How to find the power consumed by the light bulb

[Helly123]

Homework Statement

(Is the image read-able now? )

The fig 1 shows the relationship between voltage and current when voltage is applied to a certain light bulb.
Next, in fig 2, the bulb is connected in series with a 12-V battery and a 5 ohm resistor.

Question : what is the amount of power consumed by the light bulb?

Homework Equations

P = IV = V^2/R = I^2R
V = IR

The Attempt at a Solution

From the fig 1
When 12 V, the current accros the bulb is 3.5 A
So , R = V/I = 3.5 ohm

Now the bulb connected series with 5 ohm
And voltage 12 V
I = VR = 12*8.5 = 1.4 A

P = I^2R = 6.86 Wt

But I get wrong answer . Please help.

 

[tnich]

Homework Statement

View attachment 221407
(Is the image read-able now? )

The fig 1 shows the relationship between voltage and current when voltage is applied to a certain light bulb.
Next, in fig 2, the bulb is connected in series with a 12-V battery and a 5 ohm resistor.

Question : what is the amount of power consumed by the light bulb?

Homework Equations

P = IV = V^2/R = I^2R
V = IR

The Attempt at a Solution

From the fig 1
When 12 V, the current accros the bulb is 3.5 A
So , R = V/I = 3.5 ohm

Now the bulb connected series with 5 ohm
And voltage 12 V
I = VR = 12*8.5 = 1.4 A

P = I^2R = 6.86 Wt

But I get wrong answer . Please help.

You are assuming that the light bulb's resistance is a constant value. If that were true, what would be the shape of the curve in plot of current vs. voltage?

 

[CWatters]

Have you been taught the load Line method?

Involves plotting 2 lines on a graph for the source and load and noting where they cross.

Also used for solving two simultaneous equations.

 

[CWatters]

PS does the diagram show a 5R or a 3R resistor?

 

[Helly123]

PS does the diagram show a 5R or a 3R resistor?

It is the light bulb 3 ohm resistor Sir :)

 

[Helly123]

Have you been taught the load Line method?

Involves plotting 2 lines on a graph for the source and load and noting where they cross.

Also used for solving two simultaneous equations.

No

 

[Helly123]

You are assuming that the light bulb's resistance is a constant value. If that were true, what would be the shape of the curve in plot of current vs. voltage?

I see.. but how to find the right I and V?

 

[gneill]

I see.. but how to find the right I and V?

Since you aren't given a mathematical expression for the current versus voltage characteristics for the bulb you will need to work with the graph.

You were given a good hint by @CWatters when he mentioned the Load Line method. I would suggest that you look it up in wither your class textbook or course notes, or via web search.

 

[Merlin3189]

The problem is that you don't know the resistance of the light bulb. You have only the graph to show you. Some people could maybe deduce a formula from the graph, but it is not simple to do. So we non-physicists/non-mathematicians just use the graph.

For the resistor you can also draw a graph of current against voltage.
If you plot this on the same axes, you can find the point where the current is the same through both (they are in series.)
BUT you have to be careful on the voltage scale.
If there were 12V across the bulb, there would be 0V across the resistor) and vice versa.
If there were say 9V across the bulb, it would be 3V across the resistor and vice versa.
The two voltages always add up to 12V.

 

[Helly123]

The problem is that you don't know the resistance of the light bulb. You have only the graph to show you. Some people could maybe deduce a formula from the graph, but it is not simple to do. So we non-physicists/non-mathematicians just use the graph.

For the resistor you can also draw a graph of current against voltage.
If you plot this on the same axes, you can find the point where the current is the same through both (they are in series.)
BUT you have to be careful on the voltage scale.
If there were 12V across the bulb, there would be 0V across the resistor) and vice versa.
If there were say 9V across the bulb, it would be 3V across the resistor and vice versa.
The two voltages always add up to 12V.
View attachment 221479

yes.
anyway, do you mean :
current for example is 3A, V = 6V
then V for 5 resistor is I*R = 15V (this is more than the total voltage)

 

[Merlin3189]

anyway, do you mean :
current for example is 3A, Vbulb = 6V (from the graph)
then Vr for 5 Ω resistor is I*R = 15V (this is more than the total voltage) Exactly!

So the current can't be 3A as you assumed.

You could equally try to guess the current as 1 A.
Then Vb = 0.7 V (from the graph)
And Vr = I*R = 1*5 = 5 V
So total voltage = 5.7 V and this must be wrong also.

If you continue trying different currents, you will get there when you hit on the right value. (In this case it is dead easy!)
But for a more difficult case, we would draw in a line for the resistor and where the two graphs cross (ie. have equal current) will show the correct current and also the voltage across each component.

Edit: This is the "Load Line" that someone else mentioned.

 

[Helly123]

So the current can't be 3A as you assumed.

You could equally try to guess the current as 1 A.
Then Vb = 0.7 V (from the graph)
And Vr = I*R = 1*5 = 5 V
So total voltage = 5.7 V and this must be wrong also.

If you continue trying different currents, you will get there when you hit on the right value. (In this case it is dead easy!) But for a more difficult case, we would draw in a line for the resistor and where the two graphs cross (ie. have equal current) will show the correct current and also the voltage across each component.

Thank you. The answer is I = 2, R for bulb is 1ohm. So P = I^2R = 4 Watt. A bulb resistor can change as the current change?

 

[CWatters]

+1

Separate the circuit into two parts...

1) the bulb
2) the rest of the circuit

You already have a graph of the IV characteristics of the bulb. What you need is an IV graph for the rest of the circuit without the bulb.

To get that replace the bulb with a short circuit (V=0V) and calculate the current. Then replace it with an open circuit (I=0A) and calculate the voltage.

Now you have two data points to put on the graph. You can join these with a straight line ( because resistors are linear) or try inserting a resistor and calculate both I and V to get another point(s) to prove its a straight line.

Where the two lines cross is called the operating point. The values of I and V at that point is what you need to calculate the power. It's the values that satisfy both parts of the circuit (eg both equations).

 

[gneill]

It is the light bulb 3 ohm resistor Sir :)

...yet you are doing calculations for a 5 Ω resistor?

 

[CWatters]

It is the light bulb 3 ohm resistor Sir :)

Not sure what you mean. The circuit has a light bulb (which has the IV curve in the graph, so it's resistance isn't constant) and a separate 3R resistor.

 

[Merlin3189]

...The answer is I = 2, R for bulb is 1ohm. So P = I^2R = 4 Watt. A bulb resistor can change as the current change?

I agree with the power of the light bulb. P = I^2R is correct, as is P = IV. so 2V * 2A = 4W

The resistance of the bulb does change a lot as the current changes.
This is because the temperature changes from room temperature at no current to perhaps 3000 °C at 3.5 A.

PS, I've just gone along with the 5 Ω in the written question. You could try to work it out for 3 Ω as well (and some other values, just for fun and practice.)

Edit: PPS I've drawn in the load lines on your graph, for 5 Ω and 3 Ω. Let me know if you'd like to see them.

 

[Helly123]

I agree with the power of the light bulb. P = I^2R is correct, as is P = IV. so 2V * 2A = 4W

The resistance of the bulb does change a lot as the current changes.
This is because the temperature changes from room temperature at no current to perhaps 3000 °C at 3.5 A.

PS, I've just gone along with the 5 Ω in the written question. You could try to work it out for 3 Ω as well (and some other values, just for fun and practice.)

Edit: PPS I've drawn in the load lines on your graph, for 5 Ω and 3 Ω. Let me know if you'd like to see them.

Of course. I would like to see it.

 

[Merlin3189]

Ah! But have you tried it yourself yet?

 

[Merlin3189]

Wel, I guess the thread is ended now. Here are the load line diagrams.

Now the load line for a 3 Ohm resistor

Here are a range of load lines for different resistors, using a 12 V supply

And if the supply were only 10 V, the load lines would have the same slope, but start from the max voltage on the bulb of 10 V

 

[Helly123]

Wel, I guess the thread is ended now. Here are the load line diagrams.
View attachment 221547
Now the load line for a 3 Ohm resistor
View attachment 221548
Here are a range of load lines for different resistors, using a 12 V supply
View attachment 221549
And if the supply were only 10 V, the load lines would have the same slope, but start from the max voltage on the bulb of 10 V

View attachment 221550

Thank you Sir