Download Lecture 07 Load Power and Internal Resistance

1E6 Electrical Engineering
DC Circuit Analysis
Lecture 7: Load Power and Internal Resistance
6.1 Introduction
Even in the simplest of electric circuits the ultimate aim is usually to
deliver power to a load. In the circuits we have examined to date, the load has
been modelled as a pure resistance, RL. However, the load is not always a real
resistor but the resistor is used to represent any load, operation or process which
used up electrical energy and therefore power. Indeed, the simplified electric
circuits can themselves be models for power sources, instruments or equipment
which provides electrical energy to a load. If this is the case, then the factors
which influence the transfer of power to the load from the source should be
examined to ensure the optimum use of the energy consumed.
6.2 Maximum Power Transfer
The circuit shown in Fig. 1 can be considered as a model representing a
non-ideal source which has an internal resistance, RS, driving a load resistance,
RL. The power dissipated in the load is given as:
PL  VL I L
where VL where is the voltage appearing across the load and IL is the current
flowing through it.
power
source
IL
RS
VL
E
Fig. 1
RL
A Power Source Driving a Resistive Load
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The voltage across the load is essentially obtained from the potential divider
action of RS and RL which gives:
VL 
RL
E
RS  R L
The current through the load is essentially obtained from the source emf divided
by the sum of the two resistors in series.
E
IL 
RS  RL
Then:
ER L
E 2R L
E
PL 

R S  R L R S  R L R S  R L 2
It can be seen that if the voltage source were ideal with R S → 0, then the power
would become:
PL 
E2R L
RL
2
E2

RL
In this event, the full emf, E, is developed across the load resistance, R L, and the
power delivered to the load depends only the value of this emf and the value of
the load resistance. This means that all of the power drawn from the source is
dissipated in the load resistance, RL.
On the other hand, when the source is non-ideal and RS is finite, some power will
be dissipated in this resistance and is therefore essentially ‘lost’ or ‘wasted’. The
remainder of the power is delivered to the load as:
PL 
E 2R L
R S  R L 2
Usually, the internal resistance of the source or the equivalent model
representing the source is determined by factors beyond control and so cannot
always be adjusted to suit requirements. However, the resistance of the load can
often be controlled by design, at least to some degree.
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In this case, the aim is to try to optimise the load resistance, RL, so that
maximum power is transferred from the source to the load. This can be
determined by finding the derivative of the expression for the load power with
respect to the load resistance, RL and equating it to zero. Then if:
PL 
E 2R L
R S  R L 2
The quotient rule must be used:
Let
u  E2R L
with the quotient rule
so that:
v  R S  R L 
2
and
PL

R L
v
u
v
u
R L
R L
v2
PL R S  R L  E 2  2E 2 R L R S  R L 

0
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R L
R S  R L 
2
This requires:
R S  R L 2 E 2  2E 2 R L R S  R L   0
or:
So that:
R S  R L  2R L
R L  RS
Thus it can be seen that maximum power will be transferred from the source to
the load when the resistance of the load is made equal to the internal resistance
of the source.
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Note that under these conditions, the actual power delivered to the load is given
as:
PL 
E2R L
R S  R L 2
E2R L
E2


2
4R L
4R S
Recall from above that when RS = 0 the power delivered to the load is:
E2
PL 
RL
This indicates that when RS ≠ 0, but RL = RS the maximum power delivered to
the load is only one quarter of that obtained from an ideal source. This is
indicated in Fig. 2 which shows a plot of the load power as a function of the ratio
of the load and internal resistances in question. It can be seen from Fig. 2 that
when the internal resistance is not zero it has a profound influence on the power
delivered to the load.
PL
E2/RS
power drawn
from source
power delivered
to load
E2/2RL
E2/4RL
0
1
RL / Rs
Fig. 2 Power Drawn from Source and Delivered to Load
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When RL / RS → 0 which implies that RL → 0, maximum current will flow
through the load being limited only by the internal resistance RS. However, with
RL → 0 no voltage is developed across it and so no power is delivered to the load.
As the load resistance, RL, is increased the current flowing through it, IL,
decreases while the voltage across it, VL, increases. The maximum product of
voltage and current occurs as determined above when RL = RS, shown as the
peak in the curve of Fig. 2 illustrating the power delivered to the load. It is also
the case, under these conditions, that the power drawn from the source is only
half that obtained from the equivalent ideal source and that only half of this is
delivered to the load. The other half is dissipated in the internal resistance, RS,
of the non-ideal source, since it has the same value as the load resistance, RS.
When the load resistance is increased further, the current flowing through it
decreases even though the voltage developed across it increases. The product of
voltage and current decreases, resulting in a decrease in the power delivered to
the load. As the load resistance RL → ∞, the current flowing through it falls to
zero so that the power delivered also falls to zero. This gives the bell-shaped
curve for load power shown in Fig. 2.
6.2 Internal Resistance
The above analysis highlights the importance of the effect of the internal
resistance of a non-ideal voltage or current source on power transfer to the load.
Given this, it is extremely useful to have a means of measuring the internal
resistance of a source so that its value can be established and controlled if
possible. Means used on paper in the analyses of circuits studied so far, such as
using open-circuit or short-circuit load conditions, cannot always be applied as
real electric circuits or systems may not tolerate such conditions without
malfuction or damage. In this case a differential approach can be used to
measure the internal resistance.
power
source
IL
RS
VL
E
RL1 → RL2
Fig. 3 Measurement of Internal Source Resistance.
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Consider again the non-ideal voltage source having an internal resistance,
RS. A load resistance, RL is connected to the terminals, the value of which can be
changed. Consider the scenario where two different values of load resistance,
RL1 and RL2, are connected to the source in turn as indicated in Fig. 3.
With the value RL1 connected:
I L1 
E
ERL1
and VL1  I L1 RL1 
RS  RL1
RS  RL1
With the value RL2 connected:
I L2
E
ERL 2

and VL 2  I L 2 RL 2 
RS  RL 2
RS  RL 2
Taking the ratio:
R  RL 2 RL1 RS  RL 2 
VL1
ERL1

 S

VL 2 RS  RL1
ERL 2
RL 2 RS  RL1 
Cross multiplying and simplifying:
VL1RL 2 RS  RL1   VL 2 RL1 RS  RL 2 
VL1RS RL 2  VL1RL1RL 2  VL 2 RS RL1  VL 2 RL1RL 2
VL1RS RL 2  VL 2 RS RL1  VL 2 RL1RL 2  VL1RL1RL 2
RS VL1RL 2  VL 2 RL1   VL 2  VL1 RL1RL 2
Then:
RS

VL 2  VL1 RL1 RL 2

VL1RL 2  VL 2 RL1 
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Dividing by RL1RL2:
RS 
VL 2  VL1 
 VL1 VL 2 



 RL1 RL 2 
So that:

VL 2  VL1  VL
RS 

I L1  I L 2  I L
Note that if RL2 > RL1 then VL2 > VL1 and IL2 < IL1 so that RS is positive.
It should also be noted that knowledge of the two values of resistance is not
required, only the change in the output voltage and current which they produce.
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