Electrical Revision

Electrical Revision

Electrical Revision


A small electric torch is powered by a single cell which supplies 1.6 J of energy per coulomb of charge passing through the cell. When the torch is switched on, the cell supplies a constant current of 0.50 A to the bulb X. The potential difference across the bulb is 1.2 V.

(a)What is meant by “1.6 J of energy per coulomb of charge passing through the cell”?

(b)Show that the internal resistance r of the cell is 0.80 Ω.

(c)The bulb X is replaced by another bulb Y which is found to take a current of 0.30 A. Calculate

(i)the new potential difference across the internal resistance of the cell,

(ii)the potential difference across bulb Y.

(d)Show quantitatively why bulb Y is more efficient.


The circuit in Fig 3.1 below has a thermistor connected in series to a 200 Ω resistor and a 12 V battery of negligible internal resistance. Fig 3.2 shows how the resistance Rth of the thermistor varies with temperature.

(a)(i)Calculate the current in the circuit when the temperature is 25 °C.

(ii)Calculate the potential difference across the thermistor at 25 °C.

(b)Without further calculation, explain how you would expect the potential difference across the thermistor to change as temperature increases from 25 °C.

(c)The circuit in Fig 3.1 is modified by removing the 200 Ω resistance to give the circuit shown in Fig 3.3. The temperature of the thermistor is increased at a steady rate from 25 °C to 45 °C in 10 min.

(i)Calculate the power dissipated in the thermistor at 25 °C and 45 °C.

(ii)By taking the average of your answers in (c)(i), calculate an approximate value for the energy supplied by the battery during the period in which the temperature of the thermistor increases.

(iii)Explain why the energy determined in (c)(ii) is only an estimate.


A 4.0 V cell of negligible internal resistance is connected in series with two resistors of resistances 600 Ω and 400 Ω respectively (See Fig 4.1).

(a)What is the potential difference across the 400 Ω resistor?

(b)A voltmeter is connected across the 400 Ω resistor as shown in Fig 4.2.

If the voltmeter reads 1.40 V, calculate the resistance of the voltmeter.

(c)What resistance must the voltmeter have in order that it measures the potential difference across the 400 Ω resistor in Fig 4.2 to an accuracy of 1%?



Fig 8.1 shows a 15 V source of negligible internal resistance.

(i)With switch K open, find the effective resistance between point C and point D.

(ii)With switch K closed and given that the potential at point A is – 3.0 V,

1.determine the potential at point C and,

2.calculate the current I.


Fig 8.2 shows a constantan wire XY of length 75.0 cm of uniform cross-sectional area 1.2 × 108 m2. The 2.0 V source is of negligible internal resistance.

(i)Given that the resistivity of constantan is 49.0 × 10-8 Ω m, show that the resistance of the wire is 30.6 Ω.

(ii)If the galvanometer registers a null deflection when length XP is 50.0 cm,

  1. show that the potential difference across XY is 1.82 V and

2.determine the ammeter reading assuming it has negligible resistance.


The 2.00 Ω resistor is now replaced with a 4.00 Ω resistor as shown in Fig 8.3. The current shown on the ammeter is observed to be 0.382 A when the galvanometer registers a null deflection.

(i)Determine the e.m.f. and internal resistance of cell Q.

(ii)Describe and explain the qualitative change (if any) in the balance length XP if the ammeter has non-negligible resistance.


A light bulb, rated 80 W, is connected to an alternating power supply whose voltage is given by V = 110 sin (100πt)

(a)Calculate the value of the root-mean-square current in the bulb.

(b)Calculate the maximum power used by the bulb at any given instant of time.

(c)On Fig 5.1, sketch the variation of the power dissipated in the bulb with respect to time over a duration of 0.040 s. Mark appropriate numerical values along the axes.


In a 1200 W hairdryer shown in Fig 5.1, alternating current of 50 Hz passes mostly through the heating coils which act as a pure resistor.

(a)(i)Suppose the hair dryer is connected to the mains supply of 120 V r.m.s., calculate the peak current through the coils.

(ii)Write an equation in terms of twhere t is time, for the current that passes through the heating coil, given that power output is zero at t = 0.

(iii)The hairdryer is connected to a c.r.o. whose Y-plate sensitivity and time-base are set at 100 V sm-1 and 5.0 ms cm-1 respectively.

On Fig 5.2, sketch what is seen on the c.r.o. to illustrate the variation of voltage through the coil with time.

(b)The primary coil of a transformer shown in Fig 5.3 is connected to a 2.4 kV r.m.s. supply. The secondary coil is connected to the hair dryer and the current flowing through the heating coil has the same value as that calculated in (a)(i),

Given that the transformer is non-ideal and that electrical energy is converted to thermal energy in the windings of the transformer at a rate of 600 W, determine the primary current Ip.