Electric Circuits Worked ExamplesIGCSE Physics 0625

(a) P is a battery (two cells: each long thin line is the positive terminal, each short thick line the negative). Q is a switch (shown open).

R is a fuse — a small rectangle with the wire running straight through it. S is a thermistor — a resistor rectangle with a diagonal line that has a horizontal foot.

(b) For the thermistor in the syllabus (NTC), the resistance decreases as its temperature increases.

The total resistance of the series circuit decreases while the battery is unchanged, so (i) the current increases — the ammeter reading increases — and (ii) the lamp becomes brighter, because the current in it is larger.

(a) Cells in series add their e.m.f.s when they face the same way. A reversed cell subtracts its e.m.f.

(b) Resistors in series add.

Use $I = \dfrac{V}{R}$. The current is the same at every point in a series circuit, so both ammeters read the same.

(c) Use $V = IR$ for the $2.0$ Ω resistor.

(d) Now all three e.m.f.s add.

Conventional current leaves the positive terminal (long line, at the top), flows along the top wire, down through the branches and back to the negative terminal. A diode conducts only when its arrow (triangle) points in the direction of the conventional current.

Diode D points to the right, away from the positive terminal, so it conducts and current reaches the branches.

In X the diode points down — with the current — so X lights. In Y the diode points up, against the current, so there is no current in Y. In Z the LED points down, so it conducts (and glows) and the lamp in Z lights. Answer: B.

D ignores the fact that a diode blocks current in the reverse direction. A wrongly treats the LED as blocking current — an LED is a diode that also emits light, so it conducts in its forward direction.

(a) Only the motor branch is complete, so the ammeter reads the motor current: $0.40$ A.

(b) Each parallel branch has the full supply p.d. across it, so the heater has $230$ V across it.

The current from the supply is the sum of the branch currents (currents into a junction equal currents out).

(c) Use the total current with the supply p.d.

This is less than $46$ Ω, the resistance of the heater alone: adding a parallel branch gives the current another path, so the combined resistance falls.

(d) Any two: each can be switched on and off independently; each receives the full supply p.d. ($230$ V); if one fails, the other still works.

(a) First combine the two resistors in parallel.

Add the series resistor.

(b) A₁ measures the supply current.

(c) V₁ is across the $2.0$ Ω resistor, which carries the whole current.

The p.d.s in series add up to the supply p.d. V₂ reads the p.d. across the parallel section, which is the same across each branch.

(d) Each branch has $14.4$ V across it.

Check at the junction: $1.2 + 3.6 = 4.8$ A, the reading on A₁.

(a) The whole current, $0.30$ A, passes through the $12$ Ω resistor.

(b) In a series circuit the p.d.s add up to the e.m.f., so the parallel section has what is left.

(c) The $24$ Ω resistor has $2.4$ V across it.

(d) The currents in the two branches add up to $0.30$ A.

$R$ also has $2.4$ V across it.

Check: $24$ Ω and $12$ Ω in parallel give $8.0$ Ω; $12 + 8.0 = 20$ Ω; $\dfrac{6.0}{20} = 0.30$ A. ✓

(a) Convert the current to amperes first.

Use $R = \dfrac{V}{I}$ with the battery's e.m.f.

(b) The resistances in series add, so subtract the fixed resistor ($2.2$ kΩ $= 2200$ Ω).

(c) The voltmeter reads the p.d. across the $2200$ Ω resistor.

(d) Warming the thermistor decreases its resistance, so the total resistance decreases and the current increases — the milliammeter reading increases.

The p.d. across the fixed resistor is $V = IR$ with $R$ unchanged and $I$ larger, so the voltmeter reading increases (the thermistor now takes a smaller share of the $6.0$ V).

(a) The two resistances share the $9.0$ V in proportion to their resistances. The fraction across the $10$ kΩ resistor is $\dfrac{10}{10 + R_{\text{LDR}}}$.

(i) Drawer closed, $R_{\text{LDR}} = 90$ kΩ.

(ii) Drawer open, $R_{\text{LDR}} = 5.0$ kΩ.

(b) The LDR and resistor form a potential divider: the supply p.d. is shared between them in the ratio of their resistances. When light falls on the LDR its resistance decreases, so it takes a smaller share of the $9.0$ V and the fixed resistor takes a larger share.

The p.d. across the resistor (and across the alarm, which is in parallel with it) rises from $0.90$ V to $6.0$ V, above $4.0$ V, so the alarm sounds.

(c) At the threshold the resistor has $4.0$ V and the LDR has $9.0 - 4.0 = 5.0$ V. Use $\dfrac{R_1}{R_2} = \dfrac{V_1}{V_2}$.

The p.d.s in series add up to the supply p.d., so the p.d. across R₂ is:

For a potential divider the p.d.s are in the same ratio as the resistances.

The answer is C. A comes from inverting the ratio ($2.0 \times 3.0 / 6.0$). D uses the supply p.d. instead of the p.d. across R₂ ($2.0 \times 9.0 / 3.0$) — that is the total resistance, not R₂.

(a) Use $R = \dfrac{V}{I}$ for each row.

(b) One resistor: $3.30$ Ω. Two identical resistors in series have twice the resistance of one, so divide by $2$.

Two identical resistors in parallel have half the resistance of one ($\frac{1}{R} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$), so multiply by $2$.

(c) Yes. The three values, $3.20$ Ω, $3.30$ Ω and $3.41$ Ω, differ by at most about $0.2$ Ω, which is within about $6\%$ of each other — close enough to be the same within the limits of experimental accuracy (about $10\%$).

(d) To stop the resistors (and wires) heating up, which would change their resistance; or to stop the cells running down between readings.

(a) B and C in parallel: two equal resistors in parallel have half the resistance of one.

All the current passes through A.

P.d.s: A takes $12$ Ω of the $18$ Ω, so two-thirds of the $6.0$ V; B and C share the rest equally.

(b) With C broken, A and B are simply in series.

The equal lamps now share the $6.0$ V equally.

(c) Brightness depends on the power $P = IV$. Lamp A: before $0.333 \times 4.0 = 1.33$ W, after $0.25 \times 3.0 = 0.75$ W — A gets dimmer, because the total resistance increased so the current through A fell.

Lamp B: before, its current was half of $0.333$ A, so $0.167 \times 2.0 = 0.33$ W; after, $0.25 \times 3.0 = 0.75$ W — B gets brighter, because it now has a larger share of the p.d. and all of the current.

(a) From the graph, at $20$ °C the thermistor's resistance is about $4.1$ kΩ. The fixed resistor's share of the supply p.d. is:

(b) At switch-on the fixed resistor has $3.0$ V, so the thermistor has $5.0 - 3.0 = 2.0$ V. Use $\dfrac{R_1}{R_2} = \dfrac{V_1}{V_2}$.

(c) Read across from $1.33$ kΩ to the curve and down to the temperature axis: about $45$ °C.

(d) The fan switches on when $\dfrac{R_{\text{th}}}{R} = \dfrac{2.0}{3.0}$. With a larger fixed resistance $R$, this happens at a larger thermistor resistance.

The thermistor's resistance is larger at a lower temperature, so increasing the fixed resistance (or replacing it with a variable resistor and turning it up) makes the fan switch on at a lower temperature.