*Notes to a video lecture on http://www.unizor.com*

__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

*or*

**P = U·I***or*

**P = U² / R**

**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

*or*

**P = U·I***or*

**P = U² / R**

**P = I²·R**From this, knowing

*and*

**U***, we derive*

**P***and*

**I***.*

**R**

**I = P / U**

**R = U² / P**Substituting the real numbers,

*A (*

**I = 55 / 220 ≅ 0.25***amperes*)

*Ω (*

**R = 220²/55 = 880***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**

**I**_{norm}= (55W)/(110V) = 0.5A

**I**_{high}= (220V)/(220Ω) = 1A

**P**_{high}= (220V)²/(220Ω) = 220WIn 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

**to a source of electricity with voltage**

__series__**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**

**R**_{1}= 220²/55 = 880Ω

**R**_{2}= 110²/55 = 220Ω

**R**_{total}= R_{1}+ R_{2}= 1100Ω

**I = U/R**_{total}= 220/1100 = 0.2A

**P**_{1}= 0.2²·880 = 35.2WConsumed power of

**35.2W**is less than

**55W**, it will produce less light than it is manufactured for.

**P**_{2}= 0.2²·220 = 8.8WConsumed power of

**8.8W**is significantly less than

**55W**, it will be hardly lit at all.

*Problem F*

Two incandescent lamps are connected in

**to a source of electricity with voltage**

__parallel__**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**

**R**_{1}= 220²/55 = 880Ω

**R**_{2}= 110²/55 = 220Ω

**R**_{total}= 1 / (1/R_{1}+ 1/R_{2}) = 176Ω

**I**_{common}= U/R_{total}= 220/176 = 1.25A

**P**_{1}= 220²/880 = 55WConsumed 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

**P**_{2}= 220²/220 = 220WConsumed power of

**220W**is 4 times greater than normal

**55W**, it will produced a lot of heat and light, but will burn soon.

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