SOLVING Flashcards
A mercury lamp with 50-ohms hot resistance is connected to a socket with 240-volt current supply. How much current is flowing through the lamp?
4.8 amperes
I = V/R
= 240 volt/ 50 ohms
= 4.8 amperes
A piece of wire has a resistance of 50 Ω. How much power is dissipated in the wire if it carries a current of 0.50A?
P = IE or P= E2
/R or P = I2R
Where:
P = power in watt, W R = resistance in ohm, Ω E = voltage in volts, V I = current in ampere, A
P = I2R = (0.50A)2 (50 Ω) = 12.5 W
What is the current drawn by a 1000 W electric flat iron operated at 220 V?
P = IE; I = P/E = 1000W/220V = 4.54 A
A 60 Ω lamp is left connected to a 220 V source for 3 hours. How much energy is taken from the source?
Energy is the power expended or used over a period of time
W = IEt or W = E2tR
or W = I2Rt
Where:
W = energy in watt, W R = resistance in ohm, Ω E = voltage in volts, V I = current in ampere, A t = time in second, s
1 Wh = 3600watt-seconds = 3600 joules
W = E2tR= 220V 2 (3 hours) 60Ω = 2,420 Wh or 2.42 kWh or 8.712 MJ
A 12-ampere electric blower with 0.85 power factor is connected to a 240-volt convenient outlet. Calculate the power in the circuit.
P = E I cosθ = 240 volt x 12 amp x 0.85
= 2,448 watts
Two tractor headlights are connected in series to a 12-volt battery, each having a resistance of 1 ohm. What is the current flowing in the circuit?
R = R1 + R2
I = V / R
= 12 volts / 2 ohms
= 6 amperes
Determine the monthly energy consumption of a farm house having the following appliances: electric iron, 1200 watts at 2 hours; water heater, 1000 watts at 3 hours; and toaster, 1,300 watts at 30 minutes. If the average cost of energy is P10.00 per kwh, what is the total cost of the energy used in 30-day consumption?
Given:
Electric Iron - 1200 watts , 2 hours Water heater - 1000 watts, 3 hours Toaster - 1,300 watts, 30 minutes Required: Monthly energy consumption Solution:
Electric Iron = 1200 x 2 / 1000 = 2.4 kw-hr
Water heater = 1000 x 3 / 1000= 3.0 kw-hr
Toaster = 1300 x 0.5 hr / 1000= 0.65 kw-hr
Total Energy = 2.4 kw-hr + 3.0 kw-hr x 0.65 kw-hr = 6.05 kw-hr
Total Cost = 6.05 kw-hr/day x 30 days/ month x P10.00/kwhr
= P 1,815.00
All electrical equipment is off except an electric motor. The kilowatthour meter-disk revolutions are counted for a period of 6
minutes. The disk makes 20 revolutions and the Kh factor of the meter is 2.5. Determine the energy that would be used by
this motor if it were operated for 3 hours. What is the power input to the motor?
Kh factor is printed on the nameplate of the kilowatt-hour meter. Range – 1.5-5.0 Power usage (watthours) = Kh ×no.of disk revolutions time Energy (Wh) = Kh × no. of disk revolutions = 2.5 × 20 = 50 Wh
In 6 minutes: 50 Wh × 180 minutes/hour
6 minutes = 1500 W
Power input to the motor = 1500 W = 1.5 Kw
Determine the power factor and phase shift angle for a circuit where the true power is found
to be 3840 watts, the voltage is 240 volts and the current is 20 amperes.
cosθ = power factor =True power/ Apparent power = 3840 W/ 240V × 20A
= 0.8
θ = cos-1 (0.8) = 36.870
If a coil known to have an inductance of 0.1 henry is connected to a 60 Hz source, what is the inductive reactance?
XL = 2πfL Where: XL = inductive reactance in ohm, Ω f = frequency in hertz, Hz L = inductance in henry, H XL = 2πfL = 2π × 60 Hz × 0.1 H = 37.7 Ω
Find the capacitive reactance of a 13 μf capacitor connected to a 60 hertz source.
Xc = 1/2πfC
Where:
Xc = inductive reactance in ohm, Ω
f = frequency in hertz, Hz
C = inductance in henry, H
Xc = 1/2πfC =1/2π ×60Hz ×13 μf = 204 Ω
Determine the total impedance, current flow, true power and apparent power in a circuit with an 80 ohm
resistance, 90 ohm inductive reactance and 30 ohm capacitive reactance connected in series. The voltage is
100 V.
Z = (R^2 + (XC − XL)^2)^1/2 = ((80Ω)^2+ (30Ω − 90Ω)^ 2)^1/2 = 100Ω
power factor = cosθ = 80/100 = 0.8
current flow = I =E/Z = 100V/100Ω = 1A
true power = EIcosθ = 100V × 1A × 8 = 80 W
apparent power = EI = 100V × 1A = 100VA
replace a 2 hp engine with electric motor
2/3 × 2 hp = 1.333 hp. Select one 1 1⁄2 hp motor.
