Electric Heat - Problems 1: UNIZOR.COM - Physics4Teens - Electromagnetism - Direct Current
Channel: Zor Shekhtman on UNIZOR Education
Date: 2020-04-24

Unizor - Physics through a Looking Glass of Mathematician

Direct Current - Electric Heat - Problems 1

Problem A
A lamp with resistance 500Ω (ohm) is powered by a source of electricity producing a voltage of 100V (volt).
What is the power consumption of this lamp?

Solution
Power is work per unit of time. It can be expressed as
P = U·I or
P = U² / R or
P = I²·R
Substituting the real numbers,
P = 100² / 500 = 20 (watt)

Problem B
Express the work performed by a source of electricity in Problem A during 8 hours in joules and in kilowatt-hours.

Solution
By definition,
1 watt = 1 joule/sec
Power consumed by a lamp is 20 watt (that is, 20 joules per second).
During 8 hours (that is, 8·3600 seconds) the work performed by a source of electricity will be equal to
W = 20·8·3600 = 576000 J (joules)
In kilowatt-hours (kWh) the work will be a product of power in kilowatts and time in hours, that is
W = 0.020·8 = 0.160 kWh

Problem C
There are usually two characteristics written on an incandescent electric lamp: the difference in electric potential on the terminals of a source of electricity the lamp is supposed to be connected to in volts (or voltage) and the power it consumes in watts (or wattage).
Let the voltage of a lamp be U=220V and the wattage - P=55W.
Determine the electric current I going through a lamp and its electrical resistance R.

Solution
Power is work per unit of time. It can be expressed as
P = U·I or
P = U² / R or
P = I²·R
From this, knowing U and P, we derive I and R.
I = P / U
R = U² / P
Substituting the real numbers,
I = 55 / 220 ≅ 0.25A (amperes)
R = 220²/55 = 880Ω (ohms)

Problem D
What electric current is supposed to go through an incandescent lamp with characteristics 110V, 55W, and what electric current will be going through it, if we connect it to a source of electricity with voltage 220V?
How much heat will be generated by this lamp in this case of abnormally high voltage?

Solution
P = U·I = I²·R = U²/R
R = U²/P
R = (110V)²/(55V) = 220Ω
Inorm = (55W)/(110V) = 0.5A
Ihigh = (220V)/(220Ω) = 1A
Phigh = (220V)²/(220Ω) = 220W
In this case of high voltage the lamp will generate 220W of heat, which is 4 times greater amount of heat than normal. It will burn very soon.

Problem E
Two incandescent lamps are connected in series to a source of electricity with voltage 220V.
The first lamp has characteristic 220V and 55W.
The second lamp has characteristic 110V and 55W.
What electric current will be going through these lamps along a circuit?
Will the amount of heat produced by each lamp be greater or less than the one it's designed to produce?

Solution
P = U·I = I²·R = U²/R
R = U²/P
R1 = 220²/55 = 880Ω
R2 = 110²/55 = 220Ω
Rtotal = R1 + R2 = 1100Ω
I = U/Rtotal = 220/1100 = 0.2A
P1 = 0.2²·880 = 35.2W
Consumed power of 35.2W is less than 55W, it will produce less light than it is manufactured for.
P2 = 0.2²·220 = 8.8W
Consumed power of 8.8W is significantly less than 55W, it will be hardly lit at all.

Problem F
Two incandescent lamps are connected in parallel to a source of electricity with voltage 220V.
The first lamp has characteristic 220V and 55W.
The second lamp has characteristic 110V and 55W.
What electric current will be going through a common part of a circuit?
Will the amount of heat produced by each lamb be greater or less than the one it's designed to produce?

Solution
P = U·I = I²·R = U²/R
R = U²/P
R1 = 220²/55 = 880Ω
R2 = 110²/55 = 220Ω
Rtotal = 1 / (1/R1 + 1/R2) = 176Ω
Icommon = U/Rtotal = 220/176 = 1.25A
P1 = 220²/880 = 55W
Consumed power of 55W is exactly the same as normal, it will produce exactly the same amount of heat and light as it is manufactured for
P2 = 220²/220 = 220W
Consumed power of 220W is 4 times greater than normal 55W, it will produced a lot of heat and light, but will burn soon.