6.12 Rlc series ac circuits  (Page 2/9)

Calculating impedance and current

An RLC series circuit has a $\text{40.0 Ω}$ resistor, a 3.00 mH inductor, and a $\text{5.00 μF}$ capacitor. (a) Find the circuit’s impedance at 60.0 Hz and 10.0 kHz, noting that these frequencies and the values for $L$ and $C$ are the same as in [link] and [link] . (b) If the voltage source has $V_{\text{rms}}=\text{120}\text{V}$ , what is $I_{\text{rms}}$ at each frequency?

Strategy

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.

Solution for (a)

At 60.0 Hz, the values of the reactances were found in [link] to be $X_L=1\text{.}\text{13}\Omega$ and in [link] 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

Discussion for (a)

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.

Solution for (b)

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

Discussion for (a)

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