# IV graphs at A-levels...when to tangent?

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#1
My problem is that I never know when to use tangents on I-V graphs at A-level

For example, the IV graph for a filament bulb is a curve and I always thought that in order to find the resistance at a certain point you have to find the gradient of the tangent. However, several past exam questions seem to indicate that you can just read off the pair of V and I values and divide them to find resistance.

How do you know when to draw a tangent and when to use a pair of values, and why doesn't the tangent method work?

This is an example of the sort of question that prompted my thinking (attached)
Last edited by DRDANDY; 1 month ago
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1 month ago
#2
(Original post by DRDANDY)
My problem is that I never know when to use tangents on I-V graphs at A-level

For example, the IV graph for a filament bulb is a curve and I always thought that in order to find the resistance at a certain point you have to find the gradient of the tangent. However, several past exam questions seem to indicate that you can just read off the pair of V and I values and divide them to find resistance.

How do you know when to draw a tangent and when to use a pair of values, and why doesn't the tangent method work?

This is an example of the sort of question that prompted my thinking (attached)
With V / I graphs, just remember that resistance is defined as the value of V divided by I. R=V/I always.
So on such a graph, the resistance value has to be read of from the two values of V and I.

The gradient of a graph is about the rate of change of the one value with the other.

On a V against I graph the gradient at a point is dI/dV and this is certainly not the definition of resistance.

Use the gradient of a graph when you want to find the quantity that is the rate of change of the one thing with the other.
So on a velocity-time graph, the rate of change of velocity with time is acceleration. Again, this is by definition of acceleration.

At A-Level, gradients are normally drawn and relevant on graphs where the one axis is time.
I hope this helps.
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1 month ago
#3
(Original post by DRDANDY)
My problem is that I never know when to use tangents on I-V graphs at A-level

For example, the IV graph for a filament bulb is a curve and I always thought that in order to find the resistance at a certain point you have to find the gradient of the tangent. However, several past exam questions seem to indicate that you can just read off the pair of V and I values and divide them to find resistance.

How do you know when to draw a tangent and when to use a pair of values, and why doesn't the tangent method work?

This is an example of the sort of question that prompted my thinking (attached)
Let me add a few points to what Stonebridge has well explained.
If we have a straight line through the origin for a graph of V versus I, it does not matter whether you use a gradient or the ratio of V to I to find the resistance.

This can be shown using maths (direct proportion and differentiation) and I would leave it for you to prove it.

If it is an I-V graph where the graph is a straight line through the origin, then we need the reciprocal of the gradient to find the resistance.

However, if the graph is a straight line NOT through the origin, the use of gradient will not allow us to find the resistance.

This is why in general, we do not use the gradient of a V-I graph or I-V graph to find the resistance.
(Original post by Stonebridge)

On a V against I graph the gradient at a point is dI/dV and this is certainly not the definition of resistance.

At A-Level, gradients are normally drawn and relevant on graphs where the one axis is time.
There are 2 statements in Stonebridge’s post that I find “disturbing”.

On a V against I graph, the gradient at a point should be dV/dI instead of dI/dV.

As for the last statement, I don’t really think gradients are drawn on a graph. I believe it is the tangent that we draw on a graph to evaluate the gradient.

PS: Stonebridge, I am sorry for being nitpicking. 0
1 month ago
#4
(Original post by Eimmanuel)
Let me add a few points to what Stonebridge has well explained.
If we have a straight line through the origin for a graph of V versus I, it does not matter whether you use a gradient or the ratio of V to I to find the resistance.

This can be shown using maths (direct proportion and differentiation) and I would leave it for you to prove it.

If it is an I-V graph where the graph is a straight line through the origin, then we need the reciprocal of the gradient to find the resistance.

However, if the graph is a straight line NOT through the origin, the use of gradient will not allow us to find the resistance.

This is why in general, we do not use the gradient of a V-I graph or I-V graph to find the resistance.

There are 2 statements in Stonebridge’s post that I find “disturbing”.

On a V against I graph, the gradient at a point should be dV/dI instead of dI/dV.

As for the last statement, I don’t really think gradients are drawn on a graph. I believe it is the tangent that we draw on a graph to evaluate the gradient.

PS: Stonebridge, I am sorry for being nitpicking. Indeed. 0
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