Page 210 - WCS - Electrical
P. 210
Example
Example
V
Example
V
Current (I)
=
V
=
Current (I)
=
Current (I)
1 Calculate the power rating of the lamp in the
R
1 Calculate the power rating of the lamp
1 Calculate the power rating of the lamp in the in the
R
R
circuit, if 0.25 amperes of current flows and the
circuit, if 0.25 amperes of current flows and the
circuit, if 0.25 amperes of current flows and the
220
voltage is 240 volts.
=
220 220 2 amperes
voltage is 240 volts.
voltage is 240 volts.
2 amp
2 ampereseres
=
= 110
110 110
P = V x I
Power (w)
= V x I
P = V x I
P = V x I
Power (w)
=
= V x IV x I
Power (w)
V = 240 Volts
= 220 x 2
V = 240 Volts
V = 240 Volts
= 220 x 2
= 220 x 2
= 0.25 Amperes
I
= 440 watts
I
= 0.25 Amperes
I
= 0.25 Amperes
= 440 watts
= 440 watts
Therefore Power = 240 Volts x 0.25 Amperes
6 Find the total power if four 1000 W, 180 volt
Therefore Power = 240 Volts x 0.25 Amperes
Therefore Power = 240 Volts x 0.25 Amperes
6 Find the total power if four 1000 W, 180
heaters are connected in series across 240 V
= 60 Volts Ampere
= 60 Volts Ampere
heaters are connected in series across
heaters are connected in series across 240 V 240 V
= 60 Volts Ampere
supply and current carrying capacity is 15 amp.
supply and current carrying capacity is 15 amp.
But 1 Watt = 1 Volt x 1 Amphere
supply and current carrying capacity is 15 amp.
But 1 Watt = 1 Volt x 1 Ampherehere
But 1 Watt = 1 Volt x 1 Amp
Find the total power.
Find the total power.
Find the total power.
Therefore Power = 60 Watts
Series
Connection
=
Therefore Power = 60 Watts
Therefore Power = 60 Watts
Connection
=
=
Connection
Series Series
2 A current of 15 amperes flow through a resistance
heaters
4
=
2 A current of 15 amperes flow through a resistance
2 A current of 15 amperes flow through a resistance
=
heaters
4
4 =
heaters
of 10 Ohms. Calculate the power in kilowatts
of 10 Ohms. Calculate the power in kilowatts
of 10 Ohms. Calculate the power in kilowatts
(W)
=
watts
consumed.
=
watts watts
(W) (W)
=
consumed.
consumed.
=
180 V
Heater voltage
Given that R = 10 and I = 15A
=
Heater voltage
180 V 180 V
=
Heater voltage
Given that R = 10 and I =
Given that R = 10 and I = 15A 15A
Supply voltage =
240 V
Power = V x I = I x R x I = I x R
2
Supply voltage =
240 V 240 V
Supply voltage =
Power = V x I = I x R x I = I x R x R
2
Power = V x I = I x R x I = I
2
Therefore Power = 15 x 10 = 2250 Watts = 2.25 kW
2
2 V
2
Therefore Power = 15 x 10 = 2250 Watts = 2.25 kW
Therefore Power = 15 x 10 = 2250 Watts = 2.25 kW
2
2
V
2
Heater resistance (R)
V =
Heater resistance (R) =
Heater resistance (R) =
W
3 At a line voltage of 200 Volts a bulb consumes a
W
W
3 At a line voltage of 200 Volts a bulb consumes a
3 At a line voltage of 200 Volts a bulb consumes a
current of 0.91 ampheres. If the bulb is on for 12
324
180
x
180
current of 0.91 ampheres. If the bulb is on for 12
current of 0.91 ampheres. If the bulb is on for 12
x
180 180
=
180 180 x
324 324
=
hour calculate the work in Wh to find the work
=
=
=
hour calculate the work in Wh to find the
hour calculate the work in Wh to find the work work 6 Find the total power if four 1000 W, 180 volt volt
= 10
1000
given that V = 200 Volts.
given that V = 200 Volts. = = 1000 1000 10 10
given that V = 200 Volts.
ohms
=
ohms ohms
I = 0.91 Amps.
I = 0.91 Amps. Total resistance = = 32.4 x 4 = 129.6 ohms
I = 0.91 Amps.
Total resistance
32.4 x 4 = 129.6 ohms
Total resistance
=
32.4 x 4 = 129.6 ohms
t = 12 hours
t = 12 hours V V WORKSHOP CALCULATION & SCIENCE - CITS
t = 12 hours
Total current (I)
V
=
Total current (I) = = R
Total current (I)
R R
240
= 240 240 = 1.85 amperes
= 1.85 ampereseres
= = 129.6 = 1.85 amp
129.6 129.6
(W) = V x I
(W) (W) = =
V x IV x I
= = 240 x 1.85 = 444 watts
=
240 x 1.85 = 444 watts
240 x 1.85 = 444 watts
7 If a 40 watt fluorescent lamp draws a current of
7 If a 40 watt fluorescent lamp draws a current of 0.10 ampere. How much voltage will be required to illuminate
7 If a 40 watt fluorescent lamp draws a current of
7 If a 40 watt fluorescent lamp draws a current of
0.10 ampere. How much voltage will be required
it? 0.10 ampere. How much voltage will be required
0.10 ampere. How much voltage will be required
to illuminate it?
to illuminate it?
to illuminate it?
(W)
(W) (W) = = watt
watt watt
=
Therefore Power=V x I = 200 Volts x 0.91 Amps Current (I) = 0.10 ampere
Therefore Power=V x I = 200 Volts x 0.91 Amps Current (I) = = 0.10 ampere
Current (I)
0.10 ampere
Therefore Power=V x I = 200 Volts x 0.91 Amps
= 182 Watts
= 182 Watts Voltage (V) = W W
= 182 Watts
W
Voltage (V)
Therefore Work = P x t = 182 Watts x 12 hours Voltage (V) = = I
Therefore Work = P x t = 182 Watts x 12 hours
Therefore Work = P x t = 182 Watts x 12 hours I I
= 2184 Watt hour.
40
= 2184 Watt hour. = 40 40 = 400 volts
= 2184 Watt hour.
0.1
= 400
4 An adjustable resistor bears the following label: = = = 400 voltsvolts
4 An adjustable resistor bears the following label:
4 An adjustable resistor bears the following label: 0.1 0.1
1.5 k Ohms/0.08 A. What is its rated power?
8 Find the cost of running 15 HP motor for 15 days @
1.5 k Ohms/0.08 A. What is its rated power?
1.5 k Ohms/0.08 A. What is its rated power? 8 Find the cost of running 15 HP motor for 15 days @
8 Find the cost of running 15 HP motor for 15 days @ 6 hrs per day and the cost of energy is Rs. 3 per unit.
8 Find the cost of running 15 HP motor for 15 days @
Given: R = 1.5 k Ohms; I = 0.08 A
6 hrs per day and the cost of energy is Rs. 3 per unit.
Given: R = 1.5 k Ohms; I = 0.
Given: R = 1.5 k Ohms; I = 0.08 A08 A 6 hrs per day and the cost of energy is Rs. 3 per unit.
6 hrs per day and the cost of energy is Rs. 3 per unit.
Find: P
Motor power (w)
= 15 HP
Find: P Motor power (w) = 15 HP 15 HP
Find: P
=
Motor power (w)
V = R x I = 1500 Ohms.0.08 A = 120 volts
= 15 x 746 = 11,190 watts
V = R x I = 1500 Ohms.0.08 A = 120 voltsvolts = 15 x 746 = 11,190 watts
V = R x I = 1500 Ohms.0.08 A = 120
= 15 x 746 = 11,190 watts
P = V x I = 120 volts.0.08 A = 9.6 W alternatively:
Consumption per day
= 11,190 x 6
P = V x I = 120 volts.0.08 A = 9.6 W alternatively:
P = V x I = 120 volts.0.08 A = 9.6 W alternatively: Consumption per day = 11,190 x 60 x 6
Consumption per day
= 11,19
P = 1 .R = (0.08 A) .1500 Ohms = 9.6 W.
2
2
P = 1 .R = (0.08 A) .1500 Ohms = 9.6 W.
P = 1 .R = (0.08 A) .1500 Ohms = 9.6 W. = 67140 = 67.14 KWH
2
2
2
2
67.14 KWH
= 67140 =
5 Find the current and power consumed by an = 67140 = 67.14 KWH
Consumption for 15 days = 67.14 x 15
5 Find the current and power consumed
5 Find the current and power consumed by an by an Consumption for 15 days = 67.14 x 15
Consumption for 15 days = 67.14 x 15
electric iron having 110 resistance when feed
electric iron having 110 resistance when feedfeed = 1007 KWH (or) unit
electric iron having 110 resistance when
from a 220 v supply
= 1007 KWH (or) unit
from a 220 v supply = 1007 KWH (or) unit
from a 220 v supply
= Rs. 3
Cost per unit
Resistance of electric iron (R) = 110 ohms
Cost per unit
Cost per unit
=
Resistance of electric iron (R) = 110 ohms
Resistance of electric iron (R) = 110 ohms Cost for total energy = Rs. 3 Rs. 3 = Rs. 3021
= 3 x 1007
= 220 volts
Voltage (V)
= 3 x 1007
Cost for total energy
= Rs. 3021
= 220
Voltage (V)
Voltage (V) = 220 voltsvolts Cost for total energy = 3 x 1007 = Rs. 3021
Assignment
112 Workshop Calculation & Science : (NSQF) Exercise 1.7.36
112 112 Workshop Calculation & Science : (NSQF) Exercise 1.7.36
Workshop Calculation & Science : (NSQF) Exercise 1.7.36
1 Current Consumed
I = 0.136 A
Voltage ‘V’ = 220 V
P = _____ Watts
2 P = 500 Watts
I = 2.27 A
V = ______ v
3 P = 750 W
V = 220 v
I = ______ A
197
CITS : WCS - Electrical - Exercise 18
