Maths

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3. Ohm’s law states that the current flowing in a circuit varies directly as the applied voltage and inversely as the resistance. In an experiment, the resistance was increased by triple the percentage by which the voltage was increased. A student calculated the percent change in current to be 15%. Is the student correct?
• If yes, determine the percent change in voltage and resistance to three significant digits.
• If no, explain why and give an example of non-zero percent changes in V and R that would result in a 15% increase in I.

  • Maths -

    I1 = V/R.
    Let V increase by 10%; then increase R
    by 3 times as much(30%).

    I2 = (V+0.1V)/(R+0.3R).
    I2 = 1.1V/1.3R = 0.85V/R.

    I2/I1 = (0.85V/R) / (V/R) = 0.85 = 85%.
    %Change = 85%-100% = -15% = 15% Decrease
    However, this value is valid for a 10%
    increase in V only.

    If V is increased by 20% and R increased by 3 times as much(60%),the
    results is completely different:

    I = 1.2V/1.6R = 0.75V/R. This is the
    same as decreasing V by a factor of 0.75
    while holding R constant. In the first
    example, V was decreased by a factor of
    0.85. Therefore, we cannot consider the student's calculation as being correct
    unless the % increase in V was given.

    I1 = V/R.

    I2 = (V+3p*V)/(R+p*R).
    I2 = V(1+3p)/R(1+p).

    I2/I1 = (V(1+3p)/R(1+p))/(V/R).
    The Vs and Rs cancel:
    I2/I1 = (1+3p)/(1+p) = 1.15
    (1+3p)/(1+p) = 1.15
    Multiply both sides by (1+p):
    1+3p = 1.15+1.15p
    1.85p = 0.15
    P = 0.081 = 8.1 % Increase in R.

    3p = 3*.081 = 0.243 = 24.3% Increase in V.

    Check:
    I2 =(V+0.243V)/(R+0.081R)
    I2 = 1.243V/1.081R = 1.15V/R.

    I2/I1 = (1.15V/R) / (V/R) = 1.15 = 115%
    %Change=115 - !00% = 15% = 15% Increase.


























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