Electricity Worksheet-9

Electricity Worksheet-9

 

  1. A parallel combination of three resistors takes a current of 7.5 A from a 30 V supply. If the two resistors are 10 Ω and 12 Ω. Find the third one.

 

  1. A wire whose resistance is 80 Ω is cut into three pieces of equal lengths which are then arranged in parallel. Calculate the resistance of the combination.

 

  1. Five resistors, each 3 Ω, are connected as shown here. Calculate the resistance between the points (i) A and B (ii) between the point C and D.

 

  1. An electric bulb is rated at 200 V-100 W. what is its resistance? Five such bulbs burn for four hours. What is the electrical energy consumed? Calculate the cost if the rate is 50 paise per unit.

 

  1. A torch bulb is rated at 2.5 V and 750 mA. Calculate its (i) power (ii) its resistance and (iii) the energy consumed if the bulb is lighted for 4 hour.

 

Answer Key:

  1. As

Let the third resistance be x Ω. Since 10 Ω, 12 Ω or x Ω are in parallel,

                       or     

Or           or      x = 15 Ω

 

  1. Resistance of each piece

Since three pieces of (80/3) Ω are in parallel, the resultant resistance of the combination,

         

 

  1. (i) Resistance in the arm AEB, i.e., R1 = 3 Ω + 3 Ω = 6 Ω

          As R1 (= 6 Ω) and 3 Ω (in arm AB) are in parallel, effective resistance between the points A and B, i.e.,

(ii) since the effective resistance between point A and B (i.e., 2 Ω) and in the arm AC (= 3 Ω) and arm BD (= 3 Ω) are in series, effective resistance between the points C and D = 2 + 3 + 3 = 8 Ω

 

  1. Here, potential difference across the bulb, V = 200 V

Power of the bulb, P = 100 W

As  

Electric energy consumed by 1 bulb in 4 hours, i.e.,

          W = P t = 100 W × 4h = 400 Wh

Electric energy consumed by 5 bulb in 4 hours

          = 5 × 400 Wh = 2000 Wh = 2kWh = 2 unit

Cost of 2 units of electric energy = 2 × 50 paise = Re.1

 

  1. V = 2.5 V, I = 750 mA = 750 × 10–3 A = 0.75 A

Time for which the bulb is lighted, t = 4 hour

(i) power of the bulb,  P = VI = 2.5 × 0.75 = 1.875 watt

(ii) resistance of the bulb, ohm

(iii) energy consumed by the bulb in 4 hour i.e.

          W = Pt = 1.875 × 4 = 7.5 Wh = 7.5 W × 3600 s

                                                   = 27000 J