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ELECTRICITY & MAGNETISM (Fall 2011)
LECTURE # 11
BY
MOEEN GHIYAS
(Series Circuit – Chapter 5)
Introductory Circuit Analysis by Boylested (10th Edition)
TODAY’S LESSON
Today’s Lesson Contents
• Kirchhoff’s Voltage Law (KVL)
• Interchanging Series Elements
• Voltage Divider Rule
• Notation
Kirchhoff’s Voltage Law
• Kirchhoff’s voltage law (KVL) states
that the algebraic sum of the
potential rises and drops around a
closed loop (or path) is zero.
• A closed loop is any continuous
path that leaves a point in one
direction and returns to that same
point from another direction without
leaving the circuit.
Kirchhoff’s Voltage Law
• To apply Kirchhoff’s voltage law, the
summation of potential rises and
drops must be made in one
direction (whether clockwise or
anticlockwise) around the closed
loop.
• In symbolic form, where Σ
represents summation, the closed
loop
, and V the potential drops
and rises, we have
Kirchhoff’s Voltage Law
•
Reveals that, the applied
voltage of a series circuit equals the
sum of the voltage drops across the
series elements
• For clockwise direction,
• For anticlockwise direction,
Kirchhoff’s Voltage Law
• The application of Kirchhoff’s voltage law need not
follow a path that includes current-carrying elements.
• Using the clockwise direction
Kirchhoff’s Voltage Law
• Example – Determine the unknown voltages for the
network of fig.
• Solution:
• For clockwise direction,
Kirchhoff’s Voltage Law
• Example – Determine the unknown voltages for the
network of fig.
• Solution: Two possible loops; using clockwise direction,
• Loop 1 – including source E,
Kirchhoff’s Voltage Law
• Example – Determine the unknown voltages for the
network of fig.
• Solution: Two possible loops; using clockwise direction,
• Loop 2 – including R2 and R3
Kirchhoff’s Voltage Law (KVL)
• Example – Using Kirchhoff’s voltage law, determine
the unknown voltages for the network of fig.
• Solution:
Kirchhoff’s Voltage Law (KVL)
• Example – Using Kirchhoff’s voltage law, determine
the unknown voltages for the network of fig.
• Note the polarity of the unknown voltage is not
provided
• Solution:
Since the result is negative, we
know that a should be negative
and b should be positive, but the
magnitude of 18 V is correct.
Interchanging Series Elements
• The elements of a series circuit can be interchanged
without affecting the total resistance, current, or power
to each element. For instance, the network of figs.
– RT = 35 Ω, and I =70V / 35Ω = 2 A in both cases
– V2 = IR2 = (2 A)(5 ) = 10 V for both configurations
Interchanging Series Elements
• Example – Determine I and the voltage across the 7Ω resistor
for the network of Fig.
• .
Solution:
Voltage Divider Rule
• In a series circuit, the voltage
across resistive elements will divide
as the magnitude of the resistance
levels.
• The largest resistor R1 = 6Ω
captures the bulk of the applied
voltage, while the smallest resistor
R3 = 1Ω has the least.
Voltage Divider Rule
• In simple words, the voltage across
series resistors will have the same
ratio as their resistance levels.
• Note:
Since resistance level of
R1 is 6 times that of R3, the voltage
across R1 is 6 times that of R3.
• Also R2 / R3 = 3, so does V2 / V3 = 3
• Finally, since R1 is twice R2, the
voltage across R1 is twice that of R2.
Voltage Divider Rule
• Note that, if the resistance levels of all the resistors of
fig (left) are increased by the same amount, as shown
in fig (right), the voltage levels will all remain the same.
Voltage Divider Rule
• Therefore, it is the ratio of resistor values that counts
when it comes to voltage division and not the
magnitude of the resistors.
Voltage Divider Rule
• The current level of the network will be severely
affected by the change in resistance level from fig (left)
to fig (right), but the voltage levels will remain the same.
Voltage Divider Rule - Example
• Note here:
• 1 MΩ = (1000)1 kΩ = (10,000)100Ω
revealing that
• V1 = 1000V2 = 10,000V3.
• Check:
Voltage Divider Rule
• In the last slide / discussion the current was
determined before the voltages of the network
were determined
• There is, however, a method referred to as the
voltage divider rule (VDR) that permits
determining the voltage levels without first
finding the current
Voltage Divider Rule
• The rule can be derived by analyzing
the network of fig.
• Applying Ohm’s Law,
Voltage Divider Rule
• The voltage divider rule states that the voltage across a
resistor in a series circuit is equal to the value of that
resistor times the total impressed voltage across the
series elements divided by the total resistance of the
series elements.
• where Vx is the voltage across Rx, E is the impressed
voltage across the series elements, and RT is the total
resistance of the series circuit.
Voltage Divider Rule
• Example – Determine the
voltage V1 for the network.
• Solution:
• We know
• or
Voltage Divider Rule
• Example – Determine the voltage V′ in
fig across resistors R1 and R2.
• Solution:
• We know
• or
Voltage Divider Rule
• There is also no need for the voltage E in the equation
to be the source voltage of the network.
• For example, if V is the total voltage across a number
of series elements such as those shown in above fig,
then
Voltage Divider Rule
• Example – Design the voltage divider of fig such that
VR1 = 4VR2.
• Solution: The total resistance is
• Then
• .
Therefore
• .
• .
We have
and
Notation – Voltage Sources and Ground
• Three ways to sketch the same series dc circuit
• If two grounds exist in a circuit and no connection
is shown between them, even then such a
connection exists for the continuous flow of charge.
Notation – Voltage Sources and Ground
• On large schematics
where space is at a
premium and clarity is
important, voltage
sources may be
indicated as in fig (a)
rather than as
illustrated in fig (b)
Notation – Voltage Sources and Ground
• On large schematics where space is at a premium and
clarity is important, voltage sources may be indicated
as fig (a) rather than as illustrated in fig (b)
Notation – Voltage Sources and Ground
• In schematics, the potential
levels may also be indicated
to permit a rapid check of the
potential levels at various
points in a network with
respect to ground to ensure
that the system is operating
properly
Notation – Double-Subscript Notation
• Voltage is an across variable and exists between two
points resulting in a double-subscript notation
• In fig, since a is the first subscript for Vab, point a must
have a higher potential than point b if Vab = +ve value.
• If point b is at a higher potential than point a, then
Vab = -ve value.
Notation – Single-Subscript Notation
• The single-subscript notation Va specifies the voltage
at point a with respect to ground (zero volts). Thus for
voltage at point b w.r.t to ground, we have Vb
• If the voltage is less than zero volts, a negative sign
must be associated with the magnitude of Va
Notation – General Comments
• Also the voltage Vab can be determined using
Eq.
Vab = Va – Vb
• For fig below:
Notation
• Example – Find the voltage Vab for the conditions of fig
• Solution:
• Note the negative sign to reflect the fact that point b is
at a higher potential than point a.
Notation
• Example – Find voltage Va for the configuration of Fig
• Solution:
Notation
• Example – Find the voltage Vab for the configuration.
• Solution:
Notation & Voltage Divider Rule
• Example – Using the voltage divider rule, determine
the voltages V1 and V2 of fig.
• Solution: Circuit Redrawn,
• .
• .
From
voltage divider rule,
Notation & Voltage Divider Rule
• Example – For the network of fig
a) Calculate Vab.
b) Determine Vb.
c) Calculate Vc.
Notation & Voltage Divider Rule
a) Calculate Vab.
Solution:
c) Determine Vc.
Solution:
Notation & Voltage Divider Rule
b) Determine Vb.
•
Solution:
•
or
Solution to Problems
• #1d – Find the total resistance and current I for given
circuit
• Solution:
• RT = 3k Ω + 1.3k Ω + 4.5k Ω + 1.2k Ω
RT = 10kΩ
Solution to Problems
• #8b – Determine the unknown voltages using
Kirchhoff’s voltage law.
• Solution:
Loop1:
24V – 10V – V1 = 0
V1 = 14V
Loop2:
10V – V2 + 6V = 0
V2 = 16V
Solution to Problems
• #16b – Find the unknown resistance using the voltage
divider rule and the information provided for the fig.
• Solution:
• Using VDR ratio method
V3Ω = 60V
V6Ω = 120V
VR = 140 – 120V = 20V
Since V6Ω = 6 VR , therefore R6Ω = 6 R
.
R = 6Ω/6 = 1Ω
Solution to Problems
• #32b – Determine the voltages Va, Vb, and Vab for the
network
• Solution:
Va = 12 - 8 = 4V
Vb = -8V
Vab = 12V
Summary / Conclusion
• Kirchhoff’s Voltage Law (KVL)
• Interchanging Series Elements
• Voltage Divider Rule
• Notation