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

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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)

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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(Original post by

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)

**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)

**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.

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**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)

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

…

On a V against I graph the gradient at a point is dI/dV and this is certainly

At A-Level, gradients are normally drawn and relevant on graphs where the one axis is

…

**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**.…

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.

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(Original post by

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

However, if the graph is a straight line

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.

**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.

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