Some more KVL problems Simplification of Circuits Combining Sources In Series (independent voltage sources) In Parallel (independent current sources) Example problems

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Session 6: Focus
 Some more KVL problems
 Simplification of Circuits
 Combining Sources
◦ In Series (independent voltage sources)
◦ In Parallel (independent current sources)
◦ Example problems

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Some more KVL Problems

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S5_HW_Problem_2
 Find vx : 8 V
Note: This is
Example problem
is 3.4 on page 45,
Figure 3.10 in the
Ref 1 book (by
Hayt)
-60 + v8 + v10 = 0
v8 = 5 * 8 = 40 V
-60 + 40 + v10 = 0
v10 = 20 V
-v10 + v4 + vx = 0
-20 + 12 + vx = 0
vx = 8 V
i10 = v10 /10
i10 = 20 /10
i10 = 2 A
i10 + i4 = 5 A
i4 = 5 – 2 = 3 A
v4 = 3 * 4 = 12 V

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Simplification of Circuits

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Simplification of Circuits
 While trying to find out unknown quantities in a given
circuit, some techniques could be used to simplify the
solution
 Based on the quantities to be found, the given circuit can
be simplified by combining various elements
◦ Independent Sources (both voltage and current)
◦ Resistors

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Simplification of Circuits … contd.
 Note that dependent sources and the elements impacting
those dependent sources need to be untouched
 These simplifications still leave the functioning of the
rest of the circuit unaffected

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Combining Sources in Series

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Combining Voltage Sources in Series
 Various Independent voltage sources in series can be
replaced by an equivalent voltage source having a
voltage equal to the algebraic sum of the individual
sources
◦ Note that dependent sources cannot be combined and they
need to be untouched

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Combining Voltage Sources in Series
 Various Independent voltage sources in series can be
replaced by an equivalent voltage source having a
voltage equal to the algebraic sum of the individual
sources
 Combing dependent voltages sources

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Combining Current Sources in Parallel
 Various Independent current sources in parallel can
be replaced by an equivalent current source having a
current equal to the algebraic sum of the individual
sources
 The order of the parallel elements may be rearranged
as desired

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Example 1
 To be able to combine voltage sources, they must be in series.
 Since the same current (i) flows through each, this condition is
satisfied
 - 3 - 9 – 5 + 1 = -16 V (KVL starting from bottom left corner)
 -16 + 100 i + 220 * i = 0; i = 16/320 = 50 mA
 The right most diagram is also equivalent to the previous one.

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Example 2
 The value of the voltage source is taken as –ve when a –ve
terminal is seen while traversing the circuit
◦ Here in the clockwise direction
 When the resultant voltage is computed after taking the
algebraic sum of the sources, follow the same convention
while constructing the circuit with resultant voltage source
◦ Here the -1 V source’s negative terminal will be down and its +ve
terminal will be connected to 47 Ohms
-3 + 1 + 5 -4 = -1 V
Combining the sources in series
from the top right corner:
i = 1/47 = 21.3 mA +
-1 V 47 Ω
i

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Example 3
 Find v after combing current sources into one equivalent
current source: v =
3 = i1 + i2 = 2 i1
v = 5 (1.5) = 7.5 V
KCL at bottom node:
v = i1 5 = i2 5
7.5 V
i1 = i2
i1 = i2 = 3/2 = 1.5 A
i2i1
-
+
-
+
-
+

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Example 4
 Find v after combing current sources into one equivalent
current source: v =
10 = i1 = i2 + i3
v = i210 = 5 (10) = 50 V
KCL at node A:
v = i2 10 = i3 10
50 V
i2 = 10/2= 5 A
i2 = i3
10 10 10
i2
i1
i3Node A
10 = i1 = 2i2
i2= i3 = 5 A
Note: This is Practice problem 3.10 on page 54, Figure 3.23 in the Ref 1 book (by Hayt)

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Session 6: Summary
 Some more KVL problems
 Simplification of Circuits
 Combining Sources
◦ In Series (independent voltage sources)
◦ In Parallel (independent current sources)
◦ Example problems

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References
Ref 1 Ref 2