## Questions

- What current will a 500Ω resistor connected to a source of 240V draw?
- Name a device used to change light energy directly into electrical energy.
- When a current of 2.0A flows in a resistor for 10 minutes, 15000 Joules of electrical energy is dissipated. Determine the voltage across the resistor.
- An electric bulb rated 40W is operating on 240v mains. Determine the resistance of its filament.
- An electric heater rated 240V, 3000V is to be connected to a 240V mains supply, through a 10A fuse. Determine whether the fuse is suitable or not.
- A 60W bulb is used continuously for 36 hours. Determine the energy consumed, giving your answer in kilowatt hour (kwh)
- How many 1000W electric irons could be safely be connected to a 240V mains circuit fitted with a 13A fuse?
- Find the maximum number of 75W bulbs that can be connected to a 13A fuse on a mains supply of 240V.
- Determine the cost of using an electrical iron box rated 1500W, for a total of 30 hours given that the cost of electricity per kwh is Kshs. 8.
- State Ohm’s law.
- Electrical energy costs Kshs. 1 per Kwh unit. Find the cost of using an electric heater of power 1.5 Kw for a day.
- The figure below represents part of the main circuit
- Explain why it is not advisable to fix a fuse on neutral line.
- Explain why there are fuses of different rating in the distribution box.

- Calculate the power of a devise which has a p.d of 250V applied across it when a current of 0.5A passes through it.
- An electric iron box is rated 2500W and uses a voltage of 240V. Given that electricity costs Kshs. 1.10 per Kwh, what is the cost of using it for 6 hours?

## Answers

- I =
^{V}/_{R}= 240 A

500

= 0.48 A - Solar (photo) cell.
- Power = Energy used

Time taken

Power = Current x Voltage

∴ 15000 = 2 x v

10 x 60

∴ V = 15000 = 150

2 x 10 x 60 12

Voltage across resistor = 12.5V - P = V
^{2}/_{R}

40 = 240 x 240

R

R = 240 x 240 Ω

40

= 1440Ω - Current in the heater =
^{P}/_{V}=^{3000}/_{240}= 12.5A

Fuse not suitable since current exceeds the fuse value. - E = Pt

= 60 x 36 x 60 x 60J

E in Kwh = 60 x 36 x 60 x 60 KWh

1000 x 60 x 60

= 2.16 KWh - (No. of irons) x 1000 = IV

No. of irons = 13 x 240 = 3.12

100

= 3 - Maximum power = VI

= (240 x 13)W

No. of 75W bulbs =^{240}/_{75}x 13

=^{3120}/_{75}

= 41.6

Maximum number bulbs = 41 - No. of kwh =
^{1500kw}/_{1000}x 30 Kwh

= 45Kwh

Cost = Kshs. (45 x 8)

= Kshs. 360/- - Current flowing through a conductor is directly proportional to potential difference across its ends provided temperature and other physical quantities remain constant.
- No. of Kwh = (1.5 x 24)Kwh

= 36Kwh

Cost = No. of Kwh x price per Kwh

= (36 x 1) = 36Kwh. -
- When a fault occurs on an electrical appliances damage will still be done since current flows through the “live”
- There are different circuits and each carries a different amount of current.

- Power = current x voltage.

= (0.5 x 250)W

= Kshs. 125W - Energy dissipated in 6hrs = (2.5 x 6)Kwh

Cost = Kshs. (15 x 1.10)

= Kshs. 16.50

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