Go premium Contact About Us
Logo
Home › Grade10 › Grade 10 CBSE › 10 CBSE Physics › Electricity › Worksheets on Electricity › Electricity Worksheet-4

Electricity Worksheet-4

Electricity Worksheet-4

 

  1. Calculate the potential difference required across a conductor of resistance 5 Ω to make a current of 1.5 A flow through it.

 

  1. A torch bulb  has 1 Ω resistance. It draws a current of 0.3 A when glowing from a source of 3 V. calculate the resistance of the bulb when glowing and explain the reason for the difference in resistance.

 

  1. How much current does an electric heater draw from a 220 V line, if the resistance of the heater (when hot) is 50 Ω?

 

  1. How much work is done in moving a charge of 2 C across two points having a potential difference of 12 V?

 

  1. An incandescent lamp of resistance 80 Ω draws a current of 0.75 A. Find the line voltage.

 

  1. A current of 0.2 A flows through a conductor of resistance 4.5 Ω. Calculate the potential difference at the ends of the conductor.

 

  1. A bulb of resistance 400 Ω is connected to 220 V mains. Calculate the magnitude of current.

 

  1. An electric heater draws a current of 5 A when connected to 220 V mains. Calculate the resistance of its filament.

 

  1. How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?

 

  1. The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?

 

Answer:

  1. Resistance of the conductor, R = 5 Ω

Required current,     I = 1.5 A

If V is the potential difference required across the conductor, then

          V = IR = 1.5 × 5 = 7.5 V

 

  1. Potential difference, V = 3 volt

Current through the bulb when glowing, I = 0.3 A

Resistance of the bulb when glowing,

The resistance of the filament of the bulb increases from 1 Ω to 10 Ω (when it becomes hot and glows) because of  an increase in its temperature.

 

  1. Potential difference, V = 220 V

Resistance of the heater,    R = 50 Ω

Using ohm’s law,

 

 

 

 

  1. I = V/R = 220/400 = 0.55 A

 

  1. R = V/I = 220/5 = 44 Ω

 

  2. Resistance of the electric heater, 

Changed potential difference    V' = 120 V

Changed current, i.e.,

Since R remains the same, by doubling the potential difference, the current is also doubled, i.e., becomes 8 A as I ∝ V.