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

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)
(edited 2 years ago)
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.
The gradient gives acceleration here.

At A-Level, gradients are normally drawn and relevant on graphs where the one axis is time.
I hope this helps.
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. :smile:
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. :smile:

Indeed. :wink:
Reply 4
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)


as others have said, you'll never have to draw a tangent on an IV diagram; it's more of a characteristic than a graph so simply divide V by I at the given point.
yes M= V1 -V2 (-Hg +Hg)

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