Question

1.a torch bulb of 3v draws a current of 0.4a.if the bulb is switched for 5 minutes .find the energy released by the bulb.


2.an electric item of resistance 400 draws 0.5a.calculate the power

Answer 1

1. V = 3 V I = 0.4 A T= 5 min = 300 s

      E = V × I × t = 3 × 0.4 × 300 = 360 J

Torch bulb of 3v draws a current of 0.4a. The bulb is switched for 5 minutes.

So, V=3 V, T=5 min= 300 sec and the Item of resistance= 4 A

 Energy released by the bulb= V* I*T= 3*0.4*300= 360 J  So the energy released by the bulb is 360 J.

 2. R = 400 Ω I = 0.5 A (given)

 Therefore, V = I × R = 400 × 0.5 = 200 V

Answer 2

Anr: we know that v=ir so we have i=0.5 & R = 400

0.5 .400=200 KOW THAT POWER =VI WE HAVE V=200&I=0.5 THEN 200.0.5 =100W

Explanation: