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**Unformatted text preview: **150 IN1 SERIES de CIRCUITS
VOLTAGE DIVISION IN A SERIES CIRCUIT III 151
For path 2, starting at point a in a clockwise direction,
- V2 - 20 V = 0
and
V2 = - 20 V
Solutions :
Applying Kirchhoff's voltage law in the clockwise direction start-
The minus sign in the solution simply indicates that the actual pola
ties are different from those assumed.
ing at the negative terminal of the supply results in
- E+ V3 + V 2 + V 1 = 0
E = VI + V2 + V3 (as expected)
The next example demonstrates that you do not need to know why
off and
so that V2 = E - V 1 - V3 = 54 V - 18 V - 15V
elements are inside a container when applying Kirchhoff's voltage la
V2 = 21 V
They could all be voltage sources or a mix of sources and resistors.
and
doesn't matter-simply pay strict attention to the polarities encountere
V2 _ 21 V
Try to find the unknown quantities in the next examples witho
R 2
trouble.
looking at the solutions. It will help define where you may be having
12 = 3A
V1 18 V
- = 60
40 V
Example 5.1 1 emphasizes the fact that when you are applying Kirch
C, RI =
3A
hoff's voltage law, the polarities of the voltage rise or drop are the im
1 = 50
portant parameters, not the type of element involved.
with R3 =
V3 _ 15 V
13
3 A
50 X
EXAMPLE 5.11 Using Kirchhoff's voltage law, determine the un
EXAMPLE 5.14 Using Kirchhoff's voltage law and Fig, 5.12, verify
known voltage for the circuit in Fig. 5.30.
Ed. ( 5.1 ).
+ 30 V -
Solution: Note that in this circuit, there are various polarities across
Solution: Applying Kirchhoff's voltage law around the closed path:
FIG. 5.30
the unknown elements since they can contain any mixture of compo.
E = VI+ V2+ V3
Series configuration to be examined
results in
ments. Applying Kirchhoff's voltage law in the clockwise direction
in Example 5.11.
Substituting Ohm's law:
+60 V - 40 V - Vx+ 30V=0
I, RT = 1R1 + 12 R2 + 13 R3
and
V x = 60 V + 30 V-40 V =90V-40 V
but
Is = 1 1 = 12= 13
with
V , = 50 V
so that
I, RT = 1, ( R1 + R2 + R3 )
and
RT = R1+ R2 + R3
EXAMPLE 5.12 Determine the voltage Vx for the circuit in Fig. 5.31.
which is Eq. (5.1).
Note that the polarity of V, was not provided .
+ 2V -
Solution: For cases where the polarity is not included, simply make an
5.7 VOLTAGE DIVISION IN A SERIES CIRCUIT
FIG. 5.31
assumption about the polarity, and apply Kirchhoff's voltage law as be-
Applying Kirchhoff's voltage law to a circuit in
fore. If the result has a positive sign, the assumed polarity was correct. If
The previous section demonstrated that the sum of the voltages
which the polarities have not been provided for one
the result has a minus sign, the magnitude is correct, but the assumed
across the resistors of a series circuit will always equal the applied
of the voltages ( Example 5.12).
polarity must be reversed. In this case, if we assume point a to be posi-
voltage. It cannot be more or less than that value. The next question is, How
tive and point b to be negative, an application of Kirchhoff's voltage law
will a resistor's value affect the voltage across the resistor? It turns out that
n the clockwise direction results in
the voltage across series resistive elements will divide as the magnitude
V3 = 15 V
-6 V -14 V - Vx+ 2V=0
of the resistance levels.
and
1360 12 V
so that
V x = - 20 V + 2V
In other words ,
54V
V x = - 18 V
in a series resistive circuit, the larger the resistance, the more of the
2 70 V2
E =20 V
2 2 < 3 0 6V
Since the result is negative, we know that point a should be negative and
applied voltage it will capture.
point b should be positive, but the magnitude of 18 V is correct.
In addition ,
R, $ 10 2V
VI = 18 V
the ratio of the voltages across series resistors will be the same as the
EXAMPLE 5.13 For the series circuit in Fig. 5.32.
ratio of their resistance levels.
FIG. 5.32
Series configuration to be examined in
. Determine V2 using Kirchhoff's voltage law.
All of the above statements can best be described by a few simple ex-
Example 5.13.
b. Determine current 12.
amples. In Fig. 5.33, all the voltages across the resistive elements are pro-
vided. The largest resistor of 6 0 captures the bulk of the applied voltage,
FIG. 5.33
c. Find R, and R3.
while the smallest resistor, R3, has the least. In addition, note that since the
Revealing how the voltage will divide across series
Introductory, C.- 118
resistance level of R, is six times that of R3, the voltage across R, is six
resistive elements....

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- Fall '19
- SK Abid