Electric circuits for Cambridge IGCSE Physics

Reading cue. Both resistors lie between the same two junction dots, so they have the same p.d. across them; it is the current, not the voltage, that divides at the left-hand dot.

1. aState State which of the two resistors in Fig. 3.1 carries the larger current, and give a reason.

   recall2 marks

   Check answer · 2 marks

   1. the 3.0 Ω resistor
   2. both branches lie between the same two junctions and so have the same p.d. across them, and I = V/R, so the smaller resistance carries the larger current

2. bDetermine The potential difference between the two junction dots in Fig. 3.1 is 12 V. Determine the current in each resistor and the current in the single wire arriving from the left.

   routine3 marks

   Check answer · 3 marks

   1. current in the 6.0 Ω resistor = 12/6.0 = 2.0 A
   2. current in the 3.0 Ω resistor = 12/3.0 = 4.0 A
   3. current in the wire from the left = 2.0 + 4.0 = 6.0 A

3. cDetermine Each resistor in Fig. 3.1 is rated at a maximum power of 12 W. Determine the largest potential difference that may safely be applied between the two junction dots, and the current drawn at that potential difference.

   demanding4 marks

   Check answer · 4 marks

   1. for the 3.0 Ω resistor, V = √(PR) = √(12 × 3.0) = 6.0 V
   2. for the 6.0 Ω resistor, V = √(12 × 6.0) = 8.5 V, so the 3.0 Ω resistor is the one that limits the pair
   3. largest safe p.d. = 6.0 V
   4. current drawn = 6.0/6.0 + 6.0/3.0 = 3.0 A

4. dExplain A student claims that connecting a third resistor between the same two junction dots in Fig. 3.1 must raise the total resistance, because there is then more resistance present. Explain why the student is wrong, and state the largest total resistance the arrangement could ever have once a third resistor is added there.

   top of the paper4 marks

   Check answer · 4 marks

   1. a resistor connected between the same two dots is in parallel with the others and gives the current an extra path
   2. for the same p.d. the total current is therefore larger, and R = V/I, so the total resistance falls
   3. the pair alone gives 1/R = 1/6.0 + 1/3.0, so R = 2.0 Ω
   4. adding a third branch can only bring the total below 2.0 Ω, so 2.0 Ω is a value the arrangement approaches but never reaches

Reading cue. The 40 Ω sits before the first junction dot, so it carries the whole battery current; only past that dot is the current shared between the 30 Ω and the 60 Ω.

1. aDetermine The current drawn from the battery in Fig. 7.1 is 0.20 A. Determine the potential difference across the 40 Ω resistor.

   recall2 marks

   Check answer · 2 marks

   1. the 40 Ω resistor lies before the junction, so it carries the whole 0.20 A
   2. V = IR = 0.20 × 40 = 8.0 V

2. bDetermine Determine the power dissipated in the 40 Ω resistor in Fig. 7.1, and the total power supplied by the battery.

   routine2 marks

   Check answer · 2 marks

   1. power in the 40 Ω resistor = I²R = 0.20² × 40 = 1.6 W
   2. total power from the battery = VI = 12 × 0.20 = 2.4 W

3. cDetermine The 40 Ω resistor in Fig. 7.1 is replaced by a variable resistor. Determine the resistance it must be set to for the potential difference across the parallel pair to be 6.0 V.

   demanding3 marks

   Check answer · 3 marks

   1. the parallel pair has a combined resistance of (30 × 60)/90 = 20 Ω, so the current is 6.0/20 = 0.30 A
   2. p.d. across the variable resistor = 12 − 6.0 = 6.0 V
   3. R = V/I = 6.0/0.30 = 20 Ω

4. dExplain The 30 Ω resistor in Fig. 7.1 is replaced by a filament lamp whose resistance rises as it warms up. Explain what happens to the potential difference across the 40 Ω resistor as the lamp warms, and state whether the current in the 60 Ω resistor rises or falls.

   top of the paper4 marks

   Check answer · 4 marks

   1. the resistance of the lamp rises, so the combined resistance of the parallel pair rises
   2. the total resistance of the circuit rises, so the current drawn from the battery falls
   3. the p.d. across the 40 Ω resistor is that current × 40, so it falls
   4. the parallel pair therefore takes a larger share of the 12 V, so the current in the 60 Ω resistor, equal to that p.d. divided by 60, rises

Reading cue. Trace all three wires out of the supply box before answering: note which one the fuse sits in, and note that one wire ends on the metal case instead of joining the heating element.

1. aIdentify Identify the wire in Fig. 9.1 that the fuse has been placed in, and identify the wire that carries no current while the shower is working normally.

   recall2 marks

   Check answer · 2 marks

   1. the fuse is in the live wire (1)
   2. the earth wire carries no current in normal operation (1)

2. bShow (that) Show that the resistance of the heating element drawn between the live and neutral wires is about 6 Ω when the shower runs at its marked power on the marked supply.

   routine3 marks

   Check answer · 3 marks

   1. current I = P / V = 8500 / 230 (1)
   2. I = 37 A (1)
   3. R = V / I = 230 / 37 = 6.2 Ω, which is about 6 Ω (1)

3. cDetermine Fuses rated at 13 A, 30 A and 45 A are available. Determine which one should be fitted at the position drawn in Fig. 9.1, and give a reason for rejecting each of the other two.

   demanding3 marks

   Check answer · 3 marks

   1. the normal working current is about 37 A (1)
   2. a 13 A and a 30 A fuse would both melt during normal use, cutting off the shower (1)
   3. the 45 A fuse is chosen: it is the lowest rating above the working current, so it still melts on a fault (1)

4. dSuggest The earth wire in Fig. 9.1 has broken away at the solid dot on the metal case, but the shower still heats the water normally. Suggest why this fault is dangerous, and suggest how it could be found before anyone is hurt.

   top of the paper4 marks

   Check answer · 4 marks

   1. the live–neutral circuit through the element is untouched, so the shower still works and there is no sign of the fault (1)
   2. if the live wire later touches the case, the case becomes live and there is now no low-resistance path to earth (1)
   3. the current is then too small to melt the fuse, and only flows when a person touches the case, through their body (1)
   4. test the continuity between the metal case and the earth of the supply / have the appliance checked regularly by a qualified electrician (1)