Network Theorems (Part I)

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```Network Theorems (Part I)-Numerical Problems
Key points: - The problems considered in this set are involving both dependent and
independent sources. Following points may be noted
ο· Dependent sources are voltage or current sources whose output is function of
another parameter in the circuit.
ο· Dependent sources only produce a voltage or current when an independent
voltage or current source is in the circuit.
ο· Dependent sources are treated like independent sources when using nodal or mesh
analysis but not with superposition.
1.
In the following circuit find the value of vTH and RTH isc
2
2
8V
6
DC
vTH, RTH
π
π½ππ = (π+π) &times; π = ππ½
Ans.
(Replacing voltage source by it’s internal impedance)
πΉππ = (πβπ) + π = π. πβ¦
2. In the following circuit find the value of iN and RN
2
15V
DC
v1
2
4
iN, RN
πΉπ΅ = πβπ + π
Ans.
=
ππ
π
β¦
ππ΅ = short circuit current (πππ ), By source transformation the equivalent
circuit is
2
isc
2
7.5A
πππ =
π.π&times;(πβπ)
π+πβπ
=
π.π&times;π
ππ
π( )
π
4
ππ
= ππ = π ππππ
3. In the following circuit find the value of vTH and RTH
2
2A
Ans.
vTH, RTH
1
3
π&times;π
Current through 1β¦ resistor (using current division) = π+π = ππππ
π½ππ = π ππππ
πΉππ = πβπ =
π&times;π
π
π
= πβ¦
(Replacing current source by it’s internal impedance i.e. open circuit)
4.
For the following circuit find the value of vTH and RTH
30
25
P
20
5V
vTH, RTH
DC
5A
Ans. To calculate πΉππ , deactivate all independent sources i.e, replace them
by their internal impedances.
30
25
RTH
πΉππ = ππ + ππ = ππβ¦
π½ππ = πππππππ ππ ππππ π·
= π &times; ππ + π
= πππ π½
5. In the following circuit find its Thevenin and Norton equivalent
2
4V
Ans.
DC
π½ππ = π + π &times; π
= ππ½
3
2A
πΉππ = π + π = πβ¦
3
2
2A
DC
2
Isc
3
RTH
Short circuit current is due to voltage source (4 Volts) and current source(2
Amp.)
Using superposition
π
Short circuit current due to voltage source = π
π&times;π
π
Short circuit current due to current source = π+π = π
π
π
π
π°ππ = π + π = π πππ
Thevenin’s and Norton’s equivalents are shown below:
Rth
VTH
Isc
DC
Thevenin s equivalent
Rth
Norton s equivalent
6. For the following circuit find the value of vTH and RTH
i1
6
3i1
+
vTH, RTH
4
-
Ans. Circuit does not have any independent source, So, π½ππ = π.
For finding πΉππ connect test voltage source Vtest at terminals XY supplying
1A
ππ = −ππ¨
6
3i1
P
+
i1
4
-
DC
1A
Writing node equation at node P
π½ππππ
π
+
π½ππππ −πππ
π
+ ππ = π
π π½ππππ = −π ππ = π
π
π½ππππ = π
πΉππ =
π½ππππ
π
π
= π = π. πβ¦
Vtest
7. In the circuit shown in following figure find the value vTH and RTH
-+
4V
+
0.1v1
vTH, RTH
5
v1
Ans.
By source transformation
-
+
5v1 5
-+
4V
X
v1
Y
π½ππ = ππ
π. π ππ + π = ππ
π. π ππ − ππ = −π
π
ππ = π.π = π
π½ππ = π πππππ
πΉππ =
π½ππ
π°πΊπͺ
, πππ is short circuit current.
By putting short across πΏπ, ππ = π, so
π
π , πππ = π ππ πΉππ =
π½ππ
πππ
=
π
π⁄
π
= ππβ¦
dependent voltage source π. π ππ =
8. In the following circuit find the effective resistance faced by the voltage source
4
i
vS
DC
π=
Ans.
i/4
π½πΊ
π
π
+π
ππ = π½πΊ + π
ππ = π½πΊ
π=
π½πΊ
π
The effective resistance faced by the voltage source is 3β¦
9. Find the value of R (in ohms) for maximum power transfer in the network shown in
the figure.
4
5
20
3A
R
25V
Ans.
For maximum power transfer πΉπ³ = πΉπΊ (resistance looking into the
network)
Replacing independent sources by their internal impedances.
5
4
20
R
πΉπΊ = πβππ + π
=π+π
= πβ¦
πΉ = πβ¦
10. Find the Thevenin equivalent impedance ZTh between the nodes P and Q in the
following circuit
1F
1H
P
1
1
DC
1A
Q
10V
Ans.
For finding πππ deactivate all the active sources
1/S
S
P
1
1
Q
1A
πππ (Between P &amp; Q nodes)
π
= (πΊ + π) β(π + πΊ)
= πβ¦
11. Find the value of RL such that the power transferred to RL is maximum.
10
10
10
DC
RL
1A
DC
For maximum power transfer πΉπ³ = πΉππ
Ans.
10
10
10
1A
Rth
πΉππ = ππβππ + ππ
= π + ππ
= ππβ¦
πΉπ³ = ππβ¦
12. A source vs(t)=Vcos100 πt has an internal impedance of (4+j3)β¦. If an purely
resistive load connected to this source has to extract the maximum power out of the
source, find it’s value
Ans.
For pure resistive load RL to extract the maximum power
πΉπ³ = √πΉππΊ + πΏππΊ
= √ππ + ππ
= πβ¦
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