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LR SERIES CIRCUIT
An LR series circuit is a circuit which consists of an inductance, L, in series with a
resistance, R. Such an arrangement is a more realistic representation of a practical inductor. It
is impossible to wind an inductor which has no resistance. The wire used for the coil will
have some resistance. Although each turn and part of a turn has both resistance and
inductance it is convenient to separate the total inductance and the total resistance for the
purposes of circuit analysis.
In Figure F.1you will see such an LR series circuit connected to an ac supply.
i
VL
VS
VR
Figure F.1
When the above circuit is supplied with an alternating voltage as shown, a voltage is
developed across each component.
vs = vR + vL
The voltages in this equation are the instantaneous values. You could draw the sine waves
and add them instant by instant.
In a previous section you have seen that the voltage across an inductor leads the current by
90o. The current in a series circuit is the same in every component. We can use the current
phasor as reference and relate the voltages across the inductance and the resistance to it. The
supply voltage will be the phasor sum of the other two voltages. This relationship is:
VS = VR + VL
If we assume that i = IM sin wt we can draw the current phasor horizontally. You can see the
whole phasor diagram in Figure F1 below.
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As you can see in Figure F.1 VR is in phase with I and VL leads I by 90o. The resultant Vs
leads I by angle 90o.
Applying Pythagoras' theorem to the right-angled triangle we have:
V2
VR2+VL2
For the resistor
VR = IR
and for the inductance
VL = IXL
substituting we get
VS2 = (IR)2 + (IXL)2
= I2 (R2 + XL2)
VS2 = R2 + XL2
I2
Taking square roots VS = R2 + XL2
I
The expression on the right-hand side of this last equation represents the total opposition to
the flow of current. It is made up of the opposition due to the resistance and the opposition
due to the Reactance.
Impedance
The total opposition due to both the resistance and the inductance is known as the impedance
of the circuit. The symbol for the impedance is Z and it has units of ohms, .
Thus
V = Z ohms
I
and
Z = R2 +XL2
If you look at Figure F.l and apply a little trigonometry to the phase angle  you have:
sin  = VL = IXL
Vs
IZ
cos  = VR = IR
VS IZ
= R
Z
tan  = VL = IXL
VR IR
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XL = L
R
R
PROGRESS QUESTION F
A voltage, VS = 200 sin 100 t volts, is applied to a coil with an inductance of 127 mH and a
resistance of 30 .
Determine the expression for the instantaneous current.
Draw the phasor diagram for this circuit.
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CR SERIES CIRCUIT
The CR series circuit consists of a capacitance C in series with a resistance R. A good
capacitor has practically negligible series resistance: however, there are many applications in
which a CR series combination is needed. You can see such a combination in Figure F.2.
VC
VS
VR
Figure F.2
The voltages and the current shown in Figure F.2 are rms values. Kirchhoff's voltage law
applies if we use instantaneous values, then:
VS = VR + VC
However, the addition is a vector addition for the rms or peak values. You can see the phasor
diagram in Figure F.3.
Figure F.3
Again the current phasor is taken as reference since the same current flows throughout the
circuit. As you have seen, the voltage across the resistance is in phase with the current. The
voltage across the capacitance lags the current by 90O. The vector sum of VC and VR gives
the supply voltage Vs. As you can see, VS lags the current by an angle 
Applying Pythagoras' theorem to the right-angled triangle gives:
VS2 = VR2 + VC2
For the resistor
VR = I R
and for the capacitor
VC = I Xc
substituting we get:
Vs2 = (IR)2 + (1XC)2
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= I2 (R2 + XL)2
VS2
I2
= R2 + XC2
Taking square roots VS =  R2 + XC2
I
The expression on the right-hand side of this last equation represents the total opposition to
the flow of the current. It is the impedance Z.
and
Z = R2 +XC2
If you look at Figure F.3 and apply a little trigonometry to the phase angle  you have:
sin  = VC = IXC
Vs
IZ
cos  = VR = IR
VS IZ
= R
Z
tan  = VC = IXC
VR IR
XC = 1
R
CR
If you look at Figure F.18 and apply a little trigonometry to the phase angle  you have:
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PROGRESS QUESTION F
A voltage, Vs= 100 sin 200t volts, is applied to a circuit consisting of a resistance of 100k
n series with a capacitance of 0.1 F.
Draw the phasor diagram for this circuit and derive an expression for the instantaneous
current.
What are the values of the rms current and voltage?
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RLC SERIES CIRCUITS
INTRODUCTION
In this section we continue our study of series circuits. This time, however, we shall look at
circuits which contain all three types of components, that is, resistance, inductance and
capacitance.
You will learn to construct phasor diagrams which embody the ideas you have already met.
You will also find out how to calculate current and voltages and phase angles for these
circuits.
You will see that such RLC series circuits can behave inductively, capacitively or resistively
according to the values of the reactances.
You will meet the very special condition called resonance when the circuit behaves as a
resistance and the supply voltage and current are in phase. This condition gives minimum
circuit impedance and maximum current.
THE GENERAL SERIES CIRCUIT
The general series circuit contains all three possible circuit elements, R, L and C. We will
look at the behaviour of this type of circuit when connected to an ac supply. You can see such
a circuit in Figure F.25.
VL
VS
VC
VR
Figure F.25
The phasor diagram can be constructed in a similar manner to that used in Section F3 with
the other series circuits. You can see the phasor diagram in Figure F.26.
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Phasor Diagram
There are three possibilities with this circuit either:
VL  VC
VL VC
VL  VC
First Case: VL > VC
When VL is greater than Vc, the phasor diagram is as shown in Figure F.27.
Figure F.27
As you can see in Figure F.27, the current lags the supply voltage by the angle. The circuit is
said to be inductive since the phase relationship between V5 and I is the same as we found in
Section F3 for an LR series circuit.
From the full phasor diagram shown in Figure F.27 we can derive the voltage triangle. Then,
since the current is a common factor in all the voltages, we can cancel this out and find the
impedance triangle. The triangles are shown in Figure F.28.
Figure F.28
The net reactance X is that of an inductance. The value of the resultant inductive reactance is
less than that of the actual inductance because the actual inductive reactance has been
partially cancelled by the capacitive reactance.
From Figure F.28:
Vs2= Vr2 + Vx2
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The phase angle may be found from:
tan = VL – Vc
Vr
= XL- XC
R
The other trigonometrical functions are:
sin  = VL – VC
VS
X L – XC
z
cos  = VR
VS
=R
Z
Check that you have mastered this method by doing Progress Question F.7. You will
probably need two or three sheets of paper for the answer.
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PROGRESS QUESTION F.7
A supply voltage VB produces a current i = 5 sin l00t A when applied to a series circuit
consisting of a 10 F capacitor, an inductor of 3.185 H and a resistance of 100 
Determine the expression for the instantaneous supply voltage and draw the phasor diagram.
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Second Case: VL < VC
When VL is less than V0, the phasor diagram is as shown in
Figure F.29.
Figure F.29
In this case the current leads the supply voltage by an angle ~ and the circuit is capacitive.
The net reactance is capacitive and the circuit behaves as if it contained only capacitance and
resistance in series.
From Figure F.29:
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Third Case VL = VC
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PROGRESS QUESTION F.9
A resistor R = 200  a capacitor C = 0.5 F and an inductor L = 0.4 H are connected in series
across a 20 V, ac supply of frequency 1.59 kHz.
Draw the phasor diagram for the circuit and compare VC with VS. How much greater are VL
and VC than the applied voltage?
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