Question

A bulb is connected to a battery of p.d. 4 V and
internal resistance 2.5ohm. A steady current of 0.5 A
flows through the circuit. Calculate :
(i) the total energy supplied by the battery in
10 minutes,
(ii) the resistance of the bulb, and
(iii) the energy dissipated in the bulb in 10 minutes.
Ans. (i) 1200 J (ii) 5.5 2. (iii) 825 J​

Answer 1

Answer:

Power P= V*I

4*0.5=2 watt

V=IR

4=0.5(2.5+resistance of bulb)

r of bulb=8-2.5=5.5 ohm

So energy consumed in 10 minutes(=10/60 hr)

=2*(1/6)=0.333 watt hr or 20 watt-minutes

2.

 taking current as constant energy E=p*t=v*i*t

E remains also constant So for the first case 220*I*5=200*I*T

So T=(220*5)/200=5.5 minutes