Circuit Analysis

Lecture 2

BME 372 New Schesser 44

 Voltage Division
• The voltage across impedances in series divides in
proportion to the impedances.
a
a

Z1
Z1

…
b

Z2 Zn
c c

Vac = Vab + Vbc = I( Z1 + Z 2 ); KVL + Ohm' s Law

Vi Zi
Vbc = IZ 2 =
Vbc Z2
Vac Z1 + Z 2 + ! Z n
=
Vac Z1 + Z 2
BME 372 New Schesser 45
 Current Division
• The current into impedances in parallel
divides in proportion to the inverse of the
impedances.
Iac
a a …
I1 Z1 Z2 I2 I1 Z1 Zn In

c c …

I ac = I1 + I 2 = V ( 1 + 1 ); KCL + Ohm' s Law

Z1 Z 2
I1 = V Ii (1 / Z i )
Z1
=
I1 1 Z1 Z2 I ac (1 / Z1 ) + (1 / Z 2 ) + ! + (1 / Z n )
= =
I ac (1 / Z1 ) + (1 / Z 2 ) Z1 + Z 2

BME 372 New Schesser 46

 Nodal Analysis
1. Define a voltage at each node (junction point) of a
network. For example, in a 5 node network, define
5 voltage unknowns.
2. Using KCL, write an equation for each node using
the unknown voltages. In our 5 node example,
you’ll have 5 equations and 5 unknown voltage.
3. Solve for the unknown voltages and now apply
these voltages to the network to find the currents for
each impedance in the network.

BME 372 New Schesser 47

 Nodal Analysis Example
I1 V I2
Node 1
5 5
I1 + I 2 + I 3 = 0
5Vd c 5Vd c
I3 10 5 −V 5 −V V V
+ − = 0; 2 =
5 5 10 2
V =4
5 −V 1
+ 1v - - 1v + I1 = =
.2a .2a 5 5
5 + 5 5 −V 1
I2 = =
5Vd c .4a 5Vd c 5 5
10 4v
−V 4 2
- I3 = =− =−
10 10 5

BME 372 New Schesser 48

 Mesh Analysis
1. Define a current in each mesh (loop) of a network.
For example, in a 5 mesh network, define 5 current
unknowns.
2. Using KVL, write an equation for each mesh using
the unknown currents. In our 5 mesh example,
you’ll have 5 equations and 5 unknown currents.
3. Solve for the unknown currents and now apply
these currents to the network to find the voltages for
each impedance in the network.

BME 372 New Schesser 49

 Mesh Analysis Example
5 5
Mesh 1
5Vd c
I1 I2 5Vd c
10 0 = I1 5 + 10( I1 − I 2 ) − 5; 5 = 15 I1 − 10 I 2
Mesh 2
0 = I 2 5 + 5 + 10( I 2 − I1 ); - 5 = 15 I 2 − 10 I1
I1 = 1 ; I 2 = − 1
5 5
.2a + 1v - - 1v + .2a
5 + 5
5Vd c .4a 10 4v
5Vd c

BME 372 New Schesser 50

 Comparing Nodal and Mesh Analyses
Therefore, 2 Nodal Equations: NOTE: I 3 = −I 6
R1 V1 R3 V2 R5
Node 1 equation with respect to V1
I1 + I 2 + I 3 = 0
+ I1 I +
VCC1 R2 3 I6 R4 I5
Vcc1 −V1 −V1 V2 −V1
VCC2 R1 + R2 + R3 = 0
I2 I4
-- -- Node 2 equation with respect to V2
I6 + I 4 + I5 = 0 2 simultaneous equations
Nodal Analysis
V1 −V2 −V2 Vcc2 −V2 Vcc1 1 1 1 1
3 Nodes: + + =0 = V1 ( + + ) −V2
R3 R4 R5 R1 R1 R2 R3 R3
Bottom Node is reference node
Node 1 (left hand node) has a voltage with repect to the reference node V1 Vcc2 −V1 1 1 1
= +V2 ( + + )
Node 2 (right hand node) has a voltage with repect to the reference node V2 R5 R3 R3 R4 R5
Therefore, 3 Mesh Equations:
R1 R3 R5
Mesh 1 equation with respect to I1
Vcc1 = I1R1 + (I1 − I 2 )R2
+ +
R2 Mesh 2 equation with respecnt to I 2
VCC1 I2 R4 I3
--
I1 VCC2 0 = (I 2 − I1 )R2 + I 21R3 + (I 2 − I 3 )R4
--
Mesh 2 equation with respecnt to I 2
Mesh Analysis −Vcc1 = (I 3 − I 2 )R4 + I 3 R5 3 simultaneous equations
3 Meshes: Vcc1 = I1 (R1 + R2 ) + −I 2 R2 + 0
Mesh 1 has a current I1 0 = −I1R2 + I 2 (R2 + R3 + R4 ) − I 3 R4
Mesh 2 has a current I 2
−Vcc1 = 0 − I 2 R4 + I 3 (R4 + R5 )
Mesh 3 has a current I 3 BME 372 New Schesser 51
 Superposition
• Used to analyze a circuit with multiple sources.
• Steps:
1. Set all sources except for one to zero (voltage sources are
shorted-circuited, current sources are open-circuited)
2. Solve for the currents and voltages for all of the circuit
elements
3. Repeat steps 1-2 for the remaining sources.
4. Add each of the solutions to obtain the solution for the
entire circuit

BME 372 New Schesser 52

 Superposition Analysis Example
• Define all of the voltages and currents in the circuit

+ VL - - VR +
5 5
+
Source 1 5Vd c I1 VM
I3 I2 5Vd c Source 2
10
-

BME 372 New Schesser 53

 Superposition Analysis Example
+ VLs1 - - VRs1 +
Source 1 Source 2 becomes
5 + 5 a short circuit
Currents are with 5Vdc VMs1 10
I1s1 I3s1 I2s1
respect to Source 1 -

1. Simplify circuit and calculate I1s1

10 *5 50 10
Rp = 10 || 5 = = = = 3.33Ω 5
10 + 5 15 3
5 15 3 5Vdc
I1s1 = = = I1s1 3.33
10 25 5
5+
3

BME 372 New Schesser 54

 Superposition Analysis Example
2. Use current division to calculate
the remaining currents and voltages
10 10 3 2
I 2s1 = − × I1s1 = − × = − = −.4
15 15 5 5
5 5 3 1
I 3s1 = − × I1s1 = − × = − = −.2
15 15 5 5
VLs1 = I1s1 × 5 = .6 × 5 = 3v
VRs1 = I 2s1 × 5 = −.4 × 5 = −2v
VMs1 = −I 3s1 ×10 = −(−.2) ×10 = 2v + VLs1 - =3 - VRs1 + =-2
I1s1=.6 I2s1=-.4
5 + 5
5Vdc VMs1 I3s1=-.2
=2 10
-

BME 372 New Schesser 55

 Superposition Analysis Example
5 5
+ VLs2 - - VRs2 +
Now Source 1 Source 2
+
becomes a I1s2 VMs2 10 I2s2 5Vdc Currents are
short circuit I3s2
- with respect to
Source 2
1. Repeat for Source 2
5
10 *5 50 10
5Vdc Rp = 10 || 5 = = = = 3.33Ω
3.33 I2s2 10 + 5 15 3
5 15 3
I 2s2 = = =
10 25 5
5+
3

BME 372 New Schesser 56

Superposition Analysis Example
Source 2
Currents are
wth respect to
Source 2
+ VLs2 - - VRs2 +
I2s2 10 3 2
I1s2 = − × = − = −.4
15 5 5
5 + 5
I1s2 VMs2 I3s2 5Vdc I = − 5 × 3 = − 1 = −.2
3s2
- 10 15 5 5
VLs2 = I1s2 × 5 = −4 × 5 = −2v
VRs2 = I 2s2 × 5 = .6 × 5 = 3v
VMs2 = −I 3s2 ×10 = −(−.2) ×10 = 2v

BME 372 New Schesser 57

 Superposition Analysis Example
• Summing the results of each solution:
I1 = I1s1 + I1s2 = .6 − .4 = .2
I 2 = I 2s1 + I 2s2 = −.4 + .6 = .2
I 3 = I 3s1 + I 3s2 = −.2 + −.2 = −.4
VL = VLs1 +VLs2 = 3 + (−2) = 1
VR = VRs1 +VRs2 = (−2) + 3 = 1
VM = VMs1 +VMs2 = 2 + 2 = 4
+ 1v - - 1v +
5 + 5
5Vd c .2a .2a 5Vd c
.4a 10 4v
-
BME 372 New Schesser 58
 Thevenin and Norton Equivalent
Circuits
• Thevenin’s Theorem: Any circuit consisting of
passive and active components can be represented by
a voltage source in series with an equivalent set of
passive components
– The value of the voltage source equals the voltage seen at
the output terminal without any load connected to it, i.e.,
the open-circuit voltage
– The value of the equivalent set of passive components
equals the impedance looking back into the terminals with
the sources set to zero, i.e., the output impedance.

BME 372 New Schesser 59

 Thevenin and Norton Equivalent
Circuits
• Norton’s Theorem: Any circuit consisting of passive and active
components can be represented by a current source in parallel
with an equivalent set of passive components
– The value of the current source equals the current seen at the output
terminal shorted and without any load connected to it, i.e., the short-
circuit current
– The value of the equivalent set of passive components equals the
impedance looking back into the terminals with the sources set to zero,
i.e., the output impedance.
• Note that the Thevenin and Norton Equivalents Circuits are
equivalent to each other when the value of the Thevenin’s
voltage source equals the product of the equivalent impedance
times the Norton’s current source

BME 372 New Schesser 60

 Thevenin and Norton Examples
a

5 Open Circuit Voltage at terminals : ab

10 Vdc 15 VabOC = 15 ×10 = 7.5v

5 + 15
Output Impedance
b Ro = 5 || 15 = 5 ×15 = 15 = 3.75Ω
a 20 4

5 a
è Ro=3.75Ω
15 3.75

7.5Vdc
b
b

BME 372 New Schesser 61

 Thevenin and Norton Examples
a
Short Circuit Current at terminals: ab
5
10
10 Vdc 15 I abSC = = 2a
5
Output Impedance
b 5 ×15 15
Ro = 5 ||15 = = = 3.75Ω
20 4
a

5 a

10 Vdc 15
3.75
2Adc
b b

BME 372 New Schesser 62

 Homework
• Voltage and Current division
– How does the voltage divide across two capacitors in series? Show
your results.
– How does the current divide among two capacitors in parallel?
Show your results.
• Calculate the Currents and Voltages for the following
circuits:
1
1

2
1 2Vdc
2
10Adc 1 10Adc

2Vdc

2
1 25Vdc

10Adc

BME 372 New Schesser 63

 Homework
Calculate the current labeled i and the voltage labeled v in the
following circuit
R1 = 1Ω, R2 = 2Ω, R3 = 1Ω, R4 = 1Ω, R5 = 2Ω, R6 = 2Ω, R7 = 2Ω, Vcc = 4v

R1 R3

+ R2 R4 R5
VCC
--
i
+
R6 R7
v
--

BME 372 New Schesser 64

 Homework
Calculate the current labeled,
i4 i.
R1 = 2Ω, R2 = 2Ω, R3 = 2Ω, R4 = 3Ω, Vcc = 2v

R1
+ R2
VCC i R4
--

R3

BME 372 New Schesser 65

 Homework
RD 29.5k RS 500
a

1Vac CIN 5.3n

CD 53n

An electrode is connected to an oscilloscope which

has a purely capacitance input impedance, CIN. Find
and plot the output voltage Vab(jω) as function of ω.
Use Matlab to perform the plot.

66
 Homework
• Repeat the analysis of this circuit using Mesh
and Nodal Analysis. That is find and plot Vab
as a function of frequency. Use Matlab to
calculate the Bode plot.
R1 R2

1k 5k
a
1Vac
C1 C2
1n 5n

BME 372 New Schesser 67