1. 100J of heat are produced each second in a 4 ohm resistance. find the potential difference across the resistor.
2) An electric bull is connected to a 220V generator. The current is 0.50A .What is the Power of the bulb.
3) An electric refrigerator rated 400w operates 8 hour 1day.
what is the cost of the energy to operate it for 30 days at
Rs 3.00 per kwh ?
4.) Two bulls of 100 w each and two cooler of 250W each,
work on An average 6 Hours a day. If the energy costs Rs 1.75
per kwh, Calculate the monthly bill and the minimum fuse
rating when power is supplied at 250V.
5) An electric kettle is rated at 230V, 1000k . what is the
resistance of its element? What maximum current can
pass through its element?
6) How is heat produced by an electric current?
7) A Torch bulb is rated at 2.5V and 750 mA . Calculate its
1) Power 2)it's resistance 3)the energy consumed if the bulb
is lighted for 4 Hour.
Answers
1. H=I^2RT
T=1 sec, R=4 ohm, H=100 J
100=4*I^2
I^2=25
I=5 A
we know that, V=IR
V=5*4=20 V
2. power = voltage * current
P=VI=220*0.50
P=110 Watts
3. power of refrigerator = 400 w
in 1 hour it will consume= 400*3600 = 1440000 J energy
in 8 hours= 1440000*8=11520000 J energy
in 30 days= 11520000*30=345600000 J energy
which is equal to 345600 kw second= 345600/3600 = 96 kwh
so total cost = 96*3=RS 288
4. total power of all units= 100+100+250+250=700 watts
in one hour= 700*3600=2520000 J energy
in 6 hours=2520000*6=15120000 J energy
in 30 days= 15120000*30=453600000 J energy
which is equal to 453600 kw second=453600/3600=126 kwh in total 30 days
so total cost = 126*1.75= Rs 220.5
5. Power = P = 1000 W
Voltage = V = 230 v
Resistance = R
P = V²/R
1000 = 230² / R
R = 230 x 230 /1000
= 52.9 Ω
max current I=V/R=230/52.9=4.34 A
6. whenever current flows, there is a flow of electrons in the conductor and due to opposing effect of conductor on flow of electrons (usually known as resistance) comes into play and the energy lost by the electrons due to opposing effect is seen as heat produced
7. I=750 mA= 750 * 10^-3 A
V= 2.5 V
i)power = 2.5*750*10^-3=1875*10^-3=1.875 watt
ii) R=V/I=2.5/750*10^-3
R=2.5*10^3/750=2500/750=3.33 ohms
iii) energy consumed per second= 1.9 watt
in one hour= 1.9*3600=6840 J
in 4 hours= 6840*4=27360 J
hope this helps and u understand well :)