Download I 1 R 1 - I 2 R 2

Simple Circuits
&
Kirchoff’s Rules
Parallel Circuit
Series Circuit
Simple Series Circuits
 Each device occurs one after the other
sequentially.
 The Christmas light dilemma: If one light goes
out all of them go out.
R1
V+
R2
R3
Simple Series Circuit Conservation of Energy
 In a series circuit, the sum of the voltages is equal
to zero.
Vsource + V1 + V2 + V3 = 0
Where we consider the source voltage to be
positive and the voltage drops of each device to
be negative.
Vsource = V1 + V2 + V3
Since V = IR (from Ohm’s Law):
V1
Vsource = I1R1 + I2R2 + I3R3
V+
V3
V2
Simple Series Circuit Conservation of Charge
 In a series circuit, the same amount of
charge passes through each device.
IT = I 1 = I 2 = I 3
R1
V+
R2
I
R3
Simple Series Circuit – Determining
Requivalent
 What it the total resistance in a
series circuit?
 Start with conservation of energy
Vsource = V1 + V2 + V3
Vsource = I1R1 + I2R2 + I3R3
 Due to conservation of charge, ITotal = I1 = I2
= I3, we can factor out I such that
Vsource = ITotal (R1 + R2 + R3)
 Since Vsource = ITotalRTotal:
RTotal = REq = R1 + R2 + R3
Simple Parallel Circuit
 A parallel circuit exists where components are
connected across the same voltage source.
 Parallel circuits are similar to those used in
homes.
V+
R1
R2
R3
Simple Parallel Circuits
 Since each device is connected across
the same voltage source:
Vsource = V1 = V2 = V3
V+
V1
V2
V3
Simple Parallel Circuits Analogy
How Plumbing relates to current
 In parallel circuits, the total current is equal to
the sum of the currents through each individual
leg.
 Consider your home plumbing:
 Your water comes into the house under pressure.
 Each faucet is like a resistor 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 current from the voltage source
(pressurized water supply) is equal to the sum
of the currents (flow of water through faucet
and drain) in each of the resistors (faucets)
ITotal = I1 + I2 + I3
ITotal
V+
I1
ITotal
I2
I3
Simple Parallel Circuit –
Determining Requivalent
 What it the total resistance in a
parallel circuit?
 Using conservation of charge
ITotal = I1 + I2 + I3
or
V1 V2 V3
I total  

R1 R2 R3
(1)
Vsource Vsource Vsource
I total 


R1
R2
R3
(2)
 Since Vsource = V1 = V2 = V3 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 = Vsource/RTotal substitute in
(2) as follows
Vsource Vsource Vsource Vsource



Rtotal
R1
R2
R3
(2)
 Since Vsource cancels, the relationship reduces to
1
1
1
1



Req R1 R2 R3
Note: Rtotal has been replaced by Req.
(3)
Kirchoff’s Rules
 Loop Rule (Conservation of Energy):
 The sum of the potential drops
(Resistors) equals the sum of the
potential rises (Battery or cell) around
a closed loop.
 Junction Rule (Conservation of
Electric Charge):
 The sum of the magnitudes of the
currents going into a junction equals the
sum of the magnitudes of the currents
leaving a junction.
Rule #1: Voltage Rule
(Conservation of Energy)
V  0
Loop
R1
V+
ΣV
R2
R3
Vsource – V1 – V2 – V3 = 0
Rule #2: Current Rule
(Conservation of Electric Charge)

I 0
Junction
I1
I2
I3
I1 + I2 + I3 = 0
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
Voltage Rule.
 Arbitrarily choose a direction for the current to
flow in each loop and apply Kirchoff’s Junction
Rule.
Ex. (cont.)
 Apply Kirchoff’s Current Rule (Iin = Iout):
I1 + I2 = I 3
(1)
 Apply Kirchoff’s Voltage Rule to the left
loop (Σv = 0):
1 – V1 – V2 = 0
1 – I1R1 – I3R2 = 0
 Substitute (1) for I3 to obtain:
1 – I1R1 – (I1 + I2)R2 = 0
(2)
Ex. (cont.)
 Apply Kirchoff’s Voltage Rule to the right
loop:
2 – V3 – V2 = 0
2 – I2R3 – I3R2 = 0
 Substitute (1) for I3 to obtain:
2 – I2R3 – (I1 + I2)R2
=0
(3)
Ex. (cont.)
 List formulas to analyze.
I1 + I2 = I3
1 – I1R1 – (I1 + I2)R2 = 0
2 – I2R3 – (I1 + I2)R2 = 0
 Solve 2 for I1 and substitute into (3)
1 – I1R1 – I1R2 – I2R2 = 0
– I1R1 – I1R2 = I2R2 – 1
I1 (R1 + R2) = 1
I 1 = 1
- I2R2
(R1 + R2)
- I2R2
(1)
(2)
(3)
Ex. (cont.)
[
 – [
1
I2R2
+ I2 R2 = 0
[
2 – I2R3
(1 - I2R2)
(R1 + R2)
[
2 – I2R3 –
(R1 + R2)
R2 – I2R2 = 0
Multiply by (R1 + R2) to remove from denominator.
2 (R1 + R2) – I2R3 (R1 + R2) – 1R2 + I2R22 – I2R2 (R1 + R2) = 0
 Plug in known values for R1, R2, R3, 1 and 2 and
then solve for I2 and then I3.
5V(5Ω+10Ω) – I25Ω (5Ω+10Ω) – 3V(10Ω) + I2(10Ω)2 – I210Ω (5Ω+10Ω) = 0
I2 = 0.36 A
Ex. (cont.)
 Plug your answer for I2 into either
formula to find I1
 1 – I1R1 – (I1 + I2)R2 = 0
1 - I2R2
I1 =
I1 =
(R1 + R2)
3V – (0.36A)(10)
(5 + 10)
I1 = -0.04A
 What does the negative sign tell you about the
current in loop 1?
Ex. (cont.)
 Use formula (1) to solve for I3
 I1 + I2 = I3
 -0.04A + 0.36A = 0.32A
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
R1 R1
- I1 (  )
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= 5
R2=3
R3=4
and
I1=2.809 A
I2= 3.319 A, I3= -0.511 A