# Calculation of Power loss in Wiring and Power consumption in Bulb . Did I make a mistake?

## Did I solve the exercise correctly?

• ### Yes

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• ### No

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Thread Starter

#### jean.the.beginner

Joined May 29, 2016
10
Can anyone help me to check the results of this exercise that I solved by myself?
Can you think of any mistakes/ flaws I made while calculating it?

The exercise is the following: There is a Bulb, which has the specs: 12V 20W. This bulb is connected to a 12V Battery using 20 meters of wire.
this wire has the transversal area of 0,075 square mm. and it has the specific resistance (rho) of 0,017 Ohm mm^2/Meter

The exercise asks me to calculated the Dissipated power caused by the loss in the wires, as well as the power consumption by the bulb

I assumed that the resistance of a Bulb will not change (at least not considerably)
First I calculate the resistance of the wire. which can be calculated by R = Rho * Lenght / Area. Therefore 0,017 * 20 / 0,075 = 4,53333Ohm
Now I needed to calculate the Resistance of the Bulb itself. which I did using Rbulb = V^2 /P which is 144/20 = 7,2 Ohm
Now. in order to know the current which is actually flowing through the circuit (which is all in Series) I did. Current = V / Rtotal
therefore Voltage / (Resistance of wire 1 + Resistance of Wire2 + Resistance of Bulb) Therefore 12 / (453 +4,53 + 7,2) = 0,714 Amps

Now I considered that it will be like a Voltage dropper. The Rbulb will cause one drop, and the Rwire will cause another drop.

Vbulb = Vsource * (Rbulb/Rtotal) (Rtotal being = Rw1+Rw2+Rbulb)
=5,1 Volts.
Therefore the power consumption of the bulb is 5,1Volts1*0,714amps = 3,64 Watt
and to find the Voltage drop caused by the two wires, I associated them in series, and therefore multiplied their resistances by 2.

12V - 5,1V = 4,9V
therefore Power loss in the wires = 4,9V * 0,714Amps = 3,64 Watt
Please check the picture I uploaded. where all the calculations are explained.

DID I DO ANY MISTAKE?

Thanks in advance
Jean

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#### WBahn

Joined Mar 31, 2012
30,255
It's unclear from your problem statement whether the bulb and the battery are separated by 20 meters, or if they are connected via 20 meters of wire (i.e., separated by 10 meters). You might take a look at the original problem and see if it is made clear which of these it is.

#### dl324

Joined Mar 30, 2015
17,022
Didn't go through all of your calculations, but the answer has to be wrong.

You assumed 20W dissipated by the bulb; in the end you determined it was 3.64W. That makes your bulb resistance calculation wrong.

You calculated different resistances for the bulb and wire, yet you calculated the same power dissipation for them.

Thread Starter

#### jean.the.beginner

Joined May 29, 2016
10
It's unclear from your problem statement whether the bulb and the battery are separated by 20 meters, or if they are connected via 20 meters of wire (i.e., separated by 10 meters). You might take a look at the original problem and see if it is made clear which of these it is.
in the problem the battery and the bulb are separated by 20 meters. i.e. 40 meters wire total

Thread Starter

#### jean.the.beginner

Joined May 29, 2016
10
Didn't go through all of your calculations, but the answer has to be wrong.

You assumed 20W dissipated by the bulb; in the end you determined it was 3.64W. That makes your bulb resistance calculation wrong.

You calculated different resistances for the bulb and wire, yet you calculated the same power dissipation for them.
well,
but the 20W is the nominal power of the bulb, assuming it will be placed under the recommended voltage, and assuming that the supplying source is able to supply the needed current

but in the real circuit,
Didn't go through all of your calculations, but the answer has to be wrong.

You assumed 20W dissipated by the bulb; in the end you determined it was 3.64W. That makes your bulb resistance calculation wrong.

You calculated different resistances for the bulb and wire, yet you calculated the same power dissipation for them.
well,
but the 20W is the nominal power of the bulb, assuming it will be placed under the recommended voltage, and assuming that the supplying source is able to supply the needed current
a bulb has a fixed internal resistance right? (if I do not consider the temperature changes)

how should i proceed in order to solve this exercise then?
thanks a lot for the help in advance boys

#### WBahn

Joined Mar 31, 2012
30,255
in the problem the battery and the bulb are separated by 20 meters. i.e. 40 meters wire total
Look at the end of your original post.

Therefore the power consumption of the bulb is 5,1Volts1*0,714amps = 3,64 Watt
and to find the Voltage drop caused by the two wires, I associated them in series, and therefore multiplied their resistances by 2.

12V - 5,1V = 4,9V
therefore Power loss in the wires = 4,9V * 0,714Amps = 3,64 Watt
Without doing ANY math, can you see that at least one of these MUST be wrong?

Also, are you really sure that 12V - 5,1V = 4,9V ?

The math skills you are demonstrating are atrocious even at a 4th grade level. You REALLY need to build up those skills, otherwise they will continue to be a major source of errors and difficulty learning new concepts.

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