Circuit Theory/Series Resistance

Series Resistance

Two or more resistor can be connected in series to increase the total resistance . The Total Resistance is equal to the sum of all the resistor's resistance . The total resistance In Series connected circuit Current and voltage will be reduced $R_{tot}=R_{1}+R_{2}+R_{3}+...+R_{n}\,$ Series Impedance

A branch is defined as any group of resistors, capacitors and inductors that can be circled with only two wires crossing the circle boundary.

A branch connects two, non-trivial nodes or junctions.

Within a branch the components are said to be "in series."

Consider a branch containing a Resistor, Capacitor and Inductor.

Say the driving function or source is

$V_{s}=10*cos(22400t+30^{\circ })$ $V_{s}=10*e^{-5000t}cos(22400t+30^{\circ })$ There is just one current, $I$ .

Symbolic Derivation

The terminal equations are:

$\mathbb {V} _{r}=\mathbb {I} *R$ $\mathbb {V} _{L}=\mathbb {I} *j\omega L$ or $\mathbb {V} _{L}=\mathbb {I} *sL$ $\mathbb {I} =\mathbb {V} _{c}*j\omega C$ or $\mathbb {I} =\mathbb {V} _{c}*sC$ There are no junction equations and the loop equation is:

$\mathbb {V} _{r}+\mathbb {V} _{L}+\mathbb {V} _{c}-\mathbb {V} _{s}=0$ Solving the terminal equations for voltage, substituting and then dividing by $\mathbb {I}$ yields:

${\frac {\mathbb {V} _{s}}{\mathbb {I} }}=R+j\omega L+{\frac {1}{j\omega C}}$ ${\frac {\mathbb {V} _{s}}{\mathbb {I} }}=R+sL+{\frac {1}{sC}}$ In terms of impedance, if:

$Z=R+j\omega L+{\frac {1}{j\omega C}}$ $Z=R+sL+{\frac {1}{sC}}$ Then:

${\frac {\mathbb {V} _{s}}{\mathbb {I} }}=Z$ In general, impedances add in series like resistors do in the time domain:

$Z=\sum R_{i}+\sum j\omega L_{i}+\sum {\frac {1}{j\omega C_{i}}}$ $Z=\sum R_{i}+\sum sL_{i}+\sum {\frac {1}{sC_{i}}}$ Numeric Example

In rectangular form:

$Z=100+.15714j$ $Z=80.508+2.2759j$ In polar form (remember impedance is not a phasor, it is a concept in the phasor or complex frequency domain):

$Z=100.0001\angle 0.0016(0.09^{\circ })$ $Z=80.5402\angle 0.0283(1.6193^{\circ })$ 