a bulb is rated at 330V-110W. what do you think is its resistence? three such bulbs burn for 5 hrs at a stretch. what is the energy consumed? calculate the cost in rupees if the rate is 70 paise per unit
Answers
Explanation:
We have to find the resistance of bulb first, so
We have to find the resistance of bulb first, soA/c to formula of power
P=V^2/R
.•. R=V^2/P
R=330×330/110
R=990ohm
Resistance of bulb is 990 ohms
Now energy used by one bulb
E=Pt
= 110×5 Wh
= 550Wh
Total energy consumed by 3 bulbs
Total energy= 3×550 Wh
= 1650 Wh
1 unit = 1kWh= 1000 wh
Total unit= 1650/1000
= 1.65 kWh
₹1=100 paise
70 paise= 70/100
=₹0.7
Now total amount
=₹0.7× 1.65
=1155/1000
= 1.155
Answer:
Resistance of 1 bulb = 990Ω
Total cost of energy consumed = Rs 1.155
Explanation:
Power =
Work Done = Energy
Therefore ,
Power =
Energy = VIt
Time = t
Power =
Power = VI
According to the question ,
V = 330 V
P = 110 W
t = 5 hrs
Number of bulbs = 3
Cost of 1 unit of Energy = 70 paise
P = VI
I = P/V
I =
I =
By Ohm's Law
V = IR
R = V/I
R = Ω
R = Ω
R = 990 Ω
Therefore , resistance of 1 bulb is 990 Ω .
Energy = VIt
E = VIt
E = Wh
E = 550 Wh
Therefore , energy consumed by 1 bulb is 550 Wh .
Energy consumed by 3 bulbs = 3 X Energy consumed by 1 bulb
Energy consumed by 3 bulbs = 3 X 550 Wh
Energy consumed by 3 bulbs = 1650 Wh
1 kWh = 1000 Wh
1 Wh = kWh
1650 Wh = kWh
1650 Wh = 1.65 kWh
Therefore , energy consumed by 3 bulbs is 1650 Wh or 1.65 kWh .
1 Rupee = 100 paise
1 paise = Rupee
70 paise = Rupee
70 paise = Rs 0.7
Cost of 1 unit of Energy = Rs 0.7
Cost of 1 kWh of Energy = Rs 0.7
Cost of 1.65 kWh of Energy = Rs 0.7 X 1.65
Cost of 1.65 kWh of Energy = Rs 1.155