Transcript Simple Circuits & Kirchoff’s Rules

Simple Circuits
&
Kirchoff’s Rules
Simple Series Circuits
 Each device occurs
sequentially.
 The
light dilemma: If
light goes
all of them go
.
Simple Series Circuit Conservation of Energy
 In a series circuit, the
equal to
.
of the
Vsource +
Where we consider the source voltage to be
and the voltage drops of each
device to be
.
Vsource =
Since V =
(
):
Vsource =
is
Simple Series Circuit Conservation of Charge
 In a series circuit, the same amount of
passes through each device.
IT =
R1
V+
R2
R3
Simple Series Circuit – Determining
Requivalent
 What it the total
in a series circuit?
 Start with
of
Vsource =
Vsource =
 Due to conservation of charge, ITotal = I1 = I2
= I3, we can factor out I such that
Vsource =
 Since Vsource = :
RTotal = REq =
Simple Parallel Circuit
 A parallel circuit exists where components are
connected across the same
.
 Parallel circuits are similar to those used in
.
V+
Simple Parallel Circuits
 Since each device is connected across
the
:
Vsource =
V+
Simple Parallel Circuits Analogy
How Plumbing relates to current
 In parallel circuits, the
equal to the
of the
each individual leg.
is
through
 Consider your home plumbing:
 Your water comes into the house under pressure.
 Each faucet is like a
that occupies a leg in
the circuit. You turn the valve and the water flows.
 The drain reconnects all the faucets before they go
out to the septic tank or town sewer.
 All the water that flows through each of the faucets
adds up to the total volume of water coming into the
house as well as that going down the drain and into
the sewer.
 This analogy is similar to current flow through a
parallel circuit.
Simple Parallel Circuits –
Conservation of Charge & Current
 The total
from the voltage source
(pressurized water supply) is equal to the sum
of the
(flow of water through faucet
and drain) in each of the
(faucets)
ITotal =
V+
Simple Parallel Circuit –
Determining Requivalent
 What it the total resistance in a
parallel circuit?
 Using conservation of charge
ITotal =
or
 Since Vsource = V = V = V we can
substitute Vsource in (1) as follows
Simple Parallel Circuit –
Determining Requivalent
 What it the total resistance in a
parallel circuit (cont.)?
 However, since ITotal =
in (2) as follows
/
substitute
 Since Vsource cancels, the relationship reduces to
Note: Rtotal has been replaced by
.
Kirchoff’s Rules
 Loop Rule (Conservation of
):
 The sum of the
(
)equals the sum of the
(
)
around a closed loop.
 Junction Rule (Conservation of
Electric
):
 The sum of the magnitudes of the
going into a junction equals
the sum of the magnitudes of the
leaving a junction.
Rule #1: Voltage Rule
(Conservation of
R1
V+
ΣV
R2
R3
Vsource – V1 – V2 – V3 =
)
Rule #2: Current Rule
(Conservation of Electric
I1
I2
I3
I1 + I2 + I3 =
)
Example Using Kirchoff’s Laws
R1 = 5Ω
1 = 3V +
I1
I2
R2 = 10Ω
R3 = 5Ω
+
I3
2 = 5V
 Create individual loops to analyze by Kirchoff’s
.
 Arbitrarily choose a direction for the current to
flow in each loop and apply Kirchoff’s
.
Ex. (cont.)
 Apply Kirchoff’s Current Rule (
I1 + I2 =
=
):
(1)
 Apply Kirchoff’s Voltage Rule to the left
loop (Σv =
):
1 –
1 –
–
=
–
=
 Substitute (1) to obtain:
1 –
–(
)
=
(2)
Ex. (cont.)
 Apply Kirchoff’s Voltage Rule to the right
loop:
2 –
2 –
–
2 –
–(
=
–
=
 Substitute (1) to obtain:
)
=
(3)
Ex. (cont.)
 List formulas to analyze.
I1 + I2 =
1 –
2 –
–(
1 –
–
(1)
)
=
–(
) =
 Solve 2 for I1 and substitute into (3)
–
–
1 -
(
=
– 1
=
) = 1
I1 (
I1 =
–
-
)
(2)
(3)
Ex. (cont.)
(
(
) +
)
[
2 –
[
– [
–
(
)
[
2 –
=
–
=
Multiply by (R1 + R2) to remove from denominator.
2 (
)–
(
)–
+
–
(
)=0
 Plug in known values for R1, R2, R3, 1 and 2 and
then solve for I2 and then I3.
I2 =
A
Ex. (cont.)
 Plug your answer for I2 into either
formula to find I1
 1 –
I1 =
I1 =
–(
)
=
)(
)
1 (
)
3V – (
(
+
)
I1 =
 What does the
the current in loop 1?
tell you about
Ex. (cont.)
 Use formula (1) to solve for I3
 I1 + I2 =

+
=
How to use Kirchhoff’s Laws
A two loop example:
R1
I3
1
2
I2
I1
R2
R3
• Analyze the circuit and identify all circuit nodes
and use KCL.
(1) I1 = I2 + I3
• Identify all independent loops and use KVL.
(2) 1 - I1R1 - I2R2 = 0
(3) 1 - I1R1 - 2 - I3R3 = 0
(4) I2R2 - 2 - I3R3 = 0
How to use Kirchoff’s Laws
R1
I3
2
I2
I1
1
R2
R3
• Solve the equations for I1, I2, and I3:
First find I2 and I3 in terms of I1 :
I 2  (1 - I1R1 ) / R2
From eqn. (2)
I3  (1 -  2 - I1R1 ) / R3
Now solve for I1 using eqn. (1):
I1 
1
R2

1 -  2
R3
- I1 (
R1 R1
 )
R2 R3
1

R2

From eqn. (3)
1 -  2
R3
I1 
R R
1 1  1
R2 R3
Let’s plug in some numbers
R1
2
I2
I1
1
1 = 24 V
I3
R2
 2 = 12 V
Then,
R3
R1= 5W
R2=3W
R3=4W
and
I1=2.809 A
I2= 3.319 A, I3= -0.511 A