Mains Electricity Questions and Answers - Physics Form 4 Topical Revision

Questions

1. What current will a 500Ω resistor connected to a source of 240V draw?
2. Name a device used to change light energy directly into electrical energy.
3. 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.
4. An electric bulb rated 40W is operating on 240v mains. Determine the resistance of its filament.
5. 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.
6. A 60W bulb is used continuously for 36 hours. Determine the energy consumed, giving your answer in kilowatt hour (kwh)
7. How many 1000W electric irons could be safely be connected to a 240V mains circuit fitted with a 13A fuse?
8. Find the maximum number of 75W bulbs that can be connected to a 13A fuse on a mains supply of 240V.
9. 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.
10. State Ohm’s law.
11. Electrical energy costs Kshs. 1 per Kwh unit. Find the cost of using an electric heater of power 1.5 Kw for a day.
12. The figure below represents part of the main circuit
    
    
    1. Explain why it is not advisable to fix a fuse on neutral line.
    2. Explain why there are fuses of different rating in the distribution box.
13. 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.
14. 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

1. I = ^V/_R
   = 240 A
      500
   = 0.48 A
2. Solar (photo) cell.
3. 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
4. P = V²/_R
   40 = 240 x 240
                 R
   R = 240 x 240 Ω
              40
   = 1440Ω
5. Current in the heater = ^P/_V = ³⁰⁰⁰/₂₄₀ = 12.5A
   Fuse not suitable since current exceeds the fuse value.
6. 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
7. (No. of irons) x 1000 = IV
   No. of irons = 13 x 240 = 3.12
                           100
   = 3
8. Maximum power = VI
   = (240 x 13)W
   No. of 75W bulbs = ²⁴⁰/₇₅ x 13
   = ³¹²⁰/₇₅
   = 41.6
   Maximum number bulbs = 41
9. No. of kwh = ^1500kw/₁₀₀₀ x 30 Kwh
   = 45Kwh
   Cost = Kshs. (45 x 8)
   = Kshs. 360/-
10. Current flowing through a conductor is directly proportional to potential difference across its ends provided temperature and other physical quantities remain constant.
11. No. of Kwh = (1.5 x 24)Kwh
    = 36Kwh
    Cost = No. of Kwh x price per Kwh
    = (36 x 1) = 36Kwh.
12.  
    1. When a fault occurs on an electrical appliances damage will still be done since current flows through the “live”
    2. There are different circuits and each carries a different amount of current.
13. Power = current x voltage.
    = (0.5 x 250)W
    = Kshs. 125W
14. Energy dissipated in 6hrs = (2.5 x 6)Kwh
    Cost = Kshs. (15 x 1.10)
    = Kshs. 16.50