It is quite easy to calculate the total impedance of several impedances connected in series. Just add up all of the "real" parts, and then add up all of the "imaginary" parts. The answer will be in the form xR + yj. If y is zero, then the total impedance is equivalent to a resistor of value x ohms. If y is positive, then the total impedance is inductive in nature. If y is negative, then the total impedance is capacitive in nature. If x is zero, then the total impedance is either just a capacitive or an inductive reactance.

MathML formula | Formula Image | Straight text formula |
---|---|---|

${Z}_{T}={Z}_{1}+{Z}_{2}+\mathrm{.\; .\; .}+{Z}_{n}$ | Z_{T} = Z_{1} + Z_{2} + ... + Z_{n} |

Where:

Z_{T} = Total Impedance

Z_{1} = First Impedance

Z_{2} = Second Impedance

Z_{n} = n^{th} Impedance

## Example of Impedances in Series

Consider the problem below:

Stated in words the question is: "For a given sinusoidal supply of 1MHz connected in series with a capacitor of 100pF, a resistor of 2k2, an inductor of 220uH, a resistor of 470 Ohms, a capacitor of 150pF, and an inductor of 120uH what is the impedance seen by the supply? How could the circuit be simplified to present the same impedance with a series connection of fewer components ?"

The solution to the problem involves calculating the sums of the "real" and "imaginary" parts. In the solution below, the reactances for inductors (positive) and capacitors (negative) have been calculated for the 1 MHz source frequency. The solution is shown on the bottom row of the table.

Component | Value | Real part of Impedance | Imaginary part of Impedance (j, the reactance part at 1 MHz) |
---|---|---|---|

Capacitor | 100pF | 0 | - 1592 |

Resistor | 2k2 | 2200 | 0 |

Inductor | 220uH | 0 | 1382 |

Resistor | 470R | 470 | 0 |

Capacitor | 150pF | 0 | - 1061 |

Inductor | 120uH | 0 | 754 |

Total Impedance = | 2670 | -517 j |

So the answer is "The impedance seen at the terminals is 2670 - 517 j". So what does this mean in practical terms? Well a negative imaginary part is capacitive. Re-arranging the equation for capacitive reactance gives C = 1 / (2πfX_{C}). For an X_{C} of 517, this gives the capacitance as 308 pF.

Here is an image summarises the solution: