**PARALLEL AND SERIES RL and RC**

BEHAVIOUR OF INDUCTORS OR CAPACITORS WITH RESISTORS

**PARALLEL RL OR RC**

As with conventional current flow and actual electron flow, it makes little difference whether we talk about the phase angle of capacitors being +ve or -ve so long as inductors are the opposite. That is, whether we say the phase angle of a reactive component is defined as the voltage leading the current instead of the current leading the voltage. Everything simply becomes a mirror image.

Since there is only one reactive component and one resistive component in each of the circuits on this page, inductors will be the focus. Capacitors work the same way except reactance and phase angle will be negative instead of positive. Only a short comment on capacitors will follow each discussion. Simply exchange +ω for -ω

**PARALLEL RESISTOR - REACTANCE - CURRENT**

In a parallel RL circuit, the applied voltage will be the same for both. The current though the resistor will simply be V/R (Ohm's law). The current through the inductor will be simply V/X_{L}. WE CAN'T JUST ADD THESE TWO CURRENTS THOUGH as we did for two inductors in parallel or two capacitors in parallel. The current through the resistor and that through the inductor will not have the same phase angle.

Consider the following plots. The blue line represents current through the resistor, the purple line the current through the inductor and the white line the total of the two.

For X | For X | For R = ½X |

In the first, the reactance of the inductor is half the resistance of the resistor. This means twice as much current will flow through the inductor and the phase angle of the total current will be closer to that of the inductor. In the second, the impedance and resistance were chosen to be the same and so the phase angle will be half way between the two but, because the two currents are ^{Π}/2 out of phase, the total current will not be double. In the third, the resistance is half the reactance and so the resulting total current will be closer to that of the resistor.The applied voltage will have the same phase angle as the current through the resistor.

QUICK QUESTION:- Suppose the applied voltage is 12 volts. Suppose also the resistor is always 4 ohms and the reactance of the inductor is 2 ohms, 4 ohms and 8 ohms. How much power will be dissipated in each example? Answer below.

In the above plots. We could easily have said the blue line is the current through a capacitor, the purple line as the current through the resistor and the white line still the total current. We would get a mirror image of the same result.

This three plots above can be represented with a vector graphs.

I_{L} is the current through the inductor, I_{R} is the current through the resistor and I_{T} is the total current. The total current can be worked out using pythagoras.

**I _{T} = √ ( I_{L}^{2} + I_{R}^{2} )**

This isn't the end of the story because there is also a phase angle. The phase angle of the current can be worked out from the arctangent of reactance/resistance. In the first example, if the current through the inductor is twice that through the resistor, the phase angle is atan(2/1) or 1.11 radians (63 degrees). In the second example where both are the same, the phase angle will be .785 (^{Π}/4 radians or 45 degrees) and in the last example if the current through the resistor is twice that through the inductor .46 radians (27 degrees).

Current can no longer be described as a number of amps. It also has a phase angle which must also be quoted in reactive circuits.

**PARALLEL RESISTOR-REACTANCE IMPEDANCE**

The overall impedance of the circuit can be calculated from the current. This impedance will also have a phase angle. If we substitute the value for current with the source voltage and impedance, we get:-

V / |Z_{T}| = √ ( ( V_{L} / |X_{L}| )^{2} + ( V_{R} / R)^{2} )

Note this is the __magnitude of the Impedance only__ thus |Z|. It is worked out from the reactive current based on reactance |L| and resistance R. |X| is because it is only the magnitude not capacitive or inductive nature of the reactance.

Since VL, VR are the same (Parallel circuit) the voltages on both sides of the equation cancel out and we are left with:-

1 / |Z_{T}| = √ ( 1 / |Z_{L}|^{2} + 1 / R^{2} ) ... OR ... |Z_{T}| = 1 / √ ( 1 / |Z_{L}|^{2} + 1 / R^{2} )

The same formula can be used for capacitance, capacitive reactance, current and impedance except the phase angle is -ve instead of +ve.

ANSWER TO QUICK QUESTION ABOVE:- The power dissipated in each example is always 36 watts. The inductor dissipates no power. It is only the resistive component which dissipates power dependant only on the current through the resistor which will be 3 amps in each example.

**SERIES RESISTOR-REACTANCE - VOLTAGE**

In a series circuit, the current will be the same in all components. (Kirchhoff's first law.) The current will be dependant on the impedance of all. The impedances are ^{Π}/2 out of phase so the overall impedance can be worked out from a vector diagram as:-

|Z| = √ ( |X_{L}|^{2} + R^{2} )

The current flowing through the circuit will therefore be I = V / √ ( |X_{L}|^{2} + R^{2} ). The impedance has a phase angle and so does the resulting current which will be arctangent(reactance/resistance). This means that the phase angle of the current through the resistor will be the same at both ends of the resistor but this will NOT be in phase with the voltage applied to the entire circuit.

For simplicity's sake, suppose we choose components and a frequency such that the reactance and resistance are the same. In the following plot, the white line represents the voltage applied to the entire circuit, the blue line represents the voltage across the inductor and the purple line the voltage across the resistor. It is this way around because the current in the inductor lags the applied voltage and the phase of the voltage across the resistor must be in phase with this current. (Across a resistor, voltage and current are always in phase.)

Note, the voltages across the two components are ^{Π}/2 out of phase with each other which will always be the case with a resistor and perfect inductor. Because the impedances are the same, both are ^{Π}/4 out of phase with the applied voltage.

Voltages can be calcuated using ordinary trigonometry. The voltage and impedance are directly proportional so:-

The same relationship also holds for resistance as well as inductor reactance. Given any three values, the fourth can be worked out using simple algebra.

This circuit is worth further analysis because it appears to violate Kirchhoff's second law. (The sum of volatges in a loop is zero.) Using the same conditions of equal impedance and solving above equations for V_{R} and V_{L}, the voltage across the resistor and across the inductor will both be 0.7071 (√2/2) times the applied voltage. It would seem the voltage of each component added is 1.414 (√2) times the applied voltage. The answer lies in the phase angle. If two voltages of 0.7071 exactly ^{Π}/2 out of phase are added, the result is 1 with a phase angle exactly in the middle just like the above diagram.

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