Circuit Theory/Parallel Resistance

Parallel Resistance
Resistors in parallel share the same voltage. They split the current up. Giving the current multiple paths to follow means that the overall resistance decreases.




 * $$R_\mathrm{total} =\frac{1}{ \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}} = \frac{1}{\sum\frac{1}{R_i}}$$

The above equation is often called the "one over -- one over" equation.

Parallel Impedance


A group of parallel branches split up the current, but share the same voltage. The parallel branches connect the same two nodes.

The impedance of parallel branches can be combined into one impedance.

Consider these parallel branches: one with a resistor, another with a capacitor and a third with an inductor.

Now drive them with a voltage source:
 * $$V_s = 10*cos(22400t+30^\circ)$$
 * $$V_s = 10*e^{-5000t}cos(22400t+30^\circ)$$

There is just one current, $$V$$.

symbolic derivation
The terminal equations are:
 * $$\mathbb{V}_s = \mathbb{I}_r*R$$
 * $$\mathbb{V}_s = \mathbb{I}_L*j\omega L$$ or $$\mathbb{V}_s = \mathbb{I}_L*sL$$
 * $$\mathbb{I}_c = \mathbb{V}_s *j\omega C$$ or $$\mathbb{I}_c = \mathbb{V}_s *sC$$

There are no loop equations and the junction equation is:
 * $$\mathbb{I}_r + \mathbb{I}_L + \mathbb{I}_c - \mathbb{I}_s = 0$$

Solving the terminal equation for currents, substituting and dividing both sides by $$\mathbb{V}_s$$ yields:
 * $$\frac{\mathbb{I}_s}{\mathbb{V}_s} = \frac{1}{R} + \frac{1}{j\omega L} + j\omega C$$
 * $$\frac{\mathbb{I}_s}{\mathbb{V}_s} = \frac{1}{R} + \frac{1}{sL} + sC$$

In terms of impedance, if:
 * $$\frac{1}{Z} = \frac{1}{R} + \frac{1}{j\omega L} + j\omega C$$
 * $$\frac{1}{Z} = \frac{1}{R} + \frac{1}{sL} + sC$$

Then:
 * $$\frac{\mathbb{I}_s}{\mathbb{V}_s} = \frac{1}{Z}$$

In general, impedances add in series like resistors do in the time domain:
 * $$ \frac{1}{Z} = \sum\frac{1}{R_i} + \sum\frac{1}{j\omega L_i} + \sum j\omega C_i$$
 * $$ \frac{1}{Z} = \sum\frac{1}{R_i} + \sum\frac{1}{sL_i} + \sum sC_i$$

or
 * $$ Z = \frac{1}{\sum\frac{1}{R_i} + \sum\frac{1}{j\omega L_i} + \sum j\omega C_i}$$
 * $$ Z = \frac{1}{\sum\frac{1}{R_i} + \sum\frac{1}{sL_i} + \sum sC_i}$$

numeric example

 * $$Z = 99.9938 - 0.7857i = 99.9969\angle -0.0079 (-0.4502^\circ)$$
 * $$Z = 186.85 -837.07i = 857.6701\angle -1.3512 (-77.4170^\circ)$$

So the exact same components hooked in series are dominated by the inductor (large reactance of the inductor overshadows the smaller reactance of the capacitor), but when the same components are hooked in parallel, the small er capacitive reactance dominates.