chapter 2 circuit analysis techniques

1Dr.-Eng. Hisham El-SherifElectronics and Electrical Engineering Department

ELCT708: Electronics for Biotechnology

Chapter 2

Circuit Analysis Techniques



• To formulate the node-voltage equations.

• To solve electric circuits using the node – voltage method.

• To introduce the mesh – current method.

• To formulate the mesh-current equations.

• To solve electric circuits using the mesh-current method.

Objectives



Two Powerful Techniques for Circuit Analysis

Nodal Analysis, Mesh Analysis,

based on a systematic application of Kirchhoff’s current law (KCL

based on a systematic application of Kirchhoff’svoltage law (KVL)

we can analyze almost any circuit by obtaining a set of simultaneous equationsthat are then solved to obtain the required values of current or voltage.



Nodal Analysis

• So far, we have been applying KVL and KCL “as needed” to find voltages and currents in a circuit.

• Good for developing intuition, finding things quickly…

• …but what if the circuit is complicated? What if you get stuck?

• Systematic way to find all voltages in a circuit by repeatedly applying KCL: Node Voltage Method(Nodal Analysis).



Branches and Nodes ( reminder from last part)

Branch: elements connected end-to-end,nothing coming off in between (in series)

Node: place where elements are joined—includes entire wire



Steps to Determine Node Voltages Method :

Step 1Select a node as the reference node.

The reference node is commonly called the ground since it is assumed to have zero potential.

This is the reference Node



Steps to Determine Node Voltages Method ( Cont.)

Step 2Assign voltages v1, v2, . . . , vn−1 to the remaining n − 1 nodes. The voltages are referenced with respect to the reference node.

Node 0 is the reference node (v = 0), while nodes 1 and 2 are assigned voltages v1and v2.



Steps to Determine Node Voltages Method( Cont.)

Step 3Apply KCL to each of the n − 1 non reference nodes

add i1, i2, and i3 as the currents through resistors R1,R2, and R3, respectively. Atby applying KCL gives

node 1

node 2




Step 4Use Ohm’s law to express the branch currents in terms of node voltages.

The key idea to bear in mind is that, sinceresistance is a passive element, by the passive sign convention, currentmust always flow from a higher potential to a lower potential.




Step 5Solve the resulting simultaneous equations to obtain the unknownnode voltages.



Node Voltage Equations (Resistors)



Example:�The voltage drop from node X to a reference node (ground) is called the node voltage Vx.

� The current through resistors can be expressed asR

VbVaIab

−=



ExampleCalculate the node voltages

Solution



At node 1

Multiply by 4



At node 2

Multiply by 12



Example

Find the voltage at node 1, 2 & 3

Solve this example and handle it to me



Conclusion for Nodal Analysis

• Nodal analysis is simply writing KCL equations in a systematic way assuming all currents leaving.

• Nodes’ voltages’ are the circuit variables.

• Currents are expressed in terms of nodes’ voltages.

• The number of variables = Number of nodes - 1



Nodal Analysis with Voltage Sources

note that:

• A current source produces constant current in a give direction.

• A Voltage source maintains the voltage constant between its terminals.

• No need to consider Va a circuit variable.

IsVa Vb

Is is leaving Va

- Is is leaving Vb

+-

Vs

Va

Va = Vs



If a voltage source is connected between the reference node and a non-reference node.

simply set the voltage at the non-referencenode equal to the voltage of the voltage source

v1 = 10 V



If the voltage source (dependent or independent) is connected between two non-reference nodes.

we apply both KCL and KVL to determine the node voltages.

V1 and v2 are called a super-nodes as they enclose a (dependent or independent) voltage source connected between them and any elements connected in parallel.



Steps to Determine Nodal Analysis with Voltage Sources.

Step 1Choose a reference node (ground, node 0)

Step 2 Define unknown node voltages (those not connected to ground by voltage sources). Va, Vb, ……

Step 3Write KCL equation at each unknown node.

How? Each current involved in the KCL equation will either comefrom a current source (giving you the current value) or through a device like a resistor. If the current comes through a device, relate the current to the node voltages using I -V relationship (like Ohm’s law).



Step 4 Apply KCL to the supper node

Super-node



Step 5Apply KVL to the supper node

Step 6Solve the set of equations (N linear KCL equations for N unknown node voltages).



Example

• Choose a reference node.• Define the node voltages (except reference node and the one set

by the voltage source).• Apply KCL at the nodes Va and Vb with unknown voltage.

• Solve for Va and Vb in terms of circuit parameters.

0R

VVRV

RVV

3

ba

2

a

1

1a =−++−

node voltage set�

Va Vb

← reference node

R4V1 R2

+- IS

R3R1

S4

b

3

ab IRV

RVV =+−



Example Find the node voltage

Solution

Apply KCL at the super node

Multiply by 4



Apply KVL to the loop

………………………………………….1

…………………….……2

From 1 and 2



ExampleCalculate the power absorbed bythe 6 ohm resistor using nodal analysis

Apply nodal KCL at V1

062

1 211 =��

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� −−��

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�− vvv

0126121 =−��

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�+��

��

� − vvv………………1



At Node V2

0476221 =−��

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�−��

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� − vvv

0476212 =+��

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�+��

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� − vvv………………………………….2

Solve for 1 & 2

62

6 RiP =

Wvv

P 246

1446

6

221 ==��

��

� −=



ExampleFind Io using the Node-Voltage

��

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� −+��

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�=63

4 211 vvv

214 ii +=

0463

211 =−��

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� −+��

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� vvv

…..........................................................1

At node V1

…



V1 = 10.838 VV2 = 8.516 V

At node V2

43 iiIo +=

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� −+��

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�=��

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� −6

1246

2221 vvvv

06

1246

2212 =��

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� −+��

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� − vvvv………………………………..2

Solve for 1 and 2



ExampleUse nodal analysis, find vo

SolutionAt Node V1

521 += ii

521

011 +��

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� −=��

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� vvv

521

40 01 +��

��

� −=��

��

� vv



521

40 01 +��

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� −=��

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� vv

Multiply by 2

( ) ( ) 1080 01 +−= vv

( ) 7001 =− vv



At Node V0

32 5 ii =+

( )��

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� −−+��

��

�=+��

��

� −8

204

52

0001 vvvv

Multiply by 8

2074 01 −=− vv

Solving for v0 and v1

V0 = 30v

chapter 2 circuit analysis techniques

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