Page 209 - WCS - Electrical

WORKSHOP CALCULATION & SCIENCE - CITS

Therefore Work = P x t = 182 Watts x 12 hours
= 2184 Watt hour.
4 An adjustable resistor bears the following label: 1.5 k Ohms/0.08 A. What is its rated power?
Given: R = 1.5 k Ohms; I = 0.08 A
Find: P

V = R x I = 1500 Ohms.0.08 A = 120 volts
P = V x I = 120 volts.0.08 A = 9.6 W alternatively:
P = 12.R = (0.08 A)2.1500 Ohms = 9.6 W.
5 Find the current and power consumed by an electric iron having 110W resistance when feed from a 220
Example v supply V
Example Current (I) V = R
1 Calculate the power rating of the lamp in the
circuit, if 0.25 amperes of current flows and the
1 Calculate the power rating of the lamp in the Current (I) = R 220
voltage is 240 volts.
circuit, if 0.25 amperes of current flows and the = 2 amperes
220
110
voltage is 240 volts. Power (w) = 110 = V x I
2 amperes
P = V x I
P = V x I Power (w) = V x I
V = 240 Volts
= 220 x 2
V = 240 Volts = 220 x 2
= 0.25 Amperes
I
= 440 watts
I = 0.25 Amperes 6 Find the total power if four 1000 W, 180 volt
= 440 watts
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 volt heaters are connected in series across 240 V supply and
6 Find the total power if four 1000 W, 180 volt240 V
heaters are connected in series across
= 60 Volts Ampere
supply and current carrying capacity is 15 amp.
= 60 Volts Ampere current carrying capacity is 15 amp. Find the total power.
heaters are connected in series across 240 V
But 1 Watt = 1 Volt x 1 Amphere
Find the total power.
But 1 Watt = 1 Volt x 1 Amphere supply and current carrying capacity is 15 amp.
Series
Connection
=
Therefore Power = 60 Watts
Find the total power.
=
Connection
Therefore Power = 60 Watts Connection = Series Series
No. of heaters =
4
2 A current of 15 amperes flow through a resistance
4
=
heaters
of 10 Ohms. Calculate the power in kilowatts
2 A current of 15 amperes flow through a resistance Heater power (W) = 4 = 1000 watts
heaters
consumed.
of 10 Ohms. Calculate the power in kilowatts (W) = watts
Heater voltage =

Heater voltage
=
180 V
consumed. (W) = watts 180 V
Given that R = 10 and I = 15A
240 V
Supply voltage
Supply voltage = =
Given that R = 10 and I = 15A 2 Heater voltage = 180 V 240 V
Power = V x I = I x R x I = I x R
Power = V x I = I x R x I = I x R Supply voltage = 240 V V 2
2
Therefore Power = 15 x 10 = 2250 Watts = 2.25 kW
2
Heater resistance (R)
Therefore Power = 15 x 10 = 2250 Watts = 2.25 kW Heater resistance (R) = V 2 = W
2
3 At a line voltage of 200 Volts a bulb consumes a
current of 0.91 ampheres. If the bulb is on for 12
3 At a line voltage of 200 Volts a bulb consumes a W 180 x 180 324
=
hour calculate the work in Wh to find the
current of 0.91 ampheres. If the bulb is on for 12 work 180 x 180 1000 324 = 10
given that V = 200 Volts.
=
hour calculate the work in Wh to find the work = = 1000 ohms
10
given that V = 200 Volts. = ohms
I = 0.91 Amps.
32.4 x 4 = 129.6 ohms
Total resistance
=
I = 0.91 Amps. Total resistance = 32.4 x 4 = 129.6 ohms
t = 12 hours
t = 12 hours Total current (I) = V
Total current (I) = V R
R 240
=
240 129.6 = 1.85 amperes
= = 1.85 amperes
(W) 129.6 V x I
=
(W) = V x I 240 x 1.85 = 444 watts
=
196
240 x 1.85 = 444 watts
=
7 If a 40 watt fluorescent lamp draws a current of
CITS : WCS - Electrical - Exercise 18
0.10 ampere. How much voltage will be required
7 If a 40 watt fluorescent lamp draws a current of
to illuminate it?
0.10 ampere. How much voltage will be required
to illuminate it? = watt
(W)
Therefore Power=V x I = 200 Volts x 0.91 Amps (W) = watt 0.10 ampere
Current (I)
=
Therefore Power=V x I = 200 Volts x 0.91 Amps Current (I) = 0.10 ampere
= 182 Watts
W
=
= 182 Watts
Therefore Work = P x t = 182 Watts x 12 hours Voltage (V) W I
Therefore Work = P x t = 182 Watts x 12 hours Voltage (V) = I 40
= 2184 Watt hour.
=
= 2184 Watt hour. 40 0.1 = 400 volts
4 An adjustable resistor bears the following label:
= 400 volts
=
0.1
1.5 k Ohms/0.08 A. What is its rated power?
4 An adjustable resistor bears the following label: 8 Find the cost of running 15 HP motor for 15 days @
1.5 k Ohms/0.08 A. What is its rated power? 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 = 0.08 A 6 hrs per day and the cost of energy is Rs. 3 per unit.
Find: P
Motor power (w)
= 15 HP
Find: P V = R x I = 1500 Ohms.0.08 A = 120 volts Motor power (w) = 15 HP
= 15 x 746 = 11,190 watts
V = R x I = 1500 Ohms.0.08 A = 120 volts Consumption per day = 11,190 x 6
= 15 x 746 = 11,190 watts
P = V x I = 120 volts.0.08 A = 9.6 W alternatively:
P = 1 .R = (0.08 A) .1500 Ohms = 9.6 W.
2
2
P = V x I = 120 volts.0.08 A = 9.6 W alternatively: Consumption per day = 11,190 x 6 67.14 KWH
= 67140 =
P = 1 .R = (0.08 A) .1500 Ohms = 9.6 W. = 67140 = 67.14 KWH
2
2
5 Find the current and power consumed by an
electric iron having 110   resistance when
5 Find the current and power consumed by an feed Consumption for 15 days = 67.14 x 15
Consumption for 15 days = 67.14 x 15
from a 220 v supply
electric iron having 110   resistance when feed = 1007 KWH (or) unit
from a 220 v supply Cost per unit = 1007 KWH (or) unit
= Rs. 3
Resistance of electric iron (R) = 110 ohms
Resistance of electric iron (R) = 110 ohms Cost per unit = Rs. 3 = Rs. 3021
Cost for total energy
= 3 x 1007
= 220 volts
Voltage (V)
Voltage (V) = 220 volts Cost for total energy = 3 x 1007 = Rs. 3021
112 Workshop Calculation & Science : (NSQF) Exercise 1.7.36
112 Workshop Calculation & Science : (NSQF) Exercise 1.7.36