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Subject 14 · Advanced Circuits

Advanced Circuits

Move from ideal rules to practical circuits: solve any network with Kirchhoff's laws, read real component characteristics, then design sensing and measuring circuits around them. Every subsection includes a responsive interactive model.

Exam diagrams for this topic3 figures to inspect and practiseQuestions, hints and marking points in one compact subsection.

See it. Read it. Work it.

These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.

  1. 01

    InspectRead the figure comment.

  2. 02

    TraceFollow labels, arrows and axes.

  3. 03

    AnswerWork one part at a time.

  4. 04

    CheckReveal hints and marking points.

A Level

01Fig. 10.1Electricity · D.C. circuitsA Level
A battery with internal resistance in series with a variable resistor and an ammeterRAe.m.f. 12.0 VrbatteryI

Figure comment

Fig. 10.1A single series loop. Along the bottom, a cell with its long plate on the left and a small resistor labelled r sit together inside a dashed rectangle labelled battery, the cell marked e.m.f. 12.0 V. From the dashed box the wire runs to the left, up the left-hand side and along the top through a resistor labelled R that has an arrow drawn across it to show that it is variable, then down the right-hand side through a circle marked A, and back along the bottom into the box. An arrow on the top wire labelled I shows the direction of the conventional current.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

9 Electricity10 D.C. circuitsOpen A Level question · AS Level foundations Q10
Guided questions 5 parts

Reading cue. r sits inside the dashed box with the cell, so read the box as the whole source, and read 12.0 V as labelling that source, not the p.d. across its terminals.

  1. aState The ammeter reads 2.0 A. State the current in the resistor r inside the dashed box, and give the reason from Fig. 10.1.

    recall2 marks

    Check answer 2 marks
    1. 2.0 A
    2. the components form a single series loop, so the current is the same at every point in it

  2. bCalculate With R set to 4.0 Ω the ammeter reads 2.0 A. Calculate the charge that passes through the ammeter in 5.0 minutes, and the energy delivered to R in that time.

    routine3 marks

    Check answer 3 marks
    1. Q = It = 2.0 × 300 = 6.0 × 10² C
    2. p.d. across R = 2.0 × 4.0 = 8.0 V
    3. energy = VQ = 8.0 × 600 = 4.8 × 10³ J

  3. cDetermine The internal resistance of the battery is 2.0 Ω. Determine the setting of R that makes the ammeter read 1.5 A.

    demanding3 marks

    Check answer 3 marks
    1. e.m.f. = I(R + r), so 12.0 = 1.5 × (R + 2.0)
    2. R + 2.0 = 8.0 Ω
    3. R = 6.0 Ω

  4. dDeduce A second identical battery is connected in series with the first in the loop of Fig. 10.1, with R left at 4.0 Ω. Deduce whether the ammeter reading doubles.

    top of the paper4 marks

    Check answer 4 marks
    1. total e.m.f. = 24.0 V
    2. total resistance = 4.0 + 2.0 + 2.0 = 8.0 Ω
    3. I = 24.0 / 8.0 = 3.0 A
    4. the reading rises from 2.0 A to 3.0 A, not to 4.0 A, because the internal resistance in the loop has doubled as well

Transfer challenge

A 12 V car battery of internal resistance 0.020 Ω delivers 150 A to a starter motor. Calculate the terminal potential difference of the battery while the motor is turning, and explain why the headlamps dim at that moment.

Check answer 3 marks
  1. lost volts = Ir = 150 × 0.020 = 3.0 V
  2. terminal p.d. = 12 − 3.0 = 9.0 V
  3. the lamps are connected across the terminals, so the p.d. across them falls and they are less bright

AP

02Figure 6Electric CircuitsAP
Bulb X in series with the parallel combination of bulbs Y and Z across a cellXYZe.m.f. εeach bulb has resistance R

Figure comment

Figure 6A circuit diagram drawn as a rectangular loop. A cell of e.m.f. ε sits in the left-hand side of the loop. Following the wire from the cell along the top of the loop, it passes through bulb X and then reaches a junction dot where the circuit divides into two parallel branches: bulb Y lies on the upper branch, while a wire dropping from the junction carries bulb Z along a lower branch. The two branches rejoin at a second junction dot, after which a single wire runs down the right-hand side and back along the bottom to the cell. A note on the figure states that each bulb has resistance R.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

11 Electric CircuitsOpen AP question · AP Physics 2: Algebra-Based Q10
Guided questions 5 parts

Reading cue. Find the two junction dots first: everything before them carries the whole current, so X is not one of three equals — Y and Z each take half of what X carries.

  1. aIndicate Indicate which bulb carries the largest current and which two carry equal currents, using only the way the wires meet at the junction dots.

    recall3 marks

    Check answer 3 marks
    1. X carries the largest current
    2. Y and Z carry equal currents
    3. reason: X carries the sum of the two branch currents, while Y and Z are identical resistances between the same pair of junctions

  2. bDetermine The cell has negligible internal resistance. Determine the potential difference across bulb X and across bulb Y, in terms of ε.

    routine4 marks

    Check answer 4 marks
    1. parallel pair has combined resistance R/2
    2. total circuit resistance 3R/2, so current from the cell I = 2ε/3R
    3. V_X = IR = 2ε/3
    4. V_Y = ε − V_X = ε/3

  3. cDetermine Taking the current from the cell to be I, determine the fraction of the total power delivered by the cell that is dissipated in each of the three bulbs, and show that the three fractions account for all of it.

    demanding5 marks

    Check answer 5 marks
    1. current in X is I; current in each of Y and Z is I/2 by equal division between identical branches
    2. P_X = I²R and P_Y = P_Z = (I/2)²R = I²R/4
    3. total power delivered = εI = I²(3R/2), since the cell drives I through an effective resistance 3R/2
    4. fractions 2/3 in X and 1/6 in each of Y and Z
    5. the three add to exactly 1, as they must when the cell has no internal resistance

  4. dDetermine Bulb Z is removed and replaced by a thick copper wire of negligible resistance joining the same two junction dots. Determine the new power dissipated in X and in Y, and determine the factor by which the total power delivered by the cell changes.

    top of the paper5 marks

    Check answer 5 marks
    1. the wire short-circuits Y, so the potential difference across the parallel section is zero
    2. P_Y = 0 and bulb Y goes out
    3. circuit resistance becomes R, so the current becomes ε/R
    4. P_X = ε²/R, against 4ε²/9R before, so X is 2.25 times brighter
    5. total power rises from 2ε²/3R to ε²/R, a factor of 1.5

Transfer challenge

A cell of e.m.f. ε has internal resistance R and is connected to two identical bulbs, each of resistance R, joined in parallel; nothing is in series outside the cell. Determine the fraction of the power delivered by the cell that is wasted inside it, and determine the terminal potential difference before and after a third identical bulb is added in parallel with the other two.

Check answer 5 marks
  1. the internal resistance plays the part X played: it carries the whole current
  2. external resistance R/2, total 3R/2, current I = 2ε/3R
  3. fraction wasted internally = I²R/(εI) = IR/ε = 2/3
  4. with three bulbs the external resistance is R/3 and I = 3ε/4R
  5. terminal potential difference falls from ε/3 to ε/4
03Fig. 8.1Conductors and Capacitors · Electric CircuitsAP
A resistor and capacitor in series with a battery and a switchswitch, closed at t = 0RCε

Figure comment

Fig. 8.1A single-loop circuit drawn as a rectangle. A cell of e.m.f. epsilon sits in the left-hand side of the loop, with its long positive plate uppermost. Along the top wire, reading from the left, come an open switch labelled S with the note that it is closed at t = 0, and then a rectangular resistor labelled R. A capacitor labelled C, drawn as two equal parallel plates, sits in the right-hand side of the loop. Everything is in series round the one loop: there are no branches and no junction dots anywhere in the circuit.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

10 Conductors and Capacitors11 Electric CircuitsOpen AP question · AP Physics C: Electricity and Magnetism Q8
Guided questions 5 parts

Reading cue. There are no junction dots: one current passes through R and C alike, and with C uncharged at t = 0 the capacitor acts momentarily like a plain wire, putting all of ε across R.

  1. aDetermine Determine the current in the circuit immediately after the switch closes, and a long time afterwards.

    recall3 marks

    Check answer 3 marks
    1. immediately after closing, the uncharged capacitor has no potential difference across it, so all of ε is across R and I₀ = ε/R
    2. after a long time the capacitor potential difference has risen to ε
    3. no potential difference is left across R, so the current is zero

  2. bCalculate Take ε = 12 V, R = 47 kΩ and C = 22 μF. Calculate the potential difference across the resistor and across the capacitor, and the current in the circuit, at the moment t = RC.

    routine4 marks

    Check answer 4 marks
    1. RC = 47 × 10³ × 22 × 10⁻⁶ = 1.03 s
    2. V_C = ε(1 − e⁻¹) = 12 × 0.632 = 7.6 V
    3. V_R = ε − V_C = 4.4 V
    4. I = V_R/R = 9.4 × 10⁻⁵ A

  3. cDetermine For the interval from t = 0 to t = RC only, determine the energy delivered by the battery, the energy stored in the capacitor and the energy dissipated in the resistor, and comment on whether the battery's energy divides equally between the two over this interval.

    demanding5 marks

    Check answer 5 marks
    1. charge delivered q = Cε(1 − e⁻¹) = 1.67 × 10⁻⁴ C
    2. energy from the battery = εq = 2.0 × 10⁻³ J
    3. energy stored = q²/2C = 6.3 × 10⁻⁴ J
    4. energy dissipated = 2.0 × 10⁻³ − 6.3 × 10⁻⁴ = 1.4 × 10⁻³ J
    5. only about 32% is stored over this interval, so the equal split holds for the whole charging process and not for a part of it

  4. dDetermine A long time after closing, the switch is reopened. Determine what happens to the charge on the capacitor, describe the smallest change to the circuit that would let the capacitor discharge through R, and determine how long that discharge would take to reduce the charge to 10% of its initial value.

    top of the paper5 marks

    Check answer 5 marks
    1. the figure is a single loop with the switch in series, so reopening it leaves no conducting path anywhere
    2. the charge Cε = 2.6 × 10⁻⁴ C simply stays on the capacitor, apart from slow leakage
    3. a path is needed that contains C and R but not the open switch, for example a two-way switch that replaces the battery branch with a plain wire
    4. discharge then follows q = Q e^(−t/RC)
    5. t = RC ln10 = 1.03 × 2.30 = 2.4 s

Transfer challenge

A camera flash stores energy in a 350 μF capacitor charged to 300 V through a 22 kΩ resistor, and then discharges it through a flash tube of resistance 5.0 Ω. Calculate the energy stored, the time taken to reach 95% of the final charge, and the peak current through the tube.

Check answer 5 marks
  1. energy = ½CV² = ½ × 350 × 10⁻⁶ × 300² = 16 J
  2. charging time constant = 22 × 10³ × 350 × 10⁻⁶ = 7.7 s
  3. t = RC ln20 = 7.7 × 3.00 = 23 s
  4. peak discharge current = V/R = 300/5.0 = 60 A
  5. the discharge time constant is only 5.0 Ω × 350 μF = 1.8 ms, which is why the flash is brief although the charging is slow