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Homework Help: Resistor, capacitor and coil with Alternating current

  1. May 29, 2010 #1
    1. The problem statement, all variables and given/known data
    A problem from a translated Sears-Zemansky, 1965, 14-10:
    A conduction coil with inductance of 15 miliHenrys and resistance of 10 Ohms, is connected in line with a capacitor of 200 microFarads and a resistor of 12 Ohms.
    The circuit is supplied with an alternating current of 100 Volts and frequency 50 cycles/sec.
    What is the voltage between the terminals of the conduction coil?
    2. Relevant equations
    [tex]
    \begin{equation*}
    \begin{split}
    \omega &=2\pi L \\
    X_{L} &=\omega C \\
    X_{C} &=\frac{1}{{\omega} C} \\
    X &=X_{L}-X_{C} \\
    Z &=\sqrt{R^{2}+X^{2}} \\
    V_{active} &=I_{active}\times Z
    \end{split}
    \end{equation*}
    [/tex]

    3. The attempt at a solution

    [tex]
    \begin{equation*}
    \begin{split}
    X_{L} &=2\pi 50 \cdot 0.015=4.7 \\
    X_{C} &=\frac{1}{2\pi 50 \cdot 2\times 10^{-4}}=15.9 \\
    X &=4.7-15.9=-11.2 \\
    Z &=\sqrt{22^{2}+(-11.2)^{2}}=24.7
    \end{split}
    \end{equation*}
    [/tex]

    The total active current in the circuit is found from the total active voltage equation:

    [tex]100=I \cdot 24.7 \Rightarrow I=4.05[/tex]

    The "Resistance" (I don't know how it is called in English, please tell me) Z on the conduction coil itself (with its resistance) is:

    [tex]Z=\sqrt{R^{2}+X^{2}_L}=\sqrt{10^{2}+4.7^{2}}=11[/tex]

    And the voltage on the coil is, using the total current in the circuit, calculated above:

    [tex]V=I \times Z=4.05 \cdot 11=44.8[/tex]

    The answer should be: 49.4 [Volts]
    Where is the mistake?
    Thanks-Karol
     

    Attached Files:

    Last edited: May 29, 2010
  2. jcsd
  3. May 29, 2010 #2

    ehild

    User Avatar
    Homework Helper

    Your calculation is correct, but take care of rounding: use at least as many significant digits during the calculations as in the result.

    ehild
     
  4. May 29, 2010 #3
    Thanks, I'l do that.
     
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