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Stuck on A-level electronics question, please help! :-(

Hi all, I'm really stuck on understanding why the graph in this question would show a decreasing gradient. I understand why it goes from zero to the emf as X increases (there's no current in R when the slider is at B, and there's half the total current when it's at A) but I can't derive the decreasing gradient part :confused: Please can someone help??
Question:
https://imgur.com/HHSRB6H
Answer:
https://imgur.com/tIZoWQn
Thank youuu
In an exam situation id just check what would happen if x = L/2, in this case that means 2/3 of the current goes through the wire and 1/3 current through the fixed resistor so its 1/3 of 1.5V = 0.5V, and then you can draw a graph with decreasing gradient (since the q is only 2 marks id prob go with this method)

If you want to properly derive it.
The resistance of the wire, R_wire = (L-x)/L * R
so the ratio of the total current through the fixed resistor = ((L-x)/L * R )/((L-x)/L * R + R) = (L-x)/(2L-x)
and the voltage is just current*R and R is a constant.
https://www.desmos.com/calculator/n5x8js3kui
you can check this always gives a curving downwards graph.
(edited 1 month ago)
Original post by mikejenks
Hi all, I'm really stuck on understanding why the graph in this question would show a decreasing gradient. I understand why it goes from zero to the emf as X increases (there's no current in R when the slider is at B, and there's half the total current when it's at A) but I can't derive the decreasing gradient part :confused: Please can someone help??
Question:
https://imgur.com/HHSRB6H
Answer:
https://imgur.com/tIZoWQn
Thank youuu


There are 3 types of graphs that we would need to consider:
- linear graph, a straight line
- “increasing gradient”
- “decreasing gradient”

If we consider x=0.5L, the resistance of SB in parallel to the resistor R has an effective resistance of
Reff=(10.5R+1R)1=R3 R_{eff} = \left(\dfrac{1}{0.5R} + \dfrac{1}{R} \right)^{-1} = \dfrac{R}{3}

The voltage across the resistor R is
V=R/3R/3+0.5R×1.5=0.6V V = \dfrac{R/3}{R/3 + 0.5R} \times 1.5 = 0.6 \text{V}

If the graph is linear, when x=0.5L, the voltage would be 0.75 V but the calculation shows that the actual voltage is below 0.75 V so the curve will have a “decreasing gradient”.
I believe the MS is simplifying the overall curve. The exact curve does not have an overall “decreasing gradient”. You can try to derive the equation.

The equation that mosaurlodon has derived is incorrect and the explanation is also flawed.
Oh yeah actually just ignore my post this problem is a lot more complicated then I initially thought.
I tried to derive the equation but tbh im not sure if this is correct?
resistor.jpg
https://www.desmos.com/calculator/g1ccqmsy6u
Reply 4
Original post by mosaurlodon
In an exam situation id just check what would happen if x = L/2, in this case that means 2/3 of the current goes through the wire and 1/3 current through the fixed resistor so its 1/3 of 1.5V = 0.5V, and then you can draw a graph with decreasing gradient (since the q is only 2 marks id prob go with this method)
If you want to properly derive it.
The resistance of the wire, R_wire = (L-x)/L * R
so the ratio of the total current through the fixed resistor = ((L-x)/L * R )/((L-x)/L * R + R) = (L-x)/(2L-x)
and the voltage is just current*R and R is a constant.
https://www.desmos.com/calculator/n5x8js3kui
you can check this always gives a curving downwards graph.

Thanks!!
errr all of what you replied to is wrong 😅
my bad shouldve deleted it.
(edited 1 month ago)
Reply 6
Original post by Eimmanuel
There are 3 types of graphs that we would need to consider:
- linear graph, a straight line
- “increasing gradient”
- “decreasing gradient”
If we consider x=0.5L, the resistance of SB in parallel to the resistor R has an effective resistance of
Reff=(10.5R+1R)1=R3 R_{eff} = \left(\dfrac{1}{0.5R} + \dfrac{1}{R} \right)^{-1} = \dfrac{R}{3}
The voltage across the resistor R is
V=R/3R/3+0.5R×1.5=0.6V V = \dfrac{R/3}{R/3 + 0.5R} \times 1.5 = 0.6 \text{V}
If the graph is linear, when x=0.5L, the voltage would be 0.75 V but the calculation shows that the actual voltage is below 0.75 V so the curve will have a “decreasing gradient”.
I believe the MS is simplifying the overall curve. The exact curve does not have an overall “decreasing gradient”. You can try to derive the equation.
The equation that mosaurlodon has derived is incorrect and the explanation is also flawed.

Thanks very much, this makes sense! It's not a very well thought-out question for only two marks...

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