(1) A bulb is connected to a battery of p.d. 4 V and internal resistance 2.5 ohm. 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.
(2)Water in an electric kettle connected to a 220 V supply took 5 minutes to reach its boiling point.how long would it have taken if the supply voltage had fallen to 200 V ?
Answers
Answered by
53
1.
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
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
Answered by
5
Answer: 1)e=v2t/r
T=10*60=60sec
R=v/I
=4/0.5=8ohm
=(4)2*60/8
=1200j
Explanation:
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