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Home › Grade10 › Grade 10 CBSE › 10 CBSE Physics › Electricity › Worksheets on Electricity › Electricity Worksheet-25

Electricity Worksheet-25

Electricity Worksheet-25

 

  1. A wire 1.0 m long, 2.0 mm in diameter, has a resistance of 10 Ω. Calculate the resistivity of its material.

 

  1. Calculate the area of cross – section of a wire if its length is 1.0m, its resistance is 23 Ω and the resistivity of the material is 1.84 × 10–6 Ω m.

 

  1. Calculate the resistance of an electric bulb which allows a 10 A current when connected to a 220 V power source.

 

  1. A lamp rated 100 W at 220 V is connected to the mains electric supply. What current is drawn from the supply line if the voltage is 220 V ?

 

  1. Calculate the electric energy consumed by a 1200 W toaster in 20 minutes.

 

  1. A torch bulb is rated 5.0 V and 500 mA. Calculate its (i) power (ii) resistance and (iii) energy consumed when it is lighted for 4 hours.

 

  1. Two electric lamps of 100 W and 25 W respectively are joined in parallel to a supply of 200 V. Calculate the total current flowing through the circuit.

 

  1. Two identical resistors, each of resistance 2 Ω, are connected in turn (i) in series, and (ii) in parallel, to a battery of 12 V. Calculate the ratio of power consumed in the two cases.

 

  1. Two identical resistors, each of resistance 10 Ω are connected in (i) series, and (ii) in parallel, in turn to a battery of 10 V. Calculate the ratio of power consumed in the combination of resistors in the two cases.

 

  1. In the given circuit calculate (i) total resistance of the circuit, and (ii) current shown by the ammeter. 

 

Answer:

 

  1. Her;

I = 1.0 m, R = 23 Ω, p = 1.84 × 10–6 Ωm

As    

         

 

  1. I = 10A, V = 220 V. As V = IR,   

 

  1. Here; P = 100 W, V = 220 V. As P = VI,     

 

  1. P = 1200 W, t = 20 min = 20 × 60 s = 1200 s

Electric energy consumed, W = Pt = (1200 W) (1200 s) = (1200 J/s) (1200 s) = 1.44 × 106 J

 

  1. V = 5V, I = 500mA = 0.5A

(i)  P = VI = 5C × 0.5 A = 2.5 W

(ii) R = V/I = 5V/0.5 A = 10 Ω

(iii) W = P × t = 2.5 W × 4 hour = 2.5 W × (4 × 3600s) = 3600 J

 

  1. When the lamps are joined in parallel, equivalent power,

P = P1 + P2 = 100 W + 25 W = 125 W

 

  1. (i) Total resistance in series, Rs = 2 Ω + 2 Ω = 4 Ω

Power consumed,

(ii) Total resistance in parallel, i.e., Rp is given by  or Rp = 1 Ω

Power consumed in parallel, 

  or  Ps : Pp = 1:4

 

  1. (i)    

(ii)   

         

                  or     Ps : Pp = 1:4

 

  1. (i) Since R1 and R2 are in series, their resultant resistance,

          Rs = R1 + R2 = 3 Ω = 3 Ω + 2 Ω = 5 Ω

Further, Rs and R3 are in parallel, their resultant is given by

                  or         Rp = 25 Ω.

(ii)