Tutorial 11 A - Series RLC Circuits

Learning Objective

To understand phase relationships in series RLC circuits;

To draw phasor diagrams;

To calculate impedance.

To measure impedance

Key Question

What is a series RLC circuit?

What do the phasor diagrams look like for the RLC circuit?

How do we work out the impedance?

How can we deal with multiple components?

Series RLC circuits

Consider this circuit that consists of an inductor of inductance L (H), resistor of resistance R (W), and capacitor of capacitance C (F).  It is connected to an alternating voltage V that has a frequency of f. Question 1

Complete the table to show the units of the quantities.  One has been done as an example.

 Quantity Physics Code Units Resistance R Ohm (W) Voltage Volt (V) L Farad (F) Frequency f

We know that:

• In a series circuit, the current is the same throughout the circuit.

• In a capacitor circuit, the current leads the capacitor voltage by 90o.

• In an inductor circuit, the current lags the inductor voltage by 90o.

• In reactive circuits, the voltages add up as a vector sum.

• The current is always in phase with the voltage across the resistor.

Let’s look at the voltages in the circuit: Since VR is in phase with I, we can draw a phasor diagram to show the phase relationship.  We will show VL leading, and VC lagging.  Since VL is greater than VC the circuit is considered to be inductive. We can see easily that VL and VC are 180o out of phase.  This means that the vector sum of the two vectors is VLVC.  We can call this the reactive voltage:

Reactive Voltage = VLVC

So our vector diagram looks like this: We can now do the vector sum of the voltages: The impedance is calculated with the equation: We use the equation below to work out the phase angle: In these examples below we will use calculations.  You can, of course, use accurate drawing of the phase vectors to calculate the resultant voltage.

Look at the proof HERE on the extension page.

 Question 2 In an experiment to investigate the voltages in an RLC circuit, the voltage across the inductor was found to be 2.8 V, the voltage across the capacitor was found to be 1.2 V, and the voltage across the resistor was found to be 2 V.  What is the resultant voltage? Question 3 Use the data from Question 2 to work out the phase angle between the voltage V and the current.  Does V lead or lag? Question 4 The current in the circuit was found to be 0.2 A.  Use the data in, and your answer to question 2, to calculate a.      The impedance of the circuit. b.      The  resistance; c.      The reactance of the capacitor; d.      The reactance of the inductor. Question 5 Use the formula to calculate the impedance.  Is the result consistent with your previous answer? ### Capacitative Circuits

If VC were greater than VL, our phasor diagram would look like this: We can now do the vector sum of the voltages: Using a similar argument to that above we can write: And: This circuit is capacitative.

### Reactance

From previous topics we know that reactance is defined as the ratio of the voltage across a reactive component, and the current.  For a capacitor: For an inductor: Real inductors have an ohmic resistance.  We treat a real inductor as:

a perfect inductor in series with a resistor.

Often we will need to work out the reactance of these components before we are able to work out the total impedance and other numerical parameters in the series circuit.  We will look at this in the worked example.

 Worked Example A coil of resistance 5 ohms and inductance 120 mH is connected in series with a 100 mF capacitor.  They are connected to a 30 V, 50 Hz supply. Calculate: a)      The reactance of the inductor; b)      The reactance of the capacitor; c)      The impedance of the circuit; d)     The current; e)      The phase angle between the supply voltage and the current; f)       The voltage across the coil; g)      The voltage across the capacitor, Answer a)      XL = 2 × p × 50 × 120 × 10-3 = 37.70 W   b)      Xc = 1 ÷ (2 × p × 50 × 100 × 10-6) = 31.83 W   c)      The reactance of the inductor is bigger than the reactance of the capacitor.  So we write Z2 = R2 + (XL – XC)2          = 52 + ( 37.70 – 31.83)2 = 52 + 5.872  = 59.46           Z = 7.71 W   d)     I = V/Z = 30 ÷ 7.71 = 3.89 A   e)      tan f = 5.87 ÷ 5 = 1.174          f = tan-1 1.174 = 49.6o (= 0.865 rad)   f)       Impedance across the coil: Z2 = 52 + 37.702 = 1446 Þ Z = 38.0 W           V = IZ = 3.89 × 38.0 = 148 V   g)      Voltage across the capacitor = IXC = 3.89 × 31.83 = 124 V

We can draw the phasor diagram for this circuit: This phasor diagram shows what we would expect if we had a perfect inductor of 120 mH in series with a 5 ohm resistor and a 100 mF capacitor.

### Multiple Components in Series

So far we have looked at a circuit that is simply a series capacitor, inductor, and resistor.  But what happens if we have more than one resistor, or more than one inductor?  It’s actually quite simple.  We find the single equivalent resistor, or single equivalent inductor.  In series circuits they simply add up. With series capacitors, it’s a little more involved, but not too difficult: Question 6 What is the single equivalent resistor for two resistances of 5 ohms and 8 ohms? Question 7 What is the single equivalent inductor for two inductors of 80 mH and 120 mH? Question 8 What is the single equivalent capacitor for 100 mF and 40mF capacitors in series? Now we will look at adding an extra component.  We will use the same example that we looked at on page 8, but this time, we will add in an extra 10 ohm resistor.

 Worked Example A coil of resistance 5 ohms and inductance 120 mH is connected in series with a 100 mF capacitor and a 10 ohm resistor.  They are connected to a 30 V, 50 Hz supply. Calculate: a)      The reactance of the inductor; b)      The reactance of the capacitor; c)      The impedance of the circuit; d)     The current; e)      The phase angle between the supply voltage and the current; f)       The voltage across the coil; g)      The voltage across the capacitor. Answer a)      XL = 2 × p × 50 × 120 × 10-3 = 37.70 W   b)      Xc = 1 ÷ (2 × p × 50 × 100 × 10-6) = 31.83 W   c)      The resistance is 5 + 10 = 15 W.  The reactance of the inductor is bigger than the reactance of the capacitor.  So we write Z2 = R2 + (XL – XC)2 = 152 + ( 37.70 – 31.83)2 = 152 + 5.872  = 259.5  Z = 16.11 W   d)     I = V/Z = 30 ÷ 16.11 = 1.86 A   e)      tan f = 5.87 ÷ 15 = 0.391 f = tan-1 0.391 = 21.4o   f)       Impedance across the coil: Z2 = 152 + 37.702 = 1646 Þ Z = 40.6W V = IZ = 1.86 × 40.6 = 75.5 V   g)      Voltage across the capacitor = IXC = 1.86 × 31.83 = 59.2 V In real inductive components, the VL vector is never quite 90o to the VR resistor. This is because the coil has a definite resistance, so there is a phase difference that is less than 90o.

This is less of a problem with a non-electrolytic capacitor, as a capacitor has an insulating layer.  If it stops insulating, it’s not much good as a capacitor.

An electrolytic capacitor has a certain leakage current, which can be modelled as a resistor.

 Question 9 Draw the phasor diagram for the worked example. Question 10 The following four systems are complex devices which are represented as black boxes.  The wiring diagram shows them in a series circuit. Tests are carried out in each of the systems.  The conclusions are that: ·         W has an internal resistance of 10 W only; ·         X has an inductance of 230 mH and a resistance of 15 W; ·         Y has an inductance of 150 mH, a capacitance of 470 mF and a resistance of 25 W. ·         Z has a capacitance of 560 mF and a resistance of 8 W. (a) Redraw the circuit as a simple RLC circuit. (b) Work out the impedance of the whole circuit. (c) Work out the current. (d) Work out the phase angle between the current and the resulting voltage. (e) Work out the phase angle between the inductive voltages and the current (f) Sketch the phasor diagram for resistances, reactances, and impedance with values. 