Download Kirchhoff*s Rules

Kirchhoff’s Rules
• Limits of Progressive Simplification
• Kirchhoff’s Rule
• Simple Examples
• More complicated examples
• Capacitors in series/parallel
Progressive Simplification
• Simplify down
– Rtot = 10.3 Ω
– Ibat = 0.87 A
• Build back up
– Voltage across top
= 9 𝑉 − .87 𝐴 .5 Ω −
.87 𝐴 5Ω
– Current through 10 Ω
– Current through 8.7 Ω
– Current through 6 Ω
– Voltage across 4/8 Ω
Kirchhoff’s Rules
• Progressive simplification doesn’t work on some circuits.
– Can combine 1 Ω and 20 Ω, and 1 Ω and 40 Ω, but that’s it!
Kirchhoff’s Rules
Declare each branch has it’s own separate current
1. At any junction, current entering = current leaving
2. For any complete loop, sum of voltage rises/drops equals 0
• Currents I1, I2, I3
• Canal analogy
Example – Problem 23
• Only 1 current I
9 𝑉 − 𝐼 8 Ω − 𝐼 12 Ω − 𝐼 2 Ω = 0
𝐼=
9𝑉
22 Ω
= 0.41 𝐴
• Just like we’ve done all along.
Example – Problem 24
• Two batteries backwards (charging circuit)
• Only 1 current I
18 𝑉 − 𝐼 6.6 Ω − 12 𝑉 − 𝐼 2 Ω − 𝐼 1 Ω = 0
𝐼=
6𝑉
9.6 Ω
= 0.625 𝐴
• Note we went across 12 V battery backwards.
Example – Problem 24 (1)
• Branch point!
I1
• Draw currents I1, I2, I3 as shown
– Guess at direction
• Kirchhoff Rule #1 at point a
𝐼1 = 𝐼2 + 𝐼3
• Kirchhoff Rule #1 same at point b
𝐼1 = 𝐼2 + 𝐼3
a
I2
I3
b
Example – Problem 24 (2)
• Kirchhoff Rule #2 around upper loop
I1
9 𝑉 − 𝐼2 15 Ω − 𝐼1 22 Ω = 0
–
Note separate I1 I2
• Kirchhoff Rule #2 around lower loop
6 𝑉 + 𝐼2 15 Ω = 0
–
a
I2
b
Note (+) going across resistor backwards
• Kirchhoff Rule #2 around outer loop
I3
9 𝑉 + 6 𝑉 − 𝐼1 22 Ω = 0
General rule: battery (+) and resistor (-), but flip sign when you go across backwards
Example – Problem 24 (3)
•
Outer loop
I1
9 𝑉 + 6 𝑉 − 𝐼1 22 Ω = 0
15 𝑉
𝐼1 = 22 Ω = 0.68 𝐴
•
Lower loop
6 𝑉 + 𝐼2 15 Ω = 0
𝐼2 =
•
−6 𝑉
15 Ω
= −0.4 𝐴 *
Current Rule
𝐼1 = 𝐼2 + 𝐼3
•
a
𝐼3 = 𝐼1 − 𝐼2 = 1.08 𝐴
Upper loop
9 𝑉 − 𝐼2 15 Ω − 𝐼1 22 Ω = 0
9 𝑉 − −0.4 𝐴 15 Ω − 0.68 𝐴 22 Ω = 0
*assumed wrong direction for I2 but that’s OK
I2
I3
b
Example 19-3 (1)
1.
2.
3.
4.
Draw I1, I2, I3
Set node rule (Kirchhoff 1)
Set 2 loops (Kirchhoff 2)
Solve simultaneous equations
• Node rule from diagram
𝐼3 = 𝐼1 + 𝐼2
• Upper loop
45 𝑉 − 𝐼3 1Ω − 𝐼3 40 Ω − 𝐼1 30 Ω = 0
• Lower loop
80 𝑉 − 𝐼2 1Ω − 𝐼2 20 Ω + 45 𝑉 − 𝐼3 1Ω − 𝐼3 40 Ω = 0
3 variables, 3 simultaneous equations
Example 19-3 (2)
• 3 variables, 3 simultaneous equations
𝐼3 = 𝐼1 + 𝐼2
45 𝑉 − 𝐼3 1Ω − 𝐼3 40 Ω − 𝐼1 30 Ω = 0
80 𝑉 − 𝐼2 1Ω − 𝐼2 20 Ω + 45 𝑉 − 𝐼3 1Ω − 𝐼3 40 Ω = 0
•
Simplify
𝐼3 = 𝐼1 + 𝐼2
45 𝑉 − 𝐼3 41Ω − 𝐼1 30 Ω = 0
125 𝑉 − 𝐼2 21 Ω − 𝐼3 41 Ω = 0
•
My suggestion
𝐼3 = 𝐼1 + 𝐼2
45
41
𝐼1 = 30 − 30 𝐼3
𝐼2 =
125
41
−
𝐼
21
21 3
Example 19-3 (3)
•
•
Calculating
𝐼3 = 𝐼1 + 𝐼2
(1)
𝐼1 = 1.5 − 1.37 𝐼3
(2)
𝐼2 = 5.95 − 1.95 𝐼3
(3)
Substitute eqns. (2) and (3) in (1)
𝐼3 = 1.5 − 1.37 𝐼3 + 5.95 − 1.95 𝐼3
•
Simplify
4.32 𝐼3 = 7.45
𝐼3 = 1.72 𝐴
•
Plug in (2) and (3) above
𝐼1 = 1.5 − 1.37 𝐼3 = −0.86 𝐴
𝐼2 = 5.95 − 1.95 𝐼3 = 2.60 𝐴
(assumed wrong direction, that’s OK)
Example 19-3 (textbook solution)
1.
2.
3.
4.
Draw I1, I2, I3
Set node rule (Kirchhoff 1)
Set 2 loops (Kirchhoff 2)
Solve simultaneous equations
• Node rule from diagram
𝐼3 = 𝐼1 + 𝐼2
• Upper loop
45 𝑉 − 𝐼3 1Ω − 𝐼3 40 Ω − 𝐼1 30 Ω = 0
• Outer loop
80 𝑉 − 𝐼2 1Ω − 𝐼2 20 Ω + 𝐼1 30Ω = 0
This solution gives same answer!
Note sign flip, going across resistor backwards (against assumed current)
Elevation analogy
Capacitors in series and parallel
• Parallel
– Charges add
– Voltage same
𝑄𝑡𝑜𝑡 = 𝑄1 + 𝑄2 + 𝑄3 = 𝐶1 𝑉 + 𝐶2 𝑉 +𝐶3 𝑉
= 𝐶1 + 𝐶2 + 𝐶3 𝑉 = 𝐶𝑒𝑞 𝑉
– Equivalent parallel capacitance 𝐶𝑡𝑜𝑡 = 𝐶1 + 𝐶2 + 𝐶3
• Series
– Voltages add
– Charge same
𝑄
𝑄
𝑄
1
2
2
𝑉𝑡𝑜𝑡 = 𝑉1 + 𝑉2 + 𝑉3 = 𝐶 + 𝐶 + 𝐶
=𝑄
1
𝐶1
1
1
2
3
+𝐶 +𝐶
– Equivalent series capacitance
𝑄
=𝐶
𝑒𝑞
1
𝐶𝑒𝑞
1
1
1
1
2
3
=𝐶 +𝐶 +𝐶
series/parallel rules reversed from resistors!
Summary – resistors/capacitors in series/parallel
• Resistors
𝑅𝑠𝑒𝑟𝑖𝑒𝑠 = 𝑅1 + 𝑅2 + 𝑅3
1
𝑅𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
=
1
𝑅1
1
+
𝑅2
1
+
𝑅3
• Capacitors
𝐶𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 = 𝐶1 + 𝐶2 + 𝐶3
1
𝐶𝑠𝑒𝑟𝑖𝑒𝑠
=
1
𝐶1
1
+
𝐶2
+
1
𝐶3
Examples – Capacitors in series and parallel