Electricity Worksheet-10

Electricity Worksheet-10

 

  1. A 6 V-12 W lamp is connected in series with a source of 12 V supply. Calculate the value of the resistance R for the proper working of the lamp. What is the current flowing through the circuit?

 

  1. Two coils of resistances 3 Ω and 6 Ω are connected in series across a battery of emf 12 V. find the electrical energy consumed in 1 minute in each resistance when these are connected in series.

 

  1. Two bulbs A and B are rated 100 W-120 V and 10 W-120 V respectively. They are connected across a 120 V source in series. Which will consume more energy ?

 

  1. An electric iron consumes energy at the rate of (a) 840 W when heating is at the maximum rate and (b) 360 W when the heating is at the minimum. The voltage is 220 V. what are the current and the resistance in each case?

 

  1. A torch bulb of 3 V draws a current of 0.4 A. if the bulb is switched on for 5 minutes; calculate the energy released by the bulb.

 

Answer:

  1. Here, power of the lamp, P = 12 W

Voltage at which it works, V = 6 V

As     P = VI,

Resistance of the lamp,

If this lamp is connected to a source of 12 V supply directly, current through the lamp will be

= 12 /3 = 4A.

This current is double the current that the lamp can withstand. Thus, the purpose of the resistance (R) is to reduce the current to the safe value of 2 A.

Total resistance of the bulb and the resistance R = r + R. (in series)

For the efficient working of the bulb, the resistance R should be chosen in  such a way that the resistance (r + R) should allow only a current of 2 A when connected in series with the 12 V supply, i.e.,

         

Or    

Or     6 + 2 R = 12

Or     2 R = 6

Or     R = 3 Ω

 

  1. Total resistance of the circuit, R = 3 + 6 = 9 Ω

Current in the circuit,

Since the resistances are in series, same current flows in each resistance.

Electric energy consumed by R1 (= 3 Ω) in 1 minute

          W1 = I2R1 t = (4/3)2 × 3 × 60 = 320 J

Electric energy consumed by in R2 (= 6 Ω) = (4/3)2 × 6 × 60 = 640 J

 

  1. Resistance of the bulb A

                  (P = V2/R)

Resistance of the bulb B,

         

When the bulbs are connected in series, total resistance ,

          R = R1 + R2 = 144 + 1440 = 1584 Ω

When these bulbs in series are connected across 120 V, current in the circuit, i.e.,

         

Power consumed by the bulb A, i.e., P1 = I2R1 = (0.076)2 × 144 = 0.832 W

Power consumed by the bulb B, i.e., P2 = I2R2 = (0.076)2 × 1440 = 8.32 W

Hence, the bulb B consumes more energy than the bulbs A, when they are connected in series.

 

  1. (a)   

         

(b)   

         

 

  1. W = VIt = 3 × 0.4 × (5 × 60) = 360 J