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Physics Electricity Potential Dividers & Resistance [PLEASE HELP] Watch

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    Hi,
    This is a question I found in the textbook [see attached]
    Now, here are my current solutions so far.

    Part A
    i) 1.0KΩ + 1.0KΩ = 1000Ω + 1000Ω = 2000Ω [For later calculation, the answer is kept as Ω]
    ii) 1000Ω + Effective Parallel Resistance.

    \frac{1}{R_T} = \frac{1}{1000} + \frac{1}{R}

    This simplifies to

    \frac{1}{R_T} = \frac{R}{1000R} + \frac{1000}{1000R} = \frac{1000 + R}{1000R}

    so RT (parallel) =  \frac{1000R}{1000 + R}

    So the final resistance for part ii is 1000 +\frac{1000R}{1000 + R}

    Part B
    There are two ways. First I will find the current because this information becomes necessary in part C.
    I = \frac{V}{R} = \frac{6}{2000} = 3 \times 10^{-3}A\

    Then  V \ = \ IR = 3 \times 10^{-3}A \times 1000 \ohm = 3V

    ALTERNATIVE

    V_{out} = \frac{1000}{1000 + 1000} \times 6 = 3V

    Part C
    I already calculated that the current is  3\times 10^{-3}A
    Using \frac{V}{I} = R = \frac{2}{3\times 10^{-3}A} = 666.7 \Omega

    This means that
    \frac{1}{666.7} = \frac{1}{1000} + \frac{1}{R}
    or
    \frac{1}{666.7} = \frac{1000 + R}{1000R}

    How can I finish this question to find R?
    Please help.

    Answers are now included for reference.
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    I haven't read through your working, but assuming your last line is correct then it's just a linear equation in R.
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    (Original post by morgan8002)
    I haven't read through your working, but assuming your last line is correct then it's just a linear equation in R.
    I tried that, and I didn't get anywhere close to the correct answer. I got a very small negative value.
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    morgan8002 Answers are here.
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    (Original post by nwmyname)
    I tried that, and I didn't get anywhere close to the correct answer. I got a very small negative value.
    (Original post by nwmyname)
    morgan8002 Answers are here.
    You must have messed up your maths there to get a very small negative because you should get the correct answer from your working out.
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    (Original post by nwmyname)
    Hi,
    This is a question I found in the textbook [see attached]
    Now, here are my current solutions so far.



    Part C
    I already calculated that the current is  3\times 10^{-3}A
    Using \frac{V}{I} = R = \frac{2}{3\times 10^{-3}A} = 666.7 \Omega

    This means that
    \frac{1}{666.7} = \frac{1}{1000} + \frac{1}{R}
    or
    \frac{1}{666.7} = \frac{1000 + R}{1000R}

    How can I finish this question to find R?
    Please help.

    Answers are now included for reference.
    You have made an incorrect assumption.

    With the voltmeter added to the circuit, the total current can no longer be 3mA because the parallel resistance with the voltmeter added must now be less than 1K ohm.

    Rtotal = 1000 + Rparallel

    Since Rparallel is now less than 1K

    Rtotal = 1000 + (<1000) = <2000 ohms

    i.e. the current leaving and entering the supply must now be greater than 3mA.

    HINT: The p.d. developed across the top series resistor is 4V (6 - 2) with the voltmeter added to the circuit.
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    (Original post by uberteknik)
    You have made an incorrect assumption.

    With the voltmeter added to the circuit, the total current can no longer be 3mA because the parallel resistance with the voltmeter added must now be less than 1K ohm.

    Rtotal = 1000 + Rparallel

    Since Rparallel is now less than 1K

    Rtotal = 1000 + (<1000) = <2000 ohms

    i.e. the current leaving and entering the supply must now be greater than 3mA.

    HINT: The p.d. developed across the top series resistor is 4V (6 - 2) with the voltmeter added to the circuit.
    Please check the solutions as this is not stated.
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    It looks like you get the correct answer with the assumption that the current is the same, even though it isn't, possibly just a dodgy textbook question?
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    (Original post by nwmyname)
    How can I finish this question to find R?
    Your working is correct up to the point you solve for R.
    \dfrac{1}{666.7}=\dfrac{1000+R}{  1000R} \iff 1000R=666.7(1000+R)

    1000R=666700+666.7R

    333.3R=666700

    R=\dfrac{666700}{333.3} = 2000.3\Omega.

    As needed.
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    (Original post by The-Spartan)
    Your working is correct up to the point you solve for R.
    \dfrac{1}{666.7}=\dfrac{1000+R}{  1000R} \iff 1000R=666.7(1000+R)

    1000R=666700+666.7R

    333.3R=666700

    R=\dfrac{666700}{333.3} = 2000.3\Omega.

    As needed.
    Thank the lord!
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    (Original post by nwmyname)
    Please check the solutions as this is not stated.
    (Original post by JN17)
    It looks like you get the correct answer with the assumption that the current is the same, even though it isn't, possibly just a dodgy textbook question?
    (Original post by The-Spartan)
    Your working is correct up to the point you solve for R.
    \dfrac{1}{666.7}=\dfrac{1000+R}{  1000R} \iff 1000R=666.7(1000+R)

    1000R=666700+666.7R

    333.3R=666700

    R=\dfrac{666700}{333.3} = 2000.3\Omega.

    As needed.
    (Original post by nwmyname)
    Thank the lord!
    The Lord has nothing to do with it.

    The method Spartan used is correct.

    However, the textbook answer to part c) is completely in error.

    Moreover, you cannot use the 3mA current calculated in part b), because adding the voltmeter to the circuit will increase the actual current:

    You can check the textbook is incorrect by substituting the (textbook) calculated resistance values back into the potential divider circuit which produces a voltage of 2.4V across the voltmeter parallel combination (2K || 1K = 666.7 ohms) and the upper series resistor of 1K.

    V_{voltmeter} = V_{supply}(\frac{R_{parallel}}{R  _{series} + R_{parallel}})

    V_{voltmeter} = 6(\frac{666.7}{1000 + 666.7}) = 2.4V

    Ergo this is incorrect and NOT the 2V required by the question.



    The error is, as I stated originally, that the text book has used the 3mA unloaded circuit current (i.e. without the voltmeter loading the circuit) when it should have used 4mA:

    V_{series} = V_{supply} - V_{parallel} = 6 - 2 = 4V

    The total current is then given by

    I_{supply} = \frac{V_{series}}{R_{series}} = \frac{4}{1000} = 4 \rm x 10^{-3} A



    Correct solution:

    R_{parallel} = \frac{V_{parallel}}{I_{supply}}

    R_{parallel} = \frac{2}{4 \rm x 10^{-3}} = 500 \Omega

    Following through on Spartans working:

    \frac{1}{500}=\frac{1000+R}{1000  R} \iff 1000R=500(1000+R)

    1000R=500000+500R

    500R= 500000

    R=\frac{500000}{5000} = 1000\Omega.



    Check:

    Once again, check this correct working using the potential divider calculation:

    R_{parallel} = \rm 1K || 1K = 500\Omega

    V_{voltmeter} = V_{supply}(\frac{R_{parallel}}{R  _{series} + R_{parallel}})

    V_{voltmeter} = 6(\frac{500}{1000 + 500}) = 2V

    i.e. the correct p.d. developed across the lower resistor as stated in the question.
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    (Original post by uberteknik)
    The method Spartan used is correct.
    Sorry, this is my fault i should have read the question. I only saw the mathematical expressions and jumped in.
    After reading, this is infact correct. There is another component and as such there is a different current.
    This is the correct response to the question, note nwmyname i only solved the expression you gave for R
    +1 for correct solution
 
 
 
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