### Series RLC Circuits When an inductance, capacitance, and resistance are connected in series (above figure), the resistance R can be imagined as belonging entirely to the coil, when you use the preceding formulas. Then you have two vectors to add, when finding the impedance of the series RLC circuit containing three such components:
Z = (R + jXL ) + (0 + jXC)
= R + j (XL + XC)

Again, remember that XC is never positive! So, although the formulas here have addition symbols in them, you’re adding a negative number when you add in a capacitive reactance.

#### Problem 1

Suppose a resistor, a coil, and a capacitor are connected in series with R = 50 Ω, XL = 22 Ω, and XC = −33 Ω. What is the net impedance, Z?
Consider the resistor to be part of the coil, obtaining two complex vectors, 50 + j 22 and 0 − j33. Adding these gives the resistance component of 50 + 0 = 50, and the reactive component of
j22 − j33 = −j11. Therefore, Z = 50 − j11.

#### Problem 2

Consider a resistor, a coil, and a capacitor that are connected in series with R = 600 Ω, XL = 444 Ω, and XC = −444 Ω. What is the net impedance, Z?

Again, imagine the resistor to be part of the inductor. Then the complex impedance vectors are 600 + j444 and 0 − j444. Adding these, the resistance component is 600 + 0 = 600, and the reactive component is j444 − j444 = j 0. Thus, Z = 600 + j 0. This is a purely resistive impedance, and you can rightly call it 600 Ω.

#### Problem 3

Suppose a resistor, a coil, and a capacitor are connected in series. The resistor has a value of 330 Ω, the capacitance is 220 pF, and the inductance is 100 μH. The frequency is 7.15 MHz. What is the complex impedance of this series RLC circuit at this frequency?
First, calculate the inductive reactance. Remember that XL = 6.28f L and that megahertz and microhenrys go together in the formula. Multiply to obtain the following:
jXL = j(6.28 × 7.15 × 100)
= j4490

Next, calculate the capacitive reactance using the formula XC = −1/(6.28fC ). Convert 220 pF to microfarads to obtain C = 0.000220 μF. Then calculate:
jXC = −j[1/(6.28 × 7.15 × 0.000220)]
= −j101

Now, lump the resistance and the inductive reactance together, so one of the impedance vectors is 330 + j4490. The other is 0 − j101. Adding these gives Z = 330 + j 4389; this rounds off to
Z = 330 + j4390.

#### Problem 4

Suppose a resistor, a coil, and a capacitor are connected in series. The resistance is 50.0 Ω, the inductance is 10.0 μH, and the capacitance is 1000 pF. The frequency is 1592 kHz. What is the complex impedance of this series RLC circuit at this frequency?
First, calculate XL = 6.28f L. Convert the frequency to megahertz; 1592 kHz = 1.592 MHz. Then:
jXL = j(6.28 × 1.592 × 10.0)
= j100

Then calculate XC = −1/(6.28fC ). Let’s convert picofarads to microfarads, and use megahertz for the frequency. Therefore:
jXC = −j[1/(6.28 × 1.592 × 0.001000)]
= −j100

Let the resistance and inductive reactance go together as one vector, 50.0 + j100. Let the capacitive reactance be represented as 0 − j100. The sum is Z = 50.0 + j100 − j100 = 50.0 + j0. This is a pure resistance of 50.0 Ω. You can correctly say that the impedance is 50.0 Ω in this case.