v - voltage or potential difference
SI unit: volt ; symbol: V
named for Alessandro Volta (voltaic pile ~ first battery)
I - current
SI unit: ampere or amp ; symbol: A
named for André-Marie Ampère (Electromagnetism)
R - resistance
SI unit: ohm ; symbol: W
named for Georg Simon Ohm (extented Volta's work, math prof too!!)
Ohm's Law: Ohm demonstrated experimentally that over a wide range of operating conditions, that v, I and R are related by:
v = I R
v = R I
y = m x + b
Ohm's Law (version 1):
The end of the resistance at which the assumed current leaves is called the assumed low potential terminal (-);
Thus, in a resistor, current is said to flow from high potential to low potential. This also referred to as voltage drop.
Example:
Calculate vR.
Solution:
vR = R I = (12 W) (2 A)vR = 24 V
Example:
Calculate I.
Solution:
I = v / R = (16 V) / (4 W)I = 4 A
Example:
 |
| Add caption |
Calculate v.
Solution:
v = R I = (5.2 x 106 W) (2 x 10-3 A)v = 10 400 V
Ohm's Law (version 2):
Note: Useful in Nodal analysis.
Clearly, v = VA - VB
\ I = (VA - VB) / R or (VA - VB) / = I R
\ I = VAB / R or VAB = I R
VA = potential at A relative to a Common or Ground
VB = potential at B relative to Same Common or Ground
We define VAB = (VA - VB) to be the potential at A with respect to B.
Comments:
Generally,
I = [(Potential at end of reistor where assumed current enters) - (Potential at end of reistor where assumed current leaves)] / R
Example:
Calculate I (an assumed current)
Calculate VAB
Solution:
I = (VA - VB) / R = (16 V - 5 V) / (10 W) = 15 V / 10 WI = 1.5 A
\ Therefore, our true physical current is 1.5A ¯
Example:
In each case below, calculate I.
Solution:
I = (VA - VB) / R = (20 V - 0 V) / (10 W) = 20 V / 10 WI = 2 A
\ Therefore, our true physical current is 2A ¯
Example:
Solution:
I = (VA - VB) / R = (160 V - 140 V) / (10 W) = 20 V / 10 WI = 2 A
\ Therefore, our true physical current is 2A ¯
Example:
Solution:
I = (VA - VB) / R = (-20 V - (-40 V)) / (10 W) = 20 V / 10 WI = 2 A
\ Therefore, our true physical current is 2A ¯
What does the example above illustrate?
The current is determined by the difference in potentials, not the absolute potentials themselves.
Example:
(i) Calculate I1 and I2
(ii) what is the magnitude and direction of the actual physical current in each case?
Solution:
Assume I1 & I2
I1 = (VA - VB) / R = (40 V - 100 V) / (10 W) = -60 V / 10 W\ I1 = -6 A
==> Interpretation:
\ True physical current is 6 ↑ (upwards direction)
I2 = (VA - VB) / R = (100 V - 40 V) / (10 W) = 60 V / 10 W
\ I2 = 6 A
==> Interpretation:
\ True physical current is 6A ↑ (upwards direction)
Example:
Find formulas for I1 , I2 , I3 in terms of the given potentials and resistances.
Solution:
I1 = (VA - 10 V) / R1 = (40V - 10V) / 10W = 30V / 10W = 3A;
I2 = (10 V - VB) / R2 = (10V - 5V) / 10W = 5V / 10W = 0.5A;
I3 = (10 V - VC) / R3 = (10V - 5V) / 10W = 5V / 10W = 0.5Aor
Redraw the schematic:
==>
(10W + 10W) I1 - (10W) I2 - (10W) I3 = (40 V - 5 V)
-(10W) I1 + (10W + 10W) I2 + (10W) I3 = (5 V - 5 V)-(10W) I1 + (10W) I2 + (10W) I3 = (5 V - 10 V)
==>
(20W) I1 - (10W) I2 - (10W) I3 = (35 V)
-(10W) I1 + (20W) I2 + (10W) I3 = (0 V)-(10W) I1 + (10W) I2 + (10W) I3 = (-5 V)==>
I1 = 3A
I2 = 0.5A
I3 = 2A
==>I1 = I1 = 3A
I2 = I2 = 0.5A
I10V = I3 = 2A
I3 = I1 - I2 - I3 = 3A - 0.5A - 2AI3 = 0.5A
Example:
Calculate I1 , I2 and VAB in the (partial) circuit below.
Solution:
I1 = (VB - VA) / 5WW
\hysical current is 2A ®
I2 = (VB - VC) / 10W = (10V - 2V) / 10W = 8V / 10W = 0.8A
\ Physical current is 0.8A ®
VAB = 10V
Node:
In electronics, a node is a common joining point for three or more components (batteries, resistors, capacitors, etc...).
All points connected to a given node are at the same potential (relative to some Common or Ground)
Ex. Identify (circle) all nodes in the complete or partial circuits shown below
Aside: Some books refer to sometimes like as a node... not done here...
Kirchhoff's Current Law (KCL):
Basically a statement of charge conservation.KCL (version A):
The sum of all the currents entering a node = The sum of all the currents leaving same node.
"Sum of": å (currents In) = å (currents Out)
Note: å - Sigma, upper case greak letter for Sum
å (currents In) - å (currents Out) = 0
å [(currents In) + (- currents Out)] = 0
alg. å (currents In and Out) = 0
alg. å Iin/out = 0
Example:
Find I
Solution:
KCL: version A: å Iin = å Iout
==> 3A = I + 2A ==> I = 3A - 2A = 1A
KCL: version B: alg. å I = 0
==> 3A + (-2A) + (-I) = 0
or 3A - 2A - I = 0 ==> I = 3A - 2A = 1A
Example:Find I1
Solution:
KCL: version A: å Iin = å Iout
==> I1 + 2A = 3A + 6A + 10A
\ I1 = 3A + 6A + 10A - 2A = 17A
KCL: version B: alg. å I = 0
==> 2A + (-10A) + I1 + (-6A) + (-3A) = 0
\ I1 = 10A - 2A + 6A + 3A = 17A
same answer!
Example:
Find I1 and I2
Solution:
Node 2:
26A + I1 = 2A + 4A + 12A
\ I1 = -8A
or
26A - 2A + I1 - 4A - 12A = 0
I1 = 12A + 4A + 2A -26A
\ I1 = -8A
either way, physical current is 8A ¯
Node 2:
12A = I2 + 3A
\ I2 = 12A - 3A
\ I2 = 9A
either way, physical current is 9A ®
Kirchhoff's Voltage Law (KVL):
- Basically a statement of energy conservation.
travelling from '-' to '+' through a component referred to as as a voltage rise (potential rise)
KVL (version A):
The sum of all the voltages rises (potential rises) around any closed path
=
The sum of all the voltages drops (potential drops) around the same closed path.
"Sum of":
å Vrises = å Vdrops
å (voltage rises) - å (voltage drops) = 0
å [(voltage rises) + (- voltage drops)] = 0
alg. å (voltage rises and drops) = 0
alg. å Vrises/drops = 0
Example:
Find V1
Solution:
KVL version A: start at x P (clockwise): å Vrises = å Vdrops
20V = V1 + 15V
\ V1 = 5 voltsKVL version B: start at x P (clockwise): alg. å Vrises/drops = 0
20V + (-V1) + (-15V) = 0
20V - V1 - 15V = 0
\ V1 = 20V - 15V = 5 voltsExample:
Find V2
Solution:
KVL version A: start at y P (clockwise): å Vrises = å Vdrops
20V + V2 = 15V
\ V2 = 15V - 20V\ V2 = -5 volts
KVL version B: start at y P (clockwise): alg. å Vrises/drops = 0
20V + V2 - 15V = 0
\ V2 = 15V - 20V = -5 volts
KVL version B: start at y Q (counterclockwise): alg. å Vrises/drops = 0
15V - V2 - 20V = 0
\ V2 = 15V - 20V = -5 volts
Example:
Find V2
Solution:
KVL version A: start at P Q (clockwise):
25V + 25V + 5V = 20V + 6V + 16V + V2
55V = 42V + V2\ V2 = 13 volts
KVL version B: start at P Q (clockwise):
25V + 25V - 20V - 6V - 16V - V2 + 5V = 0
\ V2 = 13 volts
Example:
Find Vab
Solution:
Note: when dealing with notation like
Assume:
Vab (a: + (high); b: - (low)) or Vxy (x: + ; y: -) or Vcd (c: + ; d: -)
20V + 12V + Vab = 15V + 50V
\ Vab = 33 voltsKVL version B: start at x P (clockwise):
20V - 15V + 12V - 50V + Vab = 0
\ Vab = 33 volts==> means terminal ''a" is 33 volts higher in potential than terminal "b".
Example:
Find Vab , Vxy
Solution:
Assume:
Vab (a: high; b: low) , Vxy (x: + ; y: -)
4V + 10V - Vab - 5V = 0
\ Vab = 9 voltsApply KVL in loop 2, start at P P (clockwise):
+20V + Vxy - 4V = 0
\ Vxy = -16 voltsremember: by definition Vxy = Vx - Vy = -16V
\ Vy = 16V + Vx
\ terminal "y" is 16 volts higher in potential than terminal "x".
Example:
(i) Find V1 and V2
(ii) If R1 = 10 Wk and R2 = 2 Wk , Find I1 and I2.
Solution:
(i)
loop 1: start at Q P (clockwise): alg. å Vrises/drops = 0
40V - V1 + 10V - 18V + 10V = 0 \ V1 = 42 volts
loop 2: start at P P (clockwise): alg. å Vrises/drops = 0
18V - 10V +17V - V2 - 12V - 15V = 0
\ V2 = -2 volts
V1 = R1 . I1
--> I1 = V1 / R1
\ I1 = 42 volts / 10x103 ohms\ I1 = 4.2x10-3 amps = 4.2 mA
V2 = R2 . I2
--> I2 = V2 / R2
\ I2 = -2 volts / 2x103 ohms\ I2 = -1x10-3 A = -1 mA
In this course we will look at the following applications of systems of Linear Equations.
1) Loop Analysis of DC Circuits
2) Design of a Ring-Shunt Ammeter
3) Curve-fitting (Interpolating Polynomials)
4) Nodal Analysis of DC Circuits
5) Series / Parallel Circuits and Circuit Reduction Formulas
6) Power Dissipation in DC Circuit
7) Source Conversions
Loop Analysis (Background... Preamble)
" _ " one or more components connected...
- Define a loop current I1 , I2 , ...
- A loop current relates to actual physical currents in ways to be discussed below
- Need as many loop currents as we have indivisible small loops
- In setting up loop currents, need to ensure that every component gets included in at least one loop...
- We will arrange our direction of travel around each loop so that we can treat all resistors as "voltage drops" in our equations
- In situations where only a single loop current flows in a part of circuit, it will be easy to find actual physical current... see examples to follow...
- In situations where several loop currents flow thru a component, need to be careful...
net downward physical current is I1 + I2 = 2A + 3A = 5A
- We will generate our loop equations using KVL...
Loop Analysis Examples
Example:
(Two-Loop Circuit)
b) Determine the physical currents (magnitude & direction) in each resistor
c) Repeat problem using a different definition for loop currents (as will be described in class...)
Solution:
a) write equation...
alg. å Vrises/drops = 0
loop 1 P : 20V - 3 I1 - 4 (I1 - I2) = 0
loop 2 P : 0V - 4 (I2 - I1) - 5 I2 = 0
Simplifying:
7 I1 - 4 I2 = 20
-4 I1 + 9 I2 = 0
∆ =
7 | -4 |
-4 | 9 |
= (7) (9) - (-4) (-4) = 63 - 16 = 47
∆ = 47
∆1 =
20 | -4 |
0 | 9 |
= (20) (9) - (0) (-4) = 180 - 0 = 180
∆1 = 180
∆2 =
7 | 20 |
-4 | 0 |
= (7) (0) - (-4) (20) = 0 + 80 = 80
∆2 = 80
\ I1 = ∆1 / ∆ = 180 / 47 = 3.83A
\ I2 = ∆2 / ∆ = 80 / 47 = 1.70A
b) The physical currents (magnitude & direction) in each resistor
I3W = I1 = 3.83A ®
I4W = I1 - I2 = 3.83A - 1.70A = 2.13A ¯
I5W = I2 = 1.70A ¯
c) write equation...
alg. å Vrises/drops = 0
loop 1 P : 20V - 3 I1 - 4 (I1 + I2) = 0
loop 2 Q: 0V - 4 (I2 + I1) - 5 I2 = 0
Simplifying:
7 I1 + 4 I2 = 20
4 I1 + 9 I2 = 0
∆ =
7 | 4 |
4 | 9 |
= (7) (9) - (4) (4) = 63 - 16 = 47
∆ = 47
∆1 =
20 | 4 |
0 | 9 |
= (20) (9) - (0) (4) = 180 - 0 = 180
∆1 = 180
∆2 =
7 | 20 |
4 | 0 |
= (7) (0) - (4) (20) = 0 - 80 = -80
∆2 = -80
\ I1 = ∆1 / ∆ = 180 / 47 = 3.83A
\ I2 = ∆2 / ∆ = -80 / 47 = -1.70A
The physical currents (magnitude & direction) in each resistor
I3W = I1 = 3.83A ®
I4W = I1 + I2 = 3.83A + (-1.70A) = 2.13A ¯
I5W = I2 = 1.70A ¯
Example:
(Two-Loop Circuit)
a) Set up and solve the equations for the loop currents I1 and I2
b) Determine the physical currents (magnitude & direction) in each resistor
c) Find Vab
Solution:
a) write equation...
alg. å Vrises/drops = 0
loop 1 P: 1V - 1 I1 - 3 (I1 + I2) = 0
loop 2 Q: 2V - 3 (I2 + I1) - 2 I2 = 0
Simplifying:
4 I1 + 3 I2 = 1
3 I1 + 5 I2 = 2
∆ =
4 | 3 |
3 | 5 |
= (4) (5) - (3) (3) = 20 - 9 = 11
∆ = 11
∆1 =
1 | 3 |
2 | 5 |
= (1) (5) - (2) (3) = 5 - 6 = -1
∆1 = -1
∆2 =
4 | 1 |
3 | 2 |
= (4) (2) - (1) (3) = 8 - 3 = 5
∆2 = 5
\ I1 = ∆1 / ∆ = -1 / 11 = -0.091A
\ I2 = ∆2 / ∆ = 5 / 11 = 0.455A
b) The physical currents (magnitude & direction) in each resistor
I1W = I1 = 0.091A ¬
I3W = I1 + I2 = (-0.091A) + 0.455A = 0.364A ¯
I2W = I2 = 0.455A ¬
c) The physical currents (magnitude & direction) in each resistor
Vab = (1W) I1 = (1W) (-0.091A) = -0.091V
Example:
(Three-Loop Circuit)
a) Set up the loop equations which allow you to solve for the loop currents I1 , I2 and I3
b) Find Vab
Solution:
a) write equation...
alg. å Vrises/drops = 0
loop 1 P: 10V - 14 I1 - 6 (I1 - I2) = 0
loop 2 P: 0V - 6 (I2 - I1) - 2 I2 - 4 (I2 - I3) = 0
loop 3 P: - 4 (I3 - I2) - 4 I3 - 20 - 5 I3 = 0
Simplifying:
20 I1 - 6 I2 + 0 I3 = 10
-6 I1 + 12 I2 - 4 I3 = 0
0 I1 - 4 I2 + 13 I3 = -20
Need I3 ,... than Vab = 5W . I3
I3 , use determinants ... Cramer's...
∆ =
20 | -6 | 0 | 20 | -6 |
-6 | 12 | -4 | -6 | 12 |
0 | -4 | 13 | 0 | -4 |
= (20) (12) (13) + (-6) (-4) (0) + (0) (-6) (-4)
- (-6) (-6) (13) - (20) (-4) (-4) - (0) (12) (0)
= (3120) + (0) + (0)
- (468) - (320) - (0)
= 3120 - 788 = 2332
∆ = 2332
∆3 =
20 | -6 | 10 | 20 | -6 |
-6 | 12 | 0 | -6 | 12 |
0 | -4 | -20 | 0 | -4 |
= (20) (12) (-20) + (-6) (0) (0) + (10) (-6) (-4)
- (-6) (-6) (-20) - (20) (0) (-4) - (10) (12) (0)
= (-4800) + (0) + (240)
- (-720) - (0) - (0)
= -4560 + 720 = -3840
∆3 = -3840
Finally: I3 = ∆3 / ∆ = -3840 / 2332 = -1.65A
\ Vab = (5W) I3 = (5W) (-1.65A) = -8.25 volts
(Vba = -Vab = 8.25 volts)
Example:
For the circuit given, set up but do not solve the KVL equations for loop currents I1 , I2 and I3
Solution:
Apply KVL around each loop...
in form alg. å Vrises/drops = 0
loop 1 P: 10 - 10 I1 - 30 - 5 (I1 - I2) - 15 = 0
loop 2 P: 15 - 5 (I2 - I1) - 15 I2 + 50 - 20 I2 - 25 (I2 - I3) = 0
loop 3 P: - 30 I3 - 25 (I3 - I2) - 60 - 40 I3 = 0
...
15 I1 - 5 I2 + 0 I3 = -35
-5 I1 + 65 I2 - 25 I3 = 65
0 I1 - 25 I2 + 95 I3 = -60
Loop Analysis Continued:
Example:
Wheatstone Bridge Circuit
Solve for the circuit in Rs.
Solution:
Good choice for I1 , I2 , I3 since only need to find in order to find current in R5.
R5 is variable.
(KVL) å Vrises = å Vdrops
loop 1 P: 10 = 1 (I1 + I3) + 1 I1 + 3 (I1 - I2)
loop 2 P: 0 = 3 (I2 - I1) + 1 I2 + 1 (I2 + I3)
loop 3 P: 10 = 1 (I1 + I3) + 1 I3 + 1 (I2 + I3)
5 I1 - 3 I2 + 1 I3 = 10
-3 I1 + 5 I2 + 1 I3 = 0
1 I1 + 1 I2 + 3 I3 = 10
... Need only find I2 ...
∆ =
5 | -3 | 1 | 5 | -3 |
-3 | 5 | 1 | -3 | 1 |
1 | 1 | 3 | 1 | 1 |
= (5) (5) (3) + (-3) (1) (1) + (1) (-3) (1)
- (1) (5) (1) - (1) (1) (5) - (3) (-3) (-3)
= (75) + (-3) + (-3)
- (5) - (5) - (27)
= 69 - 37 = 32
∆ = 32
Also need
∆2 =
5 | 10 | 1 |
-3 | <><><><><><><><> 0 | 1 |
1 | 10 | 3 |
= (-1) (-3)
10 | 1 |
10 | 3 |
+ (+1) (0)
5 | 1 |
1 | 3 |
+ (-1) (1)
5 | 10 |
1 | 10 |
\∆2 = 3 (30 - 10) + 0 (15 - 1) - 1 (50 - 10)
\∆2 = 20
\ I2 = ∆2 / ∆ = 20 / 32 = 0.625A
\ Vab = R5 . I2 = 1W . (0.625) = 0.625 volts
Note: If the bridge is "balanced" then there will be no current flow in R5
In this case:
R1 R4 = R2 R3 or R1 / R3 = R2 / R4
\ if three of these resistors are known, the fourth can be determined.
Loop analysis method may be used to find values other than currents in a circuit, as in the following examples.
Example:
Solve for V and R in the following circuit
Solution:
Normally we are trying to find loop currents. But in certain design problems, we may know the current demands and want to infer other circuit component and source requirements...
two unknowns: V, R ...
Loop 1: V = 2 . I1 + R (I1 - I2)
or : V = 2 . (1) + R (1 - 0.5)
V = 2 + 0.5 R (1)
Loop 2: 0 = R (I2 - I1) + 1 . I2
or : 0 = R (0.5 - 1) + 1 . (0.5)
0 = - 0.5 R + 0.5 (2)
Any method could be employed to solve system (1) & (2):
go with substitution (2) ==> 0.5 R = 0.5 ==> R = 1 W
then sub in (1): V = 2 + (0.5) (1)
\ V = 2.5 volts
Shurt Cut Method for Loop Analysis
If all currents are chosen in the same direction (all clockwise or all counterclockwise) then certain patterns occur in the resistance coefficients:
Pattern for 4 loops would be:
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