Last month's column covered the behavior of electrical components connected in series and demonstrated that when a voltage is applied across a series circuit, the current through each component is the same. However, you'll often come across components connected in parallel, a configuration that changes the behavior of the circuit and the currents that flow through each component.
When components are connected in parallel, the voltage across each element is equal. The current flowing through each element, however, isn't necessarily equal. Current divides through the elements based on their individual impedances, seeking the path of least resistance. Elements with high impedances carry less current than elements with low impedances. Sometimes two elements put in parallel need to share the current equally, as is usually the case with paralleled transformers. To achieve equal sharing of the current, it's necessary that the impedances of each element be equal.
Inductance and capacitance
Just like with a series circuit, if a parallel circuit consists of elements other than resistances, the basic algebraic equations describing the current flowing in each branch of the circuit become differential equations. In the case of an inductance, the current (I) flowing through the inductor equals the voltage across the inductor integrated over time divided by the inductance (L) of the element in henrys, or
Analyzing the units of measurement, one can see that current flowing through an inductor represents the energy stored by the inductor per coulomb of charge passing through the inductor. Similarly, the current flowing through a capacitor equals the time rate of change of the voltage across the capacitor multiplied by the capacitance (C) in farads (F), or
As with series circuits containing resistances, inductances, and capacitances, if the inductive reactance of a parallel circuit equals the capacitive reactance, the circuit is said to be resonant at that frequency. A common example of parallel resonance is a tuning circuit, as shown in the Figure above.
In practical tuning circuits, the resistance in the capacitive branch is negligible, but the inductive branch has a considerable resistance due to the windings of the inductor. For the circuit shown in the Figure, if the inductance is 10 mH, what size capacitor must be used to cause the circuit to be resonant at 20.6 kHz?
To achieve resonance, the inductive and capacitive reactances must be equal, or
Substituting values yields
To finish, solve for C
The combination of the 10 mH inductor and the 6,000 pF capacitor is resonant at 20.6 kHz only. For any other frequency, the two reactances won't be equal, so resonance won't occur.