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Lecture: Load Power and Internal Resistance


Introduction:

Even in the easiest of electric circuits the ultimate goal is generally to deliver power to a load. In
circuits we have observed to date, the load has been modelled as a pure resistance, RL. Though, the
load is not for eternity a real resistor however the resistor is employed to symbolize any load,
operation or process that used up electrical energy and thus power. In reality, the simplified electric
circuits can themselves be models for power sources, equipment or instruments that give electrical
energy to a load. If this is the situation, then the factors that influence the transfer of power to the
load from the source must be examined to make sure the optimum use of the energy consumed.


Maximum Power Transfer:

The circuit shown in figure below can be considered as a model symbolizing a non-ideal source that
has an internal resistance, RS, driving a load resistance, RL. The power dissipated in the load is
provided as:

PL = VLIL

Where VL is the voltage appearing across load and IL is the current flowing via it.




Figure: Power Source driving a Resistive Load

The voltage across the load is basically obtained from the potential divider action of RS and RL that
offers:

VL = [(RL)/(RS +RL)]E


The current via the load is essentially received from the source emf divided by the sum of two
resistors in series.
E
VL
RS
IL
RL

Power
source


2



IL = E/(RS + RL)

Then:
(
)2
L
S
L
2
L
S
L
S
L
L
R
R
R
E
R
R
E
R
R
ER
P
+
=
+
+
=


This can be seen that if the voltage source is ideal with RS → 0, then the power would be:

L
2
2
L
L
2
L
R
E
R
R
E
P
=
=


In this event, the full emf, E, is build up across the load resistance, RL, and power delivered to the
load depends only the value of this emf and the value of load resistance. This signifies that all the
power drawn from source is dissipated in the load resistance, RL.

On other hand, whenever the source is non-ideal and RS is finite, some power will be dissipated in
this resistance and is thus essentially ‘wasted’ or ‘lost’. The remainder of power is delivered to the
load as:

PL = (E
2RL)/(RS +RL)
2

Generally, the internal resistance of the source or equivalent model symbolizing the source is
determined by the factors beyond control and therefore can’t always be adjusted to suit
requirements. Though, the resistance of the load can frequently be controlled by design, at least to
certain degree.

In this case, the goal is to try to optimise the load resistance, RL, and hence maximum power is
transferred from the source to load. This can be found out by finding the derivative of the
expression for load power in respect to the load resistance, RL and equating it to zero. Then if:

PL = (E
2RL)/(RS + RL)
2

The quotient rule should be employed:

Let u = E2RL and v = (RS + RL)
2




3





Or:

RS + RL = 2RL

So that: RL =RS

Therefore it can be seen that maximum power will be transferred from source to the load if the
resistance of the load is made equivalent to the internal resistance of the source.
Note that under such conditions, the real power delivered to load is given as:

PL = (E
2RL)/(RS + RL)
2 = (E2RL)/4RS
2 = E2/4RL

Remember from above that when RS = 0 then the power delivered to load is:

PL = E
2/RL

This points out that if RS ≠ 0, however RL = RS then the maximum power delivered to load is only
one quarter of that received from an ideal source. This is indicated in figure below which exhibits a
plot of the load power as a function of ratio of the load and internal resistances in question. This can
be seen from figure below that whenever the internal resistance is not zero it has a profound
persuade on the power delivered to load.


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Figure: Power Drawn from Source and Delivered to Load


When RL/RS → 0 that implies that RL → 0, maximum current will flow via the load being limited
merely by the internal resistance RS. Though, with RL → 0 no voltage is developed across it and
therefore no power are delivered to load. As the load resistance, RL is raised the current flowing via
it, IL, reduces while the voltage across it, VL, rises. The maximum product of current and voltage
takes place as determined above if RL = RS, shown as the peak in curve of figure above illustrating
the power delivered to load. This is also the case, under such conditions, that the power drawn from
the source is only half which obtained from the equivalent ideal source and that merely half of this
are delivered to the load. The other half is dissipated in internal resistance, RS, of the non-ideal
source, as it has similar value as the load resistance, RS. Whenever the load resistance is raised
further, the current flowing via it reduces even though the voltage developed across it rises. The
product of current and voltage reduces resultant in a reduction in the power delivered to the load.
Since the load resistance RL → ∞, the current flowing via it falls to zero and hence the power
delivered as well falls to zero. This provides the bell-shaped curve for load power shown in figure
above.


Internal Resistance:

The above analysis emphasizes the significance of the result of the internal resistance of a non-ideal
current or voltage source on power transfer to the load.

Given this, it is very helpful to have a means of measuring the internal resistance of a source and
hence its value can be established and controlled if possible. Means employed on paper in the
analyses of circuits studied so far, like using open-circuit or short-circuit load conditions, can’t
always be applied as the real electric circuits or systems might not tolerate like conditions without
RL / Rs
1
E2/RS
PL
E2/2RL
E2/4RL
Power drawn
from source
0
Power delivered
to load


5


damage or malfuction. In this case a distinct approach can be employed to compute the internal
resistance.




Figure: Measurement of Internal Source Resistance

Consider another time the non-ideal voltage source containing an internal resistance, RS. A load
resistance, RL is joined to the terminals, the value of which can be modified. Consider the scenario
where two distinct values of load resistance, RL1 and RL2, are joined to the source in turn as
indicated in figure above.

With value RL1 joined:

IL1 = E/(RS + RL1) and VL1 = IL1RL1 = [(ERL1)/(RS + RL1)]

With value RL2 joined:

IL2 = E/(RS + RL2) and VL2 = IL2RL2 = [(ERL2)/(RS + RL2)]

Taking the ratio:

(
)
(
)
1
2
2
1
2
2
1
1
2
1
L
S
L
L
S
L
L
L
S
L
S
L
L
L
R
R
R
R
R
R
ER
R
R
R
R
ER
V
V
+
+
=
+

+
=


By cross multiplying and making simpler:


E
VL
RS
IL
RL1 → RL2

Power
source

https://www.expertsminds.com/


https://www.tutorsglobe.com/



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So that:

RS = [(VL2 –VL1)/(IL1 – IL2)] = Δ VL/Δ IL

Note that when RL2 > RL1 then VL2 > VL1 and IL2 < IL1 and hence RS is positive.

It must as well be noted that the knowledge of two values of resistance is not needed, only the
change in output current and voltage which they generate.

Reference:

https://www.expertsminds.com/

https://www.tutorsglobe.com/