mmab

1. MMAC
Presentation
Group no 5.
1. Siddhant G. Sinhasane 111710112
2. Juber F. Shaikh 111710103
3. Dhananjay P. Hiwase 141810003
4.. Kapil A. Deshmukh 141810004
5. Shubham P. Kewate 141810024

2. Topic:
“Solving Electrical Circuits Problems”
(From Norman Nise Book and using Norman Nise Method )
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3. Content
• Introduction
• Problems 1
• Problems 2
• Problems 3
• Problems 4
• Problems 5
• Reference
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4. Introduction
Pg no .48 (Ref 1)
Our guiding principles are Kirchhoff’s laws.
• We sum voltages around loops or sum currents at nodes and then equate the result to zero.
• From these relationships we can write the differential equations for the circuit.
• Then we can take the Laplace transforms of the differential equations and finally solve for the
transfer function.
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5. Topic:
“Solving Electrical Circuits via Nodal Analysis”
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6. Q.1 . Find the transfer function, G(s)=
Vo(s)
Vi(s)
, for network shown in
Figure (By Nodal Analysis)
Let,
L1 = 2H , L2= 3H , C=
1
2
F, R = 1 Ω
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7. Note : 𝑉𝑅(𝑠) - 𝑉𝐶 𝑠 = 𝑉
𝑜 𝑠
Node analysis (Kirchoff’s Current law , here at node consider incoming current negative and outgoing
positive)
At Node 1
𝑉𝑅(𝑠) − 𝑉𝑖 𝑠
𝐿1 𝑠
+
𝑉𝑅(𝑠)
𝑅
+
𝑉𝑜(𝑠)
𝐿2 𝑠
=0 , let here,
1
𝑅
= 𝐺 𝑐𝑎𝑙𝑙𝑒𝑑 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒
Rearranging
1
𝐿1 𝑠
+ 𝐺 * 𝑉𝑅(𝑠) +
1
𝐿2 𝑠
* 𝑉𝑜(𝑠) =
𝑉𝑖 𝑠
𝐿1 𝑠
……..(1)
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Transformed circuit
i

8. At Node 2;
𝑉𝑐(𝑠) ∗ C∗s
1
-
𝑉𝑜(𝑠)
𝐿2 𝑠
=0
Note : 𝑉𝑅(𝑠) - 𝑉𝐶 𝑠 = 𝑉
𝑜 𝑠 , from here get 𝑉𝐶 𝑠 put in above eqn
[𝑉𝑅(𝑠) −𝑉𝑜 𝑠 ]∗ C∗s
1
-
𝑉𝑜(𝑠)
𝐿2 𝑠
=0
C* s * 𝑉𝑅(𝑠) -
1
𝐿2 𝑠
+ 𝐶 ∗ 𝑠 * 𝑉𝑜(𝑠) =0 ……..(2)
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Transformed circuit
i

9. • Equations are:
•
1
𝐿1 𝑠
+ 𝐺 * 𝑉𝑅(𝑠) +
1
𝐿2 𝑠
* 𝑉𝑜(𝑠) =
𝑉𝑖 𝑠
𝐿1 𝑠
……..(1)
• C * s * 𝑉𝑅(𝑠) -
1
𝐿2 𝑠
+ 𝐶 ∗ 𝑠 * 𝑉𝑜(𝑠) =0 ……..(2)
• Solving by Cramers rule for 𝑉𝑜(𝑠)
• 𝑉𝑜(𝑠) =
1
𝐿1 𝑠 + 𝐺
𝐶∗𝑠
𝑉𝑖 𝑠
𝐿1 𝑠
0
1
𝐿1 𝑠
+ 𝐺
C ∗ s
1
𝐿2 𝑠
−
1
𝐿2 𝑠
+𝐶∗𝑠
• Solving Determinent and putting value Of L1 ,C , L2 and R
• Solution
•
𝑉𝑜 𝑠
𝑉𝑖 𝑠
=
3∗ 𝑠2
6∗ 𝑠3+5∗ 𝑠2+4∗𝑠+2
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10. 3 ∗ 𝑠2
6 ∗ 𝑠3 + 5 ∗ 𝑠2 + 4 ∗ 𝑠 + 2
𝑉𝑖 𝑠 𝑉
𝑜 𝑠
Block diagram
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11. “Solving Complex Electrical Circuits via Mesh
Analysis”
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12. Let,
L1 = 4 H , L2= 6 H , C=
1
9
F
R1 = 2 Ω, R2 = 4 Ω, R3 = 2 Ω, R4 = 8 Ω
Laplace Transformed for transformed ckt :
L1 = 4*s H , L2= 6*s H , C=
9
𝑠
F (i.e
1
𝐶∗𝑠
𝑤𝑒 𝑤𝑟𝑖𝑡𝑒 𝑓𝑜𝑟 𝑖𝑚𝑝𝑒𝑑𝑒𝑛𝑐𝑒)
R1 = 2 Ω, R2 = 4 Ω, R3 = 2 Ω, R4 = 8 Ω
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Q ) Solve for the transfer function, G(s) =
Vo(s)
V(s)

13. 13 of 30
In Mesh 1:
(2+2+4*s)*𝐼1(𝑠) – (2+4*s)* 𝐼2 𝑠 − 2 ∗ 𝐼3 𝑠 = 𝑉 𝑠
(4+4*s)*𝐼1(𝑠) – (2+4*s)* 𝐼2 𝑠 − 2 ∗ 𝐼3 𝑠 = 𝑉 𝑠 ………(1)
In Mesh 2:
(8+6*s+2+4+4*s)*𝐼2(𝑠) – (2+4*s)* 𝐼1 𝑠 − 4 + 6 ∗ 𝑠 ∗ 𝐼3 𝑠 = 0
(14+10*s)*𝐼2(𝑠) – (2+4*s)* 𝐼1 𝑠 − 4 + 6 ∗ 𝑠 ∗ 𝐼3 𝑠 = 0
– (2+4*s)* 𝐼1 𝑠 + (14+10*s)*𝐼2(𝑠) − 4 + 6 ∗ 𝑠 ∗ 𝐼3 𝑠 = 0…….(2)
In Mesh 3:
(2 +
9
𝑠
+ 6 ∗ 𝑠 + 4) * 𝐼3 𝑠 − 2 ∗ 𝐼1 𝑠 - 4 + 6 ∗ 𝑠 ∗ 𝐼2(𝑠) =0
−2 ∗ 𝐼1 𝑠 - 4 + 6 ∗ 𝑠 ∗ 𝐼2(𝑠) + (6 +
9+6∗𝑠2
𝑠
) * 𝐼3 𝑠 =0………(3)
From Fig. it is clear that Vo(s) = 𝐼2(𝑠) *8
So we need to find: 𝐼2(𝑠)

14. 14 of 30
Equations are:
(4+4*s)*𝐼1(𝑠) – (2+4*s)* 𝐼2 𝑠 − 2 ∗ 𝐼3 𝑠 = 𝑉 𝑠 ………(1)
– (2+4*s)* 𝐼1 𝑠 + (14+10*s)*𝐼2(𝑠) − 4 + 6 ∗ 𝑠 ∗ 𝐼3 𝑠 = 0…….(2)
−2 ∗ 𝐼1 𝑠 - 4 + 6 ∗ 𝑠 ∗ 𝐼2(𝑠) + (6 +
9+6∗𝑠2
𝑠
) * 𝐼3 𝑠 =0………(3)
Solving by Cramers rule for 𝐼2(𝑠)
𝐼2(𝑠) =
(4+4∗s)
– (2+4∗s)
−2
𝑉 𝑠
0
0
−2
– (4+6∗s)
(6 + 9+6∗𝑠2
𝑠
)
(4+4∗s)
– (2+4∗s)
−2
– (2+4∗s)
(14+10∗s)∗
– (4+6∗s)
−2
– (4+6∗s)
(6 + 9+6∗𝑠2
𝑠
)
Solve using Matlab Symbolic Maths

15. 15 of 30
𝐼2(𝑠) =
(4+4∗s)
– (2+4∗s)
−2
𝑉 𝑠
0
0
−2
– (4+6∗s)
(6 + 9+6∗𝑠2
𝑠
)
(4+4∗s)
– (2+4∗s)
−2
– (2+4∗s)
(14+10∗s)∗
– (4+6∗s)
−2
– (4+6∗s)
(6 + 9+6∗𝑠2
𝑠 )

16. 16 of 30
From Output :
𝐼2(s)
1
=
2∗𝑉 𝑠 ∗
12∗𝑠3+24∗𝑠2+28∗𝑠+9
𝑠
4∗
48∗𝑠3+150∗𝑠2+220∗𝑠+117
𝑠
𝐼2(s)
𝑉 𝑠
=
12∗𝑠3+24∗𝑠2+28∗𝑠+9
1
2∗
48∗𝑠3+150∗𝑠2+220∗𝑠+117
1
Vo(s) = 𝐼2(𝑠) *8
Vo(s)
𝑉 𝑠
=
4 ∗ (12 ∗ 𝑠3
+ 24 ∗ 𝑠2
+ 28 ∗ 𝑠 + 9)
48 ∗ 𝑠3 + 150 ∗ 𝑠2 + 220 ∗ 𝑠 + 117

17. 17 of 30
4 ∗ (12 ∗ 𝑠3 + 24 ∗ 𝑠2 + 28 ∗ 𝑠 + 9)
48 ∗ 𝑠3 + 150 ∗ 𝑠2 + 220 ∗ 𝑠 + 117
V 𝑠 𝑉
𝑜 𝑠
Block diagram

18. Q.2 Find the transfer function, G(S)=
Vo(S)
Vi(S)
, for each network shown in Figure by mesh analysis.
….1 By Kirshoff’s voltage law (in loop 1)
Applying laplace transform to 1 and 2
R1
R2
Now solving 1 and 2 by cramers rule for
I2(S) =
𝑎 𝑚
𝑐 𝑛
𝑎 𝑏
𝑐 𝑑
Vi 𝑡 = 𝑅1𝐼1 𝑡 + 𝐿
𝑑𝐼1
(𝑡)
𝑑𝑡
−L
𝑑𝐼2
(𝑡)
𝑑𝑡
0 = 𝑅2𝐼2 𝑡 + 𝐿
𝑑𝐼2
(𝑡)
𝑑𝑡
−L
𝑑𝐼1
(𝑡)
𝑑𝑡
….2 By Kirshoff’s voltage law (in loop 2)
Vi 𝑠 = 𝑅1𝐼1 𝑠 + LS 𝐼1 𝑠 − 𝐿𝑆𝐼2 𝑠 …..3
0 = 𝑅2𝐼2 𝑠 + LS 𝐼2 𝑠 − 𝐿𝑆𝐼1 𝑠 …4
(𝑅1 + 𝐿𝑆) 𝑉𝑖(𝑆)
(−𝐿𝑆) 0
(𝑅1 + 𝐿𝑆) (−𝐿𝑆)
(−𝐿𝑆) (𝑅2 + 𝐿𝑆)
I2(S) =

19. I2(S) =
Vi(S)∗LS
R1∗LS+R1∗R2+R2∗LS
….3
Vo(S) = R2*I2(S) …..4
Putting eqn 3 in 4 we get
Vo(S) = ( Vi(S)∗LS
R1∗LS+R1∗R2+R2∗LS
)*R2
Vo(S)
Vi(S)
= (
LS∗R2
R1∗LS+R1∗R2+R2∗LS
)
Replacing R1=1,R2=1,L=1 in 5
Vo(S)
Vi(S)
=
S
1+2∗S
=
1
1
𝑠
+2
= transfer function
1
1
𝑠
+ 2
Vi(S) Vo(S)
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20. PROBLEM ON ELECTRIC CIRCUIT BY VOLTAGE DIVISION METHOD
Q3) Find the transfer function ,G(s)=V O(s)/ V I (s),for the network shown in figure :
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Now for the given electrical circuit , we can see there are two resistors of 1 Ω ,one inductor of 1H and a
Capacitor of 1F .

21. To solve the circuit we first reduce the two resistor in the circuit by equivalent single resistor using the Thevenin
Principle .
The steps to draw an equivalent Thevenin circuit are as follows –
Step 1 − Consider the circuit diagram by opening the terminals with respect to which the Thevenin’s equivalent
circuit is to be found.
Step 2 − Find Thevenin’s voltage VTh across the open terminals of the above circuit.
VTH =
𝑉𝑠
𝑅1+𝑅2
𝑥𝑅2
Considering our circuit , we get
V TH =
𝑉𝑖𝑆
1+1
× 1
V TH =
𝑉𝑖 𝑠
2
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22. Step 3 − Therefore equivalent resistance , R TH =
𝑅1×𝑅2
𝑅1+𝑅2
𝛺
RTH =
1×1
1+1
RTH =
1
2
𝛺
Step 4 − Draw the Thevenin’s equivalent circuit by connecting a Thevenin’s voltage VTh in series with a Thevenin’s
resistance RTh.
The equivalent circuit considering Thevenin resistance and voltage is :
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23. 17 of 30
Using the Voltage division method we can get the transfer function of the circuit :
The voltage across the capacitor is some proportion of the input voltage, namely the impedance of the capacitor divided by
the sum of the impedances. Thus,
V C (s) =
1
𝐶𝑠
𝐿𝑠+𝑅+
1
𝐶𝑠
.V (S)
Applying the same to our circuit we get ,
VO (s)=
𝑉𝑖 𝑠
2
×
1
𝑠
1
2
+𝑠+
1
𝑠
𝑉𝑜 𝑠
𝑉 𝑖 (𝑠)
=
1
2
×
1
𝑠
1+2𝑠
2
+
1
S
=
1
2𝑠
𝑆+2𝑠2+2
2𝑆
=
1
𝑠+2𝑠2+2

24. 𝟏
𝑺 + 𝟐𝑺𝟐 + 𝟐
VI (S) VO (S)
BLOCK DIAGRAM :
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Therefore the transfer function is :
𝑽𝒐 𝒔
𝑽𝒊(𝒔)
=
𝟏
𝟐𝒔𝟐+𝐬+𝟐

25. Q.4 Find the transfer function, G(S)=Vo(S)/Vi(S), for the network shown in figure, using Nodal
analysis method
Solution- • For this problem, we sum currents at the nodes rather than sum voltages around the
meshes. From Figure, the sum of currents flowing from the node A are, respectively,
𝑉1
𝑠 −𝑉(𝑠)
𝑅1+
1
𝐶1
+
𝑉1
(𝑠)
𝑅2+
1
𝐶2
= 0
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26. As R1=R2 and C1=C2 , 2 X V1(s) = V(s)
………… (1)
• As the 2 branches AB and CD are in parallel connection, the voltage drop across both will be same, therefore,
V1(s) = V2(s)
• Current flowing in branch CD will be constant, so,
𝑉𝐿
(𝑠)
𝐿(𝑠)
=
𝑉2(𝑠)
𝑅3+𝐿𝑠
and V2(s) =
𝑉(𝑠)
2
VL(s) =
𝑉(𝑠)/2
𝑅3+𝐿𝑠
X L(s) here, R3= 2 , L=2
…….. From (1)
𝑉(𝑠)
VL(𝑠)
=
2(2 + 2𝑠)
2
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27. 2( 1 + 𝑠)
𝑠
Block diagram
V(s) VL(s)
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28. Q.5 Find the transfer function, G(S)=VL(S)/Vi(S), for each network shown in Figure by mesh
analysis.
….By Kirshoff’s voltage law
(R1 +L1S) Vi(S)
(-R1) 0
(R1 + LS) (-R1)
(-R1) (L2S +R2+R1)
I2(S) =
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Vi 𝑡 = 𝑅1𝐼1 𝑡 + 𝐿1
𝑑𝐼1
(𝑡)
𝑑𝑡
− 𝑅1𝐼2 𝑡
0= 𝑅2𝐼2 𝑡 + 𝐿2
𝑑𝐼2
(𝑡)
𝑑𝑡
− 𝑅1𝐼1 𝑡
Vi 𝑠 = 𝑅1𝐼1 𝑠 + L1S 𝐼1 𝑠 − 𝑅1𝐼2(𝑆) …..1
0 = 𝑅2𝐼2 𝑠 + L2S𝐼2 𝑠 − 𝑅1𝐼1 𝑠 ..…2

29. I2(S) =
Vi(S)∗R1
R2L1S+L2S2L1+R1R2+R1L2S+R1L1S
….3
VL(S) = L2S*I2(S) …..4
Putting eqn 3 in 4 we get
VL(S) = L2s*
Vi(S)∗R1
R2L1S+L2S2L1+R1∗R2+R1L2S+R1L1S
VL(S)
Vi(S)
=
L2s∗R1
R2L1S+L2S2L1+R1R2+R1L2S+R1L1S
……5
Replacing R1=2, R2=2, L1=2, L2=2 in eq. 5
VL(S)
Vi(S)
=
𝑆
S2+3S+1
= transfer function
𝑆
S2+3S+1
Vi(S) VL(S)
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30. Reference
1. Norman S. Nise, CONTROL SYSTEMS ENGINEERING, Sixth Edition
(John Wiley & Sons, Inc)
(For Solved problem and theory refer : pg no. 47 to 57)
(Exercise problems taken from pg no. 99 and 100)
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