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
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
Attachments

131.1 KB Views: 4