- #1

Karol

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## Homework Statement

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?

## Homework 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]

## 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

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