# Resistor, capacitor and coil with Alternating current

1. May 29, 2010

### Karol

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
$$\begin{equation*} \begin{split} \omega &=2\pi L \\ X_{L} &=\omega C \\ X_{C} &=\frac{1}{{\omega} C} \\ X &=X_{L}-X_{C} \\ Z &=\sqrt{R^{2}+X^{2}} \\ V_{active} &=I_{active}\times Z \end{split} \end{equation*}$$

3. The attempt at a solution

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

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

$$100=I \cdot 24.7 \Rightarrow I=4.05$$

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:

$$Z=\sqrt{R^{2}+X^{2}_L}=\sqrt{10^{2}+4.7^{2}}=11$$

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

$$V=I \times Z=4.05 \cdot 11=44.8$$

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

#### Attached Files:

• ###### 14-10.bmp
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Last edited: May 29, 2010
2. May 29, 2010

### ehild

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

ehild

3. May 29, 2010

### Karol

Thanks, I'l do that.