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Parallel RLC Circuits

### The Circuit The schematic diagram to the right shows three components connected in parallel, and to an ac voltage source: an ideal inductance, and ideal capacitance, and an ideal resistance. In keeping with our previous examples using inductors and capacitors together in a circuit, we will use the following values for our components (Note that R is not the same as in our discussion of Series RLC Circuits, but the reactive components and frequency are the same):

• VAC = 10 vrms.
• f = 1 MHz. (ω = 6283185.3 rad/sec)
• L = 150 µh. (XL = 942.4778 Ω)
• C = 220 pf. (XC = 723.43156 Ω)
• R = 1000 Ω

According to Ohm's Law:

 iL = vL/XL = 10/942.4778 = 0.01061033 = 10.61033 mA. iC = vC/XC = 10/723.43156 = 0.013823008 = 13.823008 mA. iR = vR/R = 10/1000 = 0.01 = 10 mA.

If we measure the current from the voltage source, we find that it supplies a total of 10.503395 mA to the combined load — only about half a milliamp more than iR alone.

So we now have 10 mA of resistive current and just over 3.2 mA of reactive current, and yet the measured total current is just over 10.5 mA. By now, you probably expected this, and you're familiar with the reasons for this apparent discrepancy. Nevertheless, it is worthwhile to complete the exercise and study this circuit in detail.

### The Vectors As usual, the vectors, shown to the right, tell the story. Since this is a parallel circuit, the voltage, v, is the same across all components. It is the current that has different phases and amplitudes within the different components.

Since voltage is the same throughout the circuit, we use it as the reference, at 0°. Current through the resistor is in phase with the voltage dropped across that resistor, so iR also appears at 0°.

Current through an inductor lags the applied voltage, so iL appears at -90°. Current through a capacitor leads the applied voltage, so iC appears at +90°. Since iC is greater than iL, the net reactive current is capacitive, so its phase angle is +90°.

Now the total current, iT, is the vector sum of reactive current and resistive current. Since iR is significantly greater than the difference, iC - iL, the total impedance of this circuit is mostly resistive, and the combined vector for iT is at only a small phase angle, as shown in the diagram.

### The Mathematics

Now that we have the vectors above, we have a pretty clear idea of how the currents through the different components in this circuit relate to each other, and to the total current supplied by the source. Of course, we could draw these vectors precisely to scale and then measure the results to determine the magnitude and phase angle of the source current. However, such measurements are limited in precision. We can do far better by calculating everything and then using experimental measurements to verify that our calculations fit the real-world circuit.

 iT = (iX² + iR²)½ φ = arctan iX iR where iX = iC - iL φ = Phase angle.

Now we simply insert the values we determined earlier and solve these expressions:

 iT = (iX² + iR²)½ = ((13.823008 - 10.61033)² + 10²)½ = (3.212678² + 10²)½ = (10.3213 + 100)½ = (110.3213)½ = 10.503395 ma φ = arctan iX iR = arctan 3.212678 10 = arctan 0.3212678 = 17.81054°

These figures match the initial measured value for current from the source, and fit the rough vector diagram as well. Therefore, we can be confident that our calculations and diagram are accurate.

### When XL = XC

When a circuit of this type operates at resonance, so that XL = XC, it must also follow that iL = iC. Therefore, iC - iL = 0, and the only current supplied by the source is iR.

This is in fact the case. At resonance, current circulates through L and C without leaving these two components, and the source only needs to supply enough current to make up for losses. In this case, R represents the energy losses within the circuit, and is the only component that draws current from the source. The effective impedance of the circuit is nothing more than R, and the current drawn from the source is in phase with the voltage.

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