Download Analysis of a single-loop circuit using the KVL method

Analysis of a single-loop circuit using the KVL method
Figure 1 is our circuit to analyze. We shall attempt to determine the current through each element,
the voltage across each element, and the power delivered to or absorbed by each element. You
will note that the KVL method determines the unknown current in the loop by using a sum of
voltages in the loop.
R1
30
V2
30V
R1
30
R2
15
V1
120V
V1
120V
I
Figure 1: Single loopV2
circuit.
R1
30V
30
The first step in the analysis is to assume a reference direction for the unknown current. We
do
I=2A
not know apriori what the direction is, nor
does
it
matter.
If
our
assumption
is
wrong,
the
current
Vr1
will simply have a minus sign associated with it. The circuit below is shown with
120Van arbitrary
R2
direction for the current I. V1
Vr2
15
120V
I
voltage source with
reference current
V2
V2
R1
R1
start
here
30V
30V
direction30
as initially
30
assumed
120V
voltage s
V2
referenc
30V
computin
Vr1
R2
15
V1
120V
R1
30
I
I
R1
30
Vr1
I
R2
15
V1
120V
Vr2
I
I
Veq
Req
90
Figure
2:
Single
loop
circuit
with
I
direction.
15
V2
V2
30V
30V
The second step in the analysis is to assign the
voltage references (where
for each element
I=2A
I =needed)
−2A
I
in the circuit. We already know that the passive sign convention for resistors demands that the
sense of the current and voltage be120V
selected such that the current
enters the more positive terminal.
120V
V
R2
The choice of direction
in which we draw the current arrow is arbitrary, but once the direction is chosen, the
Vr2
15
I
withvoltages across the voltage sources and their
polarity
orientation across the resistorsvoltage
is then source
fixed. The
voltage source with current
polarities are taken as given and cannot
be current
altered as they are independent sources. Therefore,
reference
model for computing
reference arrow placed for
voltages Vr 1 and Vr 2 in this loop must
be assigned
as shown in figure 4.
direction
as initially
power dissipated
computing power dissipation
assumed
1
R1
30
R2
15
start here
V1
120V
e
V
V2
30V
R1
30
R2
15
I
q
V2
30V
R1
30
V2
30V
R1
30
Vr1
I
R2
15
V1
120V
Vr2
I
R2
15
start here
Figure 3: Single loop circuit, with V signs.
Next, we apply Kirchhoffs
voltage law to the single closed path. We may sum the voltages by
I=2A
I = −2A
I
traversing the circuit in either direction, but let’s do so in the clockwise direction, beginning at
the lower left120V
corner, (where it says start here)
120V
V and write down each voltage first encountered
at its positive reference and write down the negative of every voltage encountered at its negative
voltage source
with
terminal.
Thus we get,
voltage source with current
−120 +model
Vr1 +for
30computing
+ Vr2 = 0 .
reference current
reference arrow placed for
direction as initially
power
This is a single
equation
with
two unknowns.
Todissipated
solve this equation, we must find some way to
computing
power
dissipation
assumed
express Vr1 and Vr2 in one variable. Since this is a series circuit and the same current flows through
all elements, we may express Vr1 and Vr2 in terms of the current I and the individual resistances
using Ohms law. So we apply the Ohm’s law substitution step.
We know that for R1 and R2 , Ohms law states that:
I
Vr1 = I × 30, and Vr2 = I × 15 .
If we substitute these expressions for Vr 1 and Vr 2 into the first equation, we get:
Req
15
−120 + 30I + 30 + 15I = 0 .
Solving for I, we obtain:
45I = 90, thus I = 2A.
Since the solution for current is a positive value, we know that the assumed direction for current
is correct. If the answer had been -2A, we would know that the current would be actually flowing
the opposite direction.
Before we proceed to compute the power for the components in this circuit, let’s repeat the problem but reverse the assumed direction of current flow. Therefore we have the situation shown in
figure 4.
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DRAWN BY:
R2
15
V1
120V
V1
120V
I
V2
30V
R1
30
I=2A
Vr1
V1
120V
Vr2
I
R2
15
start here
120V
voltage source with
reference current
direction as initially
assumed
Figure 4: Single loop circuit, with V signs.
R1
R2
30
I traversing
We will now write the KVL equation by
the 15
circuit in the clockwise direction as we did
I write
before. Again we write down each voltage first encountered at its positive reference and
down the negative of every voltage
encountered at its negative terminal. Thus we get,
V1
120V
−120 − Vr1 + 30 − Vr2 = 0 .
Veq
Req
90
15
Applying the Ohms law substitution
step we get:
V2
30V
−120 − 30I + 30 − 15I = 0 .
Solving for I, we obtain:
−45I = 90, thus I = −2A.
Since the resulting current is negative, we know that the assumed direction for current is incorrect.
However, our answer is correct considering the direction of the arrow.
To compute the power dissipation, we will assume that a positive current was obtained as in the
first example. Knowing the equation for power dissipation; P = I 2 R, we can calculate the power
dissipated by the resistors.
For the 30Ω resistor: P = 2 × 2 × 30 = 120 Watts.
For the 15Ω resistor: P = 2 × 2 × 15 = 60 Watts.
Resistors always dissipate positive power. They can never generate power.
How much power is generated or consumed by the voltage sources? Remember to compute
power dissipated, we must take each element and redefine the voltage reference sign and the
current reference arrow so that the arrow points into the positive terminal of the component such
that it looks like the model for computing power dissipated shown below.
In figure 5 we see the progression of taking the 120V source and transforming its current arrow so
that the arrow points into the positive terminal just like the model for computing power.
3
120V
voltage
referen
comput
15
120V
15
I
120V
15
I
start here
V2
30V
R2
15
I=2A
Vr2
R2
15
I = −2A
I
120V
120V
voltage source with
reference current
direction as initially
assumed
V
voltage source with current
reference arrow placed for
computing power dissipation
model for computing
power dissipated
Figure 5: Orient the voltage source to match the PSC model
Note that the current
I arrow for the 120V source is now drawn as the passive sign convention
demands; it points into the positive terminal. The voltage references are identical to the model so
they were not changed. Now we can compute the power dissipation directly as:
Veq
Req
90
15 P120V = −2 × 120 = −240 Watts.
The negative power dissipation indicates that the source is actually delivering power, not dissipating it. For the 30V source, again the current arrow is drawn as the passive sign convention
demands. However in this case, the current into the source is in the same direction as the model
dictates, thus:
P30V = 2 × 30 = 60 Watts.
In this case, the 30 volt voltage source is dissipating power.
A useful check of your calculations may be accomplished by using the power check. This method
relies on the fact of the conservation of energy. In other words, the power dissipated by all the
elements must equal the power supplied by all the elements. In our example:
P120V + PR1 + P30V + PR2
= 0
−240 + 120 + 60 + 60 = 0
0 = 0.
Thus, our calculations are correct.
To summarize the KVL method:
1. Assign a reference direction for the unknown current
2. Assign voltage references to the elements
3. Apply KVL to the closed loop path
4. Substitute in Ohms law where needed to get an equation in I
5. Solve for the current I
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Remember that:
1. The assumed direction for the current flowing in the loop does not matter. If the assumed
direction is backwards to the actual flow, the magnitude will simply be negative.
2. The polarity across the resistors must conform to the passive sign convention.
3. The direction in which you sum voltages does not matter. Neither does it matter where in
the loop you start the summation process.
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