Question

1.For a heater rated at 4 KW and 220 V, calculate
(I) the current.
(ii) the resistance of the heater
(iii) the energy consumed in two hours
(iv) the cost if 1 kWh is priced at Rs. 1.5
2. An electric bulb is rated as 50 W, 220 V. Calculate the energy consumed by the bulb in 20 minuted. Express the answer in commercial unit of electrical energy.
3. Two lamps one rated 60 W ar 220 V and other 40 W at 220 V are connected in parallel to electric supply at 220 V.
(I) Draw the circuit diagram to show the connection.
(ii) Calculate the current drawn from the electric supply.
(iii) Calculate the total energy consumed by the two lamps together when they operate
for 1 hr.
4.Two room heaters are marked 220 V , 500 W and 220 V, 800 W respectively. If the heaters are connected in parallel tp 220 V main supply. Calculate
(I) the current drawn by each heater.
(ii) the resistance of each heater.
(iii, tital electrical energy in commercial unit if they operate simultaneously for 2 hours.
5. Out of 60 W and 40 W lamps which one has a higher electrical resistance when in use.
6. Two identical resistors each of resistance 10 Ohm are connected in series and in parallel in turnto a battery of 6 V. Calculate the 3atio of power consumed in the combination of resistor in two cases.

Answer 1

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

#hope helped...

Answer 2

Answer:

c number

Explanation:

E = PT = 4*2=8 this is solution of question number c of 20

.....(((((((((((((...(question number a of 20 ))))))))

p=vI

4 Kw convert into 4000w

now put this value in formula

P=VI , v equal to 220 and equal to 4000

4000 = 220 * I

I = 220/4000 = 18.18