Title: Simple Circuits
1Simple CircuitsKirchoffs Rules
2Simple Series Circuits
- Each device occurs sequentially.
- The light dilemma If light goes
all of them go .
3Simple Series Circuit - Conservation of Energy
- In a series circuit, the of the is equal to
. - Vsource
- Where we consider the source voltage to be
and the voltage drops of each device to
be . - Vsource
- Since V ( )
- Vsource
4Simple Series Circuit - Conservation of Charge
- In a series circuit, the same amount of
passes through each device. - IT
5Simple 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
6Simple Parallel Circuit
- A parallel circuit exists where components are
connected across the same . - Parallel circuits are similar to those used in
.
V
7Simple Parallel Circuits
- Since each device is connected across the
- Vsource
V
8Simple Parallel Circuits AnalogyHow Plumbing
relates to current
- In parallel circuits, the is equal to
the of the through each individual
leg. - 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.
9Simple 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
10Simple 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
11Simple Parallel Circuit Determining Requivalent
- What it the total resistance in a parallel
circuit (cont.)? - However, since ITotal / substitute in (2)
as follows - Since Vsource cancels, the relationship reduces to
Note Rtotal has been replaced by .
12Kirchoffs 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.
13Rule 1 Voltage Rule (Conservation of )
Vsource V1 V2 V3
14Rule 2 Current Rule (Conservation of Electric
)
I1
I2
I3
I1 I2 I3
15Example Using Kirchoffs Laws
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- Create individual loops to analyze by Kirchoffs
. - Arbitrarily choose a direction for the current to
flow in each loop and apply Kirchoffs .
16Ex. (cont.)
- Apply Kirchoffs Current Rule ( )
- I1 I2 (1)
- Apply Kirchoffs Voltage Rule to the left loop
(Sv ) - ?1
- ?1
- Substitute (1) to obtain
- ?1 ( ) (2)
17Ex. (cont.)
- Apply Kirchoffs Voltage Rule to the right loop
- ?2
- ?2
- Substitute (1) to obtain
- ?2 ( ) (3)
18Ex. (cont.)
- List formulas to analyze.
- I1 I2 (1)
- ?1 ( ) (2)
- ?2 ( ) (3)
- Solve 2 for I1 and substitute into (3)
- ?1
- ?1
- I1 ( ) ?1 -
- ?1 -
- ( )
19Ex. (cont.)
- ( ) ( )
- Plug in known values for R1, R2, R3, ?1 and ?2
and then solve for I2 and then I3.
Multiply by (R1 R2) to remove from denominator.
?2 ( ) ( )
( ) 0
I2 A
20Ex. (cont.)
- Plug your answer for I2 into either formula to
find I1 - ?1 ( )
- What does the tell you about the
current in loop 1?
I1
21Ex. (cont.)
- Use formula (1) to solve for I3
- I1 I2
-
22How to use Kirchhoffs Laws
A two loop example
- Analyze the circuit and identify all circuit
nodes and use KCL.
(1) I1 I2 I3
- Identify all independent loops and use KVL.
(2) e1 - I1R1 - I2R2 0 (3) e1 - I1R1 - e2 -
I3R3 0 (4) I2R2 - e2 - I3R3 0
23How to use Kirchoffs Laws
- Solve the equations for I1, I2, and I3
- First find I2 and I3 in terms of I1
Now solve for I1 using eqn. (1)
24Lets plug in some numbers
e1 24 V
e 2 12 V
R1 5W R23W R34W
Then, and
I12.809 A I2 3.319 A, I3
-0.511 A