d.s)

ΣE = Σ(IR)

Key term

Electromotive Force (e.m.f.): The energy supplied by a source (like a battery) per unit charge that passes through it, measured in volts (V).

Examiner insight

Examiners award marks for correctly identifying a valid loop, summing the e.m.f.s (with correct signs), and summing the p.d.s (IR drops) before equating them.

Common pitfall

Students often forget to account for the direction of e.m.f.s. If two batteries are connected in opposition (e.g., positive to positive), their e.m.f.s must be subtracted, not added.

Worked example 13 marks

A circuit consists of a 12.0 V battery connected in series with a 2.0 Ω resistor and a 4.0 Ω resistor. Calculate the current flowing in the circuit.

  1. 1

    Step 1: Identify a closed loop. The entire circuit is a single closed loop.

  2. 2

    Step 2: Apply Kirchhoff's Second Law: ΣE = Σ(IR).

  3. 3

    Step 3: Sum the e.m.f.s in the loop. There is only one battery, so ΣE = 12.0 V.

  4. 4

    Step 4: Sum the p.d.s across the resistors. The resistors are in series, so the same current I flows through both. The total p.d. is the sum of the individual p.d.s: Σ(IR) = (I × 2.0 Ω) + (I × 4.0 Ω).

  5. 5

    Σ(IR) = I × (2.0 + 4.0) = 6.0I

  6. 6

    Step 5: Equate the sum of e.m.f.s and the sum of p.d.s and solve for I.

  7. 7

    12.0 V = 6.0I

  8. 8

    I = 12.0 / 6.0 = 2.0 A

Worked example 23 marks

In the loop shown, two batteries are connected in opposition. Calculate the current I.

  1. 1

    Step 1: Choose a direction for the current, let's assume clockwise. Apply Kirchhoff's Second Law: ΣE = Σ(IR).

  2. 2

    Step 2: Sum the e.m.f.s, taking direction into account. The 9.0 V battery drives current clockwise, while the 3.0 V battery opposes it. So, the net e.m.f. is ΣE = 9.0 V - 3.0 V = 6.0 V.

  3. 3

    Step 3: Sum the p.d.s around the loop. The p.d.s are across the 10 Ω and 20 Ω resistors. Σ(IR) = (I × 10 Ω) + (I × 20 Ω) = 30I.

  4. 4

    Step 4: Equate and solve for I.

  5. 5

    6.0 V = 30I

  6. 6

    I = 6.0 / 30 = 0.20 A

  7. 7

    Since the result is positive, our assumed clockwise direction is correct.

Recap

  • Kirchhoff's Second Law deals with the conservation of energy in a circuit loop.
  • The sum of e.m.f.s around any closed loop equals the sum of p.d.s in that loop.
  • E.m.f. is energy supplied per unit charge; p.d. is energy dissipated per unit charge.
  • Pay careful attention to the direction of batteries; opposing e.m.f.s are subtracted.
  • A closed loop is any complete path for current to flow around and return to its starting point.

Quick check

  1. A 6V battery is connected to a resistor. If the p.d. across the resistor is 6V, what can you say about the internal resistance of the battery?1 mark
  2. State the difference between e.m.f. and p.d. in terms of energy transfer.2 marks

3. Resistors in Series

When resistors are connected in series, they are placed one after another in a single line, forming a single path for the current. This means the current is the same through each resistor. The total potential difference from the power source is shared between the resistors; the bigger the resistance of a component, the larger its share of the p.d. The total resistance of the combination, called the equivalent resistance (R_T), is simply the sum of the individual resistances.

R_T = R_1 + R_2 + R_3 + ...

Key term

Series Circuit: A circuit in which components are connected end-to-end, providing only one path for the current to flow.

Common pitfall

A common mistake is assuming the voltage is the same across each resistor in series. Remember, voltage is shared, not constant.

Worked example 14 marks

Three resistors with resistances 5.0 Ω, 10.0 Ω, and 15.0 Ω are connected in series to a 6.0 V battery with negligible internal resistance. Calculate(a) the total resistance of the circuit,(b) the current flowing from the battery, and(c) the potential difference across the 10.0 Ω resistor.

  1. 1

    Part (a): Calculate the total resistance.

  2. 2

    Step 1: For resistors in series, the total resistance is the sum of individual resistances.

  3. 3

    R_T = R_1 + R_2 + R_3

  4. 4

    R_T = 5.0 Ω + 10.0 Ω + 15.0 Ω = 30.0 Ω

  5. 5

    Part (b): Calculate the current.

  6. 6

    Step 2: Use Ohm's Law for the whole circuit, V_T = I_T * R_T.

  7. 7

    I_T = V_T / R_T = 6.0 V / 30.0 Ω = 0.20 A

  8. 8

    Part (c): Calculate the p.d. across the 10.0 Ω resistor.

  9. 9

    Step 3: Use Ohm's Law for the specific resistor. The current is the same (0.20 A) through all components in series.

  10. 10

    V_10 = I * R_10 = 0.20 A × 10.0 Ω = 2.0 V

Recap

  • In a series circuit, there is only one path for the current.
  • The current is the same through all components in series.
  • The total p.d. is shared between the components.
  • Total resistance is the sum of the individual resistances: R_T = R_1 + R_2 + ...
  • Adding a resistor in series always increases the total resistance of the circuit.

Quick check

  1. Two resistors, 100 Ω and 200 Ω, are in series. What is their total resistance?1 mark
  2. If the current through the 100 Ω resistor in the question above is 0.05 A, what is the current through the 200 Ω resistor?1 mark

4. Resistors in Parallel

When resistors are connected in parallel, they are arranged in separate branches. This provides multiple paths for the current. The current from the source splits at a junction, with the amount of current in each branch depending on its resistance (more current flows through the path of least resistance). A key feature of parallel circuits is that the potential difference across each branch is the same. The total (equivalent) resistance of a parallel combination is always less than the smallest individual resistance in the combination, because adding more paths makes it easier for current to flow.

1/R_T = 1/R_1 + 1/R_2 + 1/R_3 + ...

For two resistors: R_T = (R_1 * R_2) / (R_1 + R_2)

Key term

Parallel Circuit: A circuit in which components are connected across the same two points, providing multiple paths for the current.

Examiner insight

Examiners often test understanding by asking which resistor in a parallel combination dissipates the most power. Since P = V²/R and V is constant, the resistor with the lower resistance will dissipate more power.

Common pitfall

The most frequent error when calculating parallel resistance is forgetting to take the reciprocal at the end. Students often calculate 1/R_T and give that as the final answer, instead of R_T.

Worked example 15 marks

A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel to a 12 V power supply. Calculate(a) the total resistance of the combination,(b) the total current from the supply, and(c) the current through the 6.0 Ω resistor.

  1. 1

    Part (a): Calculate the total resistance.

  2. 2

    Step 1: Use the formula for parallel resistors.

  3. 3

    1/R_T = 1/R_1 + 1/R_2 = 1/6.0 + 1/3.0

  4. 4

    Step 2: Find a common denominator to add the fractions.

  5. 5

    1/R_T = 1/6.0 + 2/6.0 = 3/6.0 = 1/2.0

  6. 6

    Step 3: Invert the result to find R_T. Do not forget this step!

  7. 7

    R_T = 2.0 Ω

  8. 8

    Part (b): Calculate the total current.

  9. 9

    Step 4: Use Ohm's Law for the whole circuit, V_T = I_T * R_T.

  10. 10

    I_T = V_T / R_T = 12 V / 2.0 Ω = 6.0 A

  11. 11

    Part (c): Calculate the current in the 6.0 Ω resistor.

  12. 12

    Step 5: The p.d. across each parallel branch is the same as the supply voltage, 12 V. Use Ohm's Law for that branch.

  13. 13

    I_6 = V / R_6 = 12 V / 6.0 Ω = 2.0 A

Recap

  • In a parallel circuit, there are multiple paths for the current.
  • The potential difference is the same across all components in parallel.
  • The total current is the sum of the currents in the individual branches.
  • The reciprocal of the total resistance is the sum of the reciprocals of individual resistances.
  • Adding a resistor in parallel always decreases the total resistance.

Quick check

  1. Two identical 10 Ω resistors are connected in parallel. What is their total resistance?1 mark
  2. A 12V battery is connected to a parallel combination of a 4Ω and a 6Ω resistor. What is the p.d. across the 6Ω resistor?1 mark

5. Ammeters and Voltmeters

Ammeters and voltmeters are instruments used to measure current and potential difference. For a meter to be useful, it must not significantly change the values it is trying to measure. An ammeter measures the current flowing through a component, so it must be connected in series with it. To avoid reducing the current it's measuring, an ideal ammeter has zero resistance. In practice, real ammeters have a very low resistance. A voltmeter measures the potential difference across a component, so it must be connected in parallel with it. To avoid drawing significant current from the main circuit (which would alter the p.d. it's measuring), an ideal voltmeter has infinite resistance. In practice, real voltmeters have a very high resistance.

Key term

Loading Effect: The alteration of a circuit's behaviour and readings caused by connecting a measuring instrument with non-ideal resistance.

Examiner insight

For full marks when explaining why meters have high/low resistance, you must link the connection method (series/parallel) to the need to avoid altering the quantity being measured (current/p.d.).

Fun fact

Early voltmeters were galvanometers with a large series resistor. The higher the series resistance, the more voltage was needed to produce a certain deflection, allowing the scale to be calibrated in volts.

Worked example 14 marks

A 12 V battery with negligible internal resistance is connected to two 600 Ω resistors in series. A voltmeter with a resistance of 1200 Ω is connected in parallel across one of the 600 Ω resistors. What is the reading on the voltmeter?

  1. 1

    Step 1: First, determine the p.d. across the resistor without the voltmeter. The two 600 Ω resistors act as a potential divider. The p.d. across one is (600 / (600+600)) * 12 V = 6.0 V. This is the 'true' value.

  2. 2

    Step 2: Now consider the effect of the voltmeter. The voltmeter (1200 Ω) is in parallel with one of the 600 Ω resistors. Calculate the equivalent resistance of this parallel combination.

  3. 3

    1/R_p = 1/600 + 1/1200 = 2/1200 + 1/1200 = 3/1200

  4. 4

    R_p = 1200 / 3 = 400 Ω

  5. 5

    Step 3: The circuit is now effectively a 400 Ω resistor in series with the other 600 Ω resistor. Calculate the total resistance of this new circuit.

  6. 6

    R_T = 400 Ω + 600 Ω = 1000 Ω

  7. 7

    Step 4: The voltmeter reading is the p.d. across the parallel combination (R_p). This combination is part of a new potential divider. Use the potential divider formula.

  8. 8

    V_reading = (R_p / R_T) * V_supply = (400 Ω / 1000 Ω) * 12 V

  9. 9

    V_reading = 0.4 * 12 V = 4.8 V

  10. 10

    Step 5: Note that the voltmeter reading (4.8 V) is lower than the true p.d. (6.0 V) because its non-infinite resistance 'loaded' the circuit.

Recap

  • Ammeters are connected in series and must have very low resistance.
  • Voltmeters are connected in parallel and must have very high resistance.
  • An ideal ammeter has zero resistance; an ideal voltmeter has infinite resistance.
  • A real voltmeter draws a small current, which can alter the p.d. it is intended to measure (the loading effect).
  • A real ammeter has a small p.d. across it, which can reduce the total current in the circuit.

Quick check

  1. To measure the current through a lamp, where should you connect an ammeter?1 mark
  2. Why would a voltmeter with a resistance of 100 Ω be a very poor instrument for measuring p.d. in typical school lab circuits?2 marks

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. State Kirchhoff's first and second laws, and name the physical principle that each law is a consequence of.4 marks
  2. A 4.0 Ω resistor and an 8.0 Ω resistor are connected in parallel. This combination is then connected in series with a 3.0 Ω resistor and a 12 V battery of negligible internal resistance. Calculate the total current drawn from the battery.4 marks
  3. Explain, with reference to its connection in a circuit, why an ammeter must have a very low resistance.2 marks
  4. In the circuit shown, the current passing through the 6.0 Ω resistor is 1.5 A. Calculate (a) the potential difference across the parallel combination, (b) the current through the 4.0 Ω resistor, and (c) the current I from the source.5 marks
  5. A circuit loop contains a 12 V cell and a 4V cell connected in opposition. The total resistance of the loop is 16 Ω. Calculate the power dissipated in the circuit.3 marks
  6. A potential divider is made by connecting a 10 kΩ resistor and a 20 kΩ resistor in series with a 9 V supply. A voltmeter is connected across the 20 kΩ resistor. If the voltmeter has infinite resistance, what is its reading?2 marks
  7. Referring to the previous question, the voltmeter is replaced with one that has a resistance of 20 kΩ. Calculate the new reading on this voltmeter.4 marks
  8. For the circuit below, use Kirchhoff's laws to find the current in the 5.0 Ω resistor. (Diagram shows a 10V battery connected to a junction. One branch has a 5.0 Ω resistor. The other branch has a 2.0 Ω resistor in series with a 4V battery opposing the 10V battery. The branches then rejoin).6 marks
  9. Show, using Kirchhoff's laws, that the total resistance R_T for two resistors R_1 and R_2 in parallel is given by R_T = (R_1 * R_2) / (R_1 + R_2).4 marks
  10. A student claims that in a series circuit, the resistor with the highest resistance dissipates the most power, while in a parallel circuit, the resistor with the lowest resistance dissipates the most power. Justify whether the student is correct for both cases using relevant equations.4 marks

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