Calculating Impedance and Current

For each frequency, we use $Z=\sqrt{R^2+(X_L-X_C{)}^2}$ to find the impedance and then Ohm’s law to find current. We can take advantage of the results of the previous two examples rather than calculate the reactances again.

At 60.0 Hz, the values of the reactances were found in Example 23.10 to be $X_L=1\text{.}\text{13}\Omega$ and in Example 23.11 to be $X_C=\text{531}\Omega$. Entering these and the given $\text{40.0 Ω}$ for resistance into $Z=\sqrt{R^2+(X_L-X_C{)}^2}$ yields

Similarly, at 10.0 kHz, $X_L=\text{188}\Omega$ and $X_C=3\text{.}\text{18}\Omega$, so that

In both cases, the result is nearly the same as the largest value, and the impedance is definitely not the sum of the individual values. It is clear that $X_L$ dominates at high frequency and $X_C$ dominates at low frequency.

The current $I_{\text{rms}}$ can be found using the AC version of Ohm’s law in Equation $I_{\text{rms}}=V_{\text{rms}}/Z$:

$I_{\text{rms}}=\frac{V_{\text{rms}}}{Z}=\frac{\text{120}\text{V}}{\text{531}\Omega }=0\text{.}\text{226}\text{A}$ at 60.0 Hz

Finally, at 10.0 kHz, we find

$I_{\text{rms}}=\frac{V_{\text{rms}}}{Z}=\frac{\text{120}\text{V}}{\text{190}\Omega }=0\text{.}\text{633}\text{A}$ at 10.0 kHz

The current at 60.0 Hz is the same (to three digits) as found for the capacitor alone in Example 23.11. The capacitor dominates at low frequency. The current at 10.0 kHz is only slightly different from that found for the inductor alone in Example 23.10. The inductor dominates at high frequency.