Heating Effect of an Electric Current Questions and Answers - Physics Form 3 Topical Revision

Questions

1. An electric bulb rated 40W is operating on 240V mains. Determine the resistance of its filament
2. When a current of 2A flows in a resistor for 10 minutes, 15 KJ of electrical energy is dissipated. Determine the voltage across the resistor.
3. How many 100W electric irons could be safely connected to a 240V moving circuit fitted with a 13A fuse?
4. A heater of resistance R₁ is rated P watts, V volts while another of resistance R₂ is rated 2P watts, ^V/₂ volts. Determine ^R1/_R2
5. State THREE factors which affect heating by an electric current.
6. What is power as it relates to electrical energy?
7. An electrical appliance is rated as 240V, 200W. What does this information mean?
8. An electrical heater is labelled 120W, 240V.
   Calculate;
   1. The current through the heating element when the heater is on.
   2. The resistance of the element used in the heater.
9. An electric toy is rated 100W, 240V. Calculate the resistance of the toy when operating normally.

Answers

1. P = V²
         R
   40 = 240²
            R
   R= 1440 Ω
2. P= VI
   ^W/_t = VI
   ¹⁵⁰⁰⁰/_{10 x 60} = V x 2
   V=     15000    
        10 x 60 x 2
   = 150
       12     
   = 12.5V                      
   Voltage across resistor is 12.5V
3. Solution
   (No. of ions) x 1000 = IV
   No. of ions      = 13 x 240
                              1000
   = 3.12             
   = 3A
4. Solution
   R₁ =V² ………….. (i)
          P
   R₂ = (^V/₂)² ÷ 2P
   = V² x 1 = V² ……. (ii)
      4     2P   8P
   
   R₁ = V² ÷ V²
   R₂    P     8P  
   R₁ = V² × 8P
   R₂    P       V²      
   = 8
5.  
   • Amount of current, I
   • Resistance, R of the conductor
   • Time t for which the current flows
6. Is the rate at which electrical energy is converted to useful work per unit time?
7. That the appliance operates at a voltage of 240 volts. When it is operating normally, the electrical power outputs is 200 watts i.e. 200J of electrical energy is converted to other useful energy per unit time
8.  
   1. 
      Electrical power P = VI
                            120 = 240I
                             ∴ I = ¹²⁰/₂₄₀               
                                 = 0.5A
   2. From Ohm’s law
      V= IR
      R= ^V/_I
      = ²⁴⁰/_0.5               
      = 480Ω
      The resistance of element = 480Ω
9. Solution
   P= V²/_R
   R= V²/_P
   = 240 x 240
         100
   = 24 x 24        
   = 576Ω