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    Hi for Ohmic conductors i know the gradient should be a straight line but the problem is, how do you plot the graph? What is on the axis? Is current on the x-axis or voltage?

    A batter has emf 12V and internal resistance 0.5 ohms connected to two resistros R1 and R2 or resistances 3.5 ohms and 2.0 ohms respectively, in series.

    a) calculate the p.d across the terminals of the battery and across the resistor R1

    b) why is the p.d across the terminals of the battery less than the emf of the batter?


    Please help
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    Hi for Ohmic conductors i know the gradient should be a straight line but the problem is, how do you plot the graph? What is on the axis? Is current on the x-axis or voltage?
    usually on the x-axis u put the "dependant" variable, i.e. the value that YOU can control, and since you control the voltage, u put the voltage on the x-axis.
    to plot the graph: u have to make an experiment, find values of I at variable values of V ( which might be given in the question), and plot V v.s. I, you will get a straight line:
    comparing :
    y = mx + c and
    V = RI
    then c = 0 and m = R
    so the gradient is R
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    (Original post by yazan_l)
    usually on the x-axis u put the "dependant" variable, i.e. the value that YOU can control, and since you control the voltage, u put the voltage on the x-axis.
    to plot the graph: u have to make an experiment, find values of I at variable values of V ( which might be given in the question), and plot V v.s. I, you will get a straight line:
    comparing :
    y = mx + c and
    V = RI
    then c = 0 and m = R
    so the gradient is R

    This is contradictory. You are correct in saying that USUALLY you put the dependent variable on the x-axis, but you have gone on to compare V=IR with y=mc+c...

    ...In order for characteristic graphs to have resistance as the gradient, there is a convention where we put voltage on the y-axis. I would draw it as such if asked, but double check any graphs which they present you with.
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    A batter has emf 12V and internal resistance 0.5 ohms connected to two resistros R1 and R2 or resistances 3.5 ohms and 2.0 ohms respectively, in series.

    a) calculate the p.d across the terminals of the battery and across the resistor R1
    V = IR
    12 = I ( 3.5+2+0.5) = I . 6
    I = 2 A

    V = IR
    V = 2 (0.5)
    Vinternal res. = 1 V

    Vacross terminals = 11V


    VR1 = 2 ( 3.5) = 7 V

    VR2 = 4V
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    (Original post by Spenceman_)
    This is contradictory. You are correct in saying that USUALLY you put the dependent variable on the x-axis, but you have gone on to compare V=IR with y=mc+c...

    ...In order for characteristic graphs to have resistance as the gradient, there is a convention where we put voltage on the y-axis. I would draw it as such if asked, but double check any graphs which they present you with.

    SORRY!!!
    it should be: comparing y = mx+c
    with I = V/R
    then y = I, x = V, and gradient is 1/R
    SORRY FOR THAT MISTAKE


    Edit: this depends on what u use for x-axis and what for y-axis. Sorry again
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    Ok no problem!

    But i dont understand what difference it will make if we plot either way as the gradient would still be a constant?
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    (Original post by swifty)
    Ok no problem!

    But i dont understand what difference it will make if we plot either way as the gradient would still be a constant?
    as i said: if u ploted V on x-axis and I on y-axis :
    y = mx + c
    I = V/R
    then gradient m = 1/R

    but if u plotted I on x-axis and V on y-axis:
    y = mx + c
    V = RI
    then gradient m = R

    so they ARE constants, but not equal!
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    b) why is the p.d across the terminals of the battery less than the emf of the batter?
    because there is internal resistace : i.e. between the terminals, there is a bettery that supplies voltage and a resistance that takes voltage
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    Another question

    A batter has an e.m.f of 12V and an internal resistance fo 0.8ohms. A load of resistance 5ohms is connected across the battery.

    This circuit is to be altered by the addition of another resistor to the load, to make the total power transferred to the adapted external circuit from the battery a maximum.

    Find the maximum power the battery can deliver.
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    (Original post by swifty)
    Another question

    A batter has an e.m.f of 12V and an internal resistance fo 0.8ohms. A load of resistance 5ohms is connected across the battery.

    This circuit is to be altered by the addition of another resistor to the load, to make the total power transferred to the adapted external circuit from the battery a maximum.

    Find the maximum power the battery can deliver.
    hmmm, yum yum electricity. lets c if i can do it
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    assuming the second resistance is 5 ohms too.

    to get max power P= VI

    erm struggling
 
 
 
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