Electrical Quantities Worked ExamplesIGCSE Physics 0625

(a) Before rubbing, the strip and cloth are uncharged: each has equal amounts of positive and negative charge. Friction between them transfers electrons (negative charge) from the strip to the cloth.

The strip has lost electrons, so it now has more positive charge than negative charge — it is positively charged. Only electrons move; the positive charges stay where they are.

(b) The cloth has gained the electrons, so it is negatively charged (with a charge equal in size to the strip's).

(c) Both strips are positively charged. Like charges repel, so the hanging strip is pushed away and moves away from the second strip.

Potential difference is the work done per unit charge, $V = \dfrac{W}{Q}$. Work is in joules and charge in coulombs, so $1 \text{ V} = 1 \text{ J/C}$: B.

A is the ratio upside down. C, joule per second, is the watt — the unit of power. D, ampere second, is the coulomb, because $Q = It$.

(a) Convert the time to seconds first.

Current is the charge passing per unit time, so $Q = It$.

(b) The p.d. is the work done per unit charge, $V = \dfrac{W}{Q}$, so $W = QV$.

(c) Write the equation first, then substitute.

Check: energy per second $= \dfrac{18\,000}{2400} = 7.5$ W, which agrees.

(a) An electric field is a region in which an electric charge experiences a force.

(b) Between the plates the field lines are straight, parallel and equally spaced, running at right angles to the plates. The direction of a field is the direction of the force on a positive charge, so the lines point from the positive plate to the negative plate — downwards.

(c) A negative charge feels a force opposite to the field direction. The field points downwards, so the force on the dust particle is upwards, towards the positive plate (unlike charges attract).

(d) Around the sphere the field lines are radial (like the spokes of a wheel), starting at the surface and spreading outwards, with arrows pointing inwards, towards the sphere — because a positive charge would be attracted towards it.

Resistance is directly proportional to length: tripling the length multiplies $R$ by $3$.

Resistance is inversely proportional to cross-sectional area. Area of a circle $= \pi r^2$, so area is proportional to $d^2$. Doubling the diameter multiplies the area by $2^2 = 4$, which divides $R$ by $4$.

Combine the two effects.

The answer is B. C ($18$ Ω) comes from dividing by $2$ instead of $4$ — treating $R$ as inversely proportional to the diameter. D multiplies by the area factor instead of dividing, and A divides by the length factor.

(a) X is a straight line through the origin: current is proportional to p.d., so it is a resistor of constant resistance (a fixed resistor, or a metal wire at constant temperature). Y is a curve that gets less steep: a filament lamp. Z conducts in one direction only: a diode.

(b) Read the current in Y at $3.0$ V from the graph: $0.16$ A. Use $R = \dfrac{V}{I}$.

At $6.0$ V the current in Y is $0.25$ A.

(c) As the p.d. increases, the current increases, so the filament gets hotter. The resistance of the filament increases with temperature ($19$ Ω to $24$ Ω above), so the current rises less and less for each extra volt — the graph curves over towards the $V$-axis. The shape is the same for negative p.d. because the lamp works either way round.

(d) For negative p.d. (reverse direction) the current in Z is zero — Z has a very high resistance. For positive p.d. the current stays almost zero until about $0.6$ V, then rises very steeply. The current flows in one direction only.

(a) Find the value of one small division first. Voltmeter: $1$ V is split into $10$ divisions, so each is $0.1$ V. The needle is $3$ divisions past $2$: $V = 2.3$ V.

Ammeter: $0.2$ A is split into $10$ divisions, so each is $0.02$ A. The needle is $3$ divisions before $0.40$: $I = 0.34$ A.

(b) Use $R = \dfrac{V}{I}$.

To 2 significant figures this rounds up to $6.8$ Ω (the third figure is $6$, so the second figure goes up).

(c) The ammeter is connected in series with the resistor; the voltmeter is connected in parallel (across) the resistor.

(a) $R = \dfrac{V}{I}$ for the last two rows, to 3 significant figures like the rest of the column.

(b) Plot $(0.80,\ 4.22)$ and $(1.00,\ 5.26)$ as neat crosses. All five points lie close to a straight line through the origin; draw one thin ruled line with the points balanced either side of it.

(c) $R$ is directly proportional to $L$, because the graph is a straight line that passes through the origin.

(d) Read across from $R = 3.5$ Ω to the line and down to the $L$-axis. The line's gradient is about $5.26$ Ω per metre.

(e) Any two of: the diameter (cross-sectional area) of the wire; the material of the wire; the temperature of the wire (for example by switching off between readings so it does not heat up).

(a) Write the equation, convert kW to W, rearrange and substitute.

(b) One kilowatt-hour is the energy transferred by an appliance of power $1$ kW in $1$ hour.

(c) Keep the power in kW and find the total time in hours.

Use $E = Pt$ with $P$ in kW and $t$ in h.

(d) Cost = energy in kW h × price per kW h.

(a) The e.m.f. is the electrical work done by a source in moving unit charge around a complete circuit.

(b) The cells are in series, all the same way round, so their e.m.f.s add.

(c) Convert the time to seconds and use $Q = It$.

(d) From $E = \dfrac{W}{Q}$, the work done by the battery is $W = EQ$.

(e) From $V = \dfrac{W}{Q}$, the energy transferred in X is $W = VQ$.

In a series circuit the p.d.s across the components add up to the e.m.f. (the cells and wires have negligible resistance).

(a) Find the current from $P = IV$.

Then the resistance from $R = \dfrac{V}{I}$.

(b) Resistance is directly proportional to length. Each metre of this wire has $5.5$ Ω.

(c) Halving the diameter divides the cross-sectional area by $2^2 = 4$, so the resistance of each metre is multiplied by $4$.

Now find the length needed.

(a) Metals contain free electrons that can move through the metal. The negative rod repels these electrons, so electrons move from A into B, away from the rod.

A is left with fewer electrons than positive charges, so A's side near the rod becomes positive; B has extra electrons, so it becomes negative. The positive charges do not move.

(b) Separating the spheres traps the electrons on B. Afterwards A is positive and B is negative, with charges of equal size (the total charge is still zero).

(c) The spheres are conductors, so electrons can move freely through them and across from A to B. The stands are insulators, so the charge cannot flow away to the ground; with conducting stands, the spheres would not keep their charge.