Distinguish between kilowatt and kilowatt hour for a heater rated at 4.4KW, 220 V, calculate the
a) current drawn by heater
b) Resistance of heater element
c)Energy consumed by the heater in 5 hrs.
d)cost of running the 1 kilowatt cost Rs 6.5.
Answers
Answer:
Hey frnd...
1)Power, P = 4 kW = 4000 W
Voltage, V = 220 V
Now, current is, I = P/V = 4000/220 = 18.18 A
Resistance is, R = V2/P = 2202/4000 = 12.1 Ω
Energy consumed in 2 h is = 4000 × 2/1000 = 2 kWh
So, cost of energy consumed is = 4.60 × 1.5 = Rs 6.9
2)P=50W V=220V TIME =20 min=20/60hrs
E=P X T
=50/1000 X 20/60
=0.016KWH
3)
we have given two lamps in such a way :
Power of 1st lamp, P₁ = 40W , voltage of 1st lamp, V₁ = 220V
power of 2nd lamp , P₂ = 60W , voltage of 2nd lamp ,V₂ = 220V
we know, one things ,
Power = V²/R [ when potential difference is same then consider P = V²/R ]
R = V²/P
so, resistance of 1st lamp , R₁ = V₁²/P₁ = (220)²/60 = 4840/6 = 2420/3Ω
resistance of 2nd lamp , R₂ = V₂²/P₂ = (220)²/40 = 48400/40 = 1210Ω
Now, question said ,
(a) Both the lamps are in parallel ,
So, 1/Req = 1/R₁ + 1/R₂
1/Req = 3/2420 + 1/1210 = 5/2420 = 1/484
Req = 484Ω
Now, Current drawn from electrical supply ,i = potential difference/Req
= 220V/484 = 20/44 = 5/11 A
(b) energy consumed by lamps in one hour = Energy consumed by 1st lamps in one hour + energy consumed by 2nd lamps in one hour
= P₁ × 1hour + P₂ × 1 hour
=(P₁ + P₂) × 1 hour
= (60W + 40W) × 1 hour
= 100 Wh
= 100/1000 KWh [ ∵ 1 KWh = 10²Wh ]
= 0.1 KWh
4)(i)
Using, P = VI => I = P/V
For the heater marked 220 V, 500 W; the current drawn will be, I = 500/220 = 2.3 A
For the heater marked 220 V, 800 W; the current drawn will be, I/ = 800/220 = 3.6 A
(ii)
Using, P = V2/R => R = V2/P
For the heater marked 220 V, 500 W; the resistance will be, R = 2202/500 = 96.8 Ω
For the heater marked 220 V, 800 W; the resistance will be, R = 2202/800 = 60.5 Ω
(iii)
The net power of the system of heaters is = 500 + 800 = 1300 W
When used for 2 hours the energy utilised in kWh is = 1300 × 2/1000 = 2.6 kWh
5)I=P/V
I1=60/v
I2=40/V
I1/I2=60/V/40/V =60/40
=> I1>I2
SINCE, R IS INDIRECTLY PROPORTIONAL TO I
I2 HAS HIGH RESISTANCE
SO, 40W LAMP HAS GREATER RESISTANCE
6)case 1: when the resistance is connected in series
then the total resistance is 10 + 10 = 20 ohm and the given voltage is 6 volt
then acc. to the formula I = V/R we find that the current is 0.3 A
case 2: when the resistance is connected in parallel
again applying the formula...I= V/R we find that the current is 1.2 A
now let the current in series connection be I1 and in the parallel connection be I2
volttage is same in both the cases
we know that P = VI ,
taking out the ratio of both the cases appyling the above formula,we get
6 * 0.3 / 6 * 1.2
we get, 1/4,therefore the ratio is 1:4
Explanation: