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Circuits
What are circuits?
• You can think of a circuit as a pathway through
which electricity can travel.
• Circuits can be as simple or complex
– Your Night light
– Computer chip
• But they all work on the same principle
Flow of Electricity
• The electricity must
leave one end of the
power source and
return to the opposite
end in an unbroken
loop, or "circle." In the
case of a battery, the
electricity leaves the
negative (-) end and
returns to the positive
(+) end.
Flow of Electricity
• Called Current
– Symbol is I
• Units of Amperes
– 1 Amp = 1 C / s
• Current is the rate of flow of charge per unit
time

Q
I
t
dQ

dt
Components of Circuits
• Power source
– This supplies the voltage that causes the flow of
electrons from higher to lower potential
• Voltage is also known as Electric Potential Difference
– Think of voltage as lifting something up off the
ground. As you lift the object you are increasing
its potential energy. The higher the voltage the
higher the potential energy.
– Battery, Wall Outlet, Car Battery, etc.,
Other components
•
•
•
•
•
•
•
•
Switches
Resistors
Capacitors
LED’s
Integrated Circuits - - Op Amps
Inductors
Transistors
Transformers
Resistors
• Component used to control the current level in
various parts of the circuit
• Like the word implies, it resists the flow of
current
• All materials have a property called resistivity
• Resistance of a uniform material is given by:
R = ρL/A
Where R is the resistance measured in Ohms Ω, L is the
length of the material and A is the area
• Temperature dependant
Resistors
Capacitors
• Component of circuits that store electric
charges
• Two conductors separated by some distance
carrying opposite charge on their surface
• Ratio of the magnitude of the charge on either
conductor to the magnitude of the potential
difference between them:
Q
C
V
Capacitors
• Symbol is
• Measured in Farads which are C/V
• Material can be placed in between plates to
increase the Capacitance of the capacitor
– Dielectric
• The new capacitance is proportional to the original
capacitance
C '  Co
Capacitors
What a simple Circuit would look like
• This is called a RC circuit because it is
composed of a resistor and capacitor. (we’ll
get to this later)
Rules for Resistors in Circuits
• Resistors can be configured in series
– Series
• All resistors are lined up in a row. The current leaving
one resistor enters the next resistor
R1
R2
V
• To find the equivalent resistance you simply add the
values of the resistors
Req   Ri
1
Rules for Resistors in Circuits
• Resistors can be configured in Parallel
– Parallel
• Resistors are configured in different branches, so that
part of the current runs through one resistor and the
rest flows through the others.
R1
R2
V
Series Example
• 4 resistors are wired in series. The
resistor values are as follow:
– R1= 5 Ω R2=10 Ω R3=15 Ω R4 = 1
1) Draw the circuit
2) Find the equivalent resistance.
Rules for Resistors in Parallel
• To find the equivalent Resistance for Resistors
in parallel we take the inverse of the sum of
the reciprocal of the resistors

1
Req    
R
1
i 

1
For resistors in parallel the Req is always
less than the smallest resistor value
Parallel Example
• 3 resistors are wired in parallel. The
resistor values are as follow:
– R1= 5 Ω R2=10 Ω R3=15 Ω
1) Draw the circuit
2) Find the equivalent resistance.
Quick Review
1. Two 3W resistors placed in series would provide a resistance
which is equivalent to one _____W resistor.6
2. Three resistors with resistance values of 5W , 6W , and 7W
are placed in series. These would provide a resistance which
is equivalent to one _____W resistor.
18
3. Two resistors with resistance values of 3W and 6W are
placed in parallel. These would provide a resistance which is
equivalent to one _____W resistor.
2
4. Three resistors with resistance values of 3W , 4W , and 12W
are placed in parallel. These would provide a resistance
which is equivalent to one ____ W resistor.
1.5
Parallel and Series Combined
• In most circuits resistors are added in both
series and parallel
• In this case we have simplify all components
to find the equivalent resistance
Series and Parallel Example
Series and Parallel Example
Step (1):
Simplify the circuit in a step-by-step fashion by
combining groups of resistors in series or parallel to an
equivalent single resistor, thereby producing an
equivalent circuit which can be more easily solved.
For the example shown, two combinations will be
required.
R234
R234
• Now we have a simple resistors in series
problem
Parallel Series Example 2
R1= 15W R2 = 15W R3 = 10W R4 = 4W
Find the equivalent Resistance
Step 1: Find R23
Parallel: R23 = 6W
Parallel Series Example 2
R1= 15W R23 = 6W R4 = 4W
Find the equivalent Resistance
Step 2: Find R234
Series: R234 = 10W
Parallel Series Example 2
R2341
R1= 15W and R234 = 10W
Find the equivalent Resistance
Step 3: Find R2341 Parallel: R2341 = 6W
Very Important Rule
• Voltage = Current x Resistance
•
V
= I
x R
• If we know the voltage and resistance of a
circuit, we can find the total current running
through the circuit
– I = V/R
• After we find the equivalent Resistance, Req,
we use this to find the total current
Very Important Rule
• Current through resistors in series are the
same
– There is a voltage drop across each resistor!!
• Voltage across parallel resistors are the same
– A different amount of current runs through each
resistor in parallel!!
Example
Req=
9W
3W
6W
18V
Find the Req:
18V
Req = 3 + 6 = 9 W
Find the Total Current:
I = V/Req = 2 A
Find the Current through each Resistor:
2 A (series circuit)
Find the voltage drop across each resistor:
V3W  IR3 = (2)(3) = 6 V
V6W = IR6 = (2)(6) = 12 V
Example 2
Req=
2W
3W
18V
Find the Req:
6W
18V
Req = (1/3+1/6)-1= 2W
Find the Total Current:
I = V/Req = 9 A
Find the voltage drop across each resistor:
Find the Current through each Resistor:
18V (parallel circuit)
I3W  V/R3 = (18)/(3) = 6 A
I6W = V/R6 = (18)/(6) = 3 A
Power
• Power dissipated by a resistor is given by
Power = Current x Voltage
P=IxV
But V = I x R so we can also say that
2
IR
P=
P = V2/R
Example
3W
18V
6W
I = V/Req = 2 A
2 A through all resistors
both resistors
V3W  IR3 = (2)(3) = 6 V
V6W = IR6 = (2)(6) = 12 V
• What is the Power consumed by the resistors
• P3W = IV = (2)(6) = 12 Watts
• P6W = IV = (2)(12) = 24 Watts
Power Example 2
Req: 2W
Total Current: 9 A
voltage drop across each resistor:18V
Current through each Resistor: I3W 6 A
I6W = 3 A
• What is the Power consumed by the resistors
• P3W = IV = (6)(18) = 108 Watts
• P6W = IV = (3)(18) = 54 Watts
Solutions to HW#1
1.
2.
3.
4.
5.
6.
7.
8.
Req = 30 W
Req = 2.7273 W
Req = 42 W
Req = 2.03 W
Req = 12 W
Req = 10 W
Req = 20 W
Req = 5 W
Warm Up Review
To sum it all up
R1= 15W R2 = 15W R3 = 10W
R4 = 4W
Total Current: 2 Amps
R1:
R2:
R3:
I through R1 = 0.8 A
I through R2 = 0.48 A
I through R3 = 0.72 A
12 Volt Drop across R1
7.2 Volt Drop across R2
R4:
7.2 Volt Drop across R2
I through R4 = 1.2 A
4.8 Volt Drop across R4
LAB CHART
Ohm’s Law Lab Activity
Voltage
Current
Resistance
using Ohms
Law
Actual
Resistor
Value
% Error of
Resistance
Power
Consumed
by Resistor
Capacitors
• Component of circuits that store electric
charges
• Two conductors separated by some distance
carrying opposite charge on their surface
• Ratio of the magnitude of the charge on either
conductor to the magnitude of the potential
difference between them:
Q
C
V
Capacitors
• Symbol is
• Measured in Farads which are C/V
• Material can be placed in between plates to
increase the Capacitance of the capacitor
– Dielectric
• The new capacitance is proportional to the original
capacitance
C '  Co
Capacitors
What about capacitors
• Similar to resistors we can combine capacitors
in parallel and in series
• We can add capacitors to find an equivalent
capacitance
Capacitors in Parallel
• For capacitors in parallel we can simply just add
the value of capacitors.
Ceq   Ci
Capacitor in Parallel Example
• What is the equivalent capacitance of the
following circuit if C1 = 4mF C2 = 3mF and C3 =
12mF
C3
Ceq = 19mF
Capacitors is Series
• Capacitors in series, we find the
inverse of the sum of the reciprocal
of the capacitor values

1
Ceq    
 1 Ci 
1
Capacitor in Series Example
• Find the equivalent resistance if C1= 5 mF and
C2 = 20 mF
Ceq = 4mF
Parallel and Series Combined
• In most circuits capacitors are added in both
series and parallel
• In this case we have simplify all components
to find the equivalent capacitance
• Very Similar to Combined Resistor Circuits
Example
Find the equivalent
capacitor
C = 4mF
C = 3mF
C = 2mF
C = 3mF
C = 6mF
• First we simplify the capacitors in parallel to
get two capacitors in series
• Finally we can the inverse of the summation of
the reciprocals of the two capacitors
Ceq=2mF
Example
10F
18F
10F
8F
9F
15F
Find the equivalent capacitance
Ceq= 30F
And there’s more…..
• We can find the charge and voltage drop
across each capacitor
Example
• Find the total charge on each capacitor
C = 4mF
C = 3mF
C = 2mF
18 V
We Use…….
Q = VC
Example
C = 4mF
C = 3mF
C = 2mF
18 V
• First we need to find the equivalent
capacitance
• Ceq = 2 mF
Example
Equivalent Circuit
C = 2mF
18 V
• Then we use Q = VC to find the total charge in
the circuit
• Q = (18V)(2mF) = 36mC
Series Capacitors
Charge on Series Capacitors
Since charge cannot be added or
taken away from the conductor
between series capacitors, the
net charge there remains zero.
This charge Q is the charge you
get by calculating the
equivalent capacitance
Q = V x Ceq
Example
Both have
36mC on
them
C = 6mF
C = 3mF
18 V
Equivalent
capacitance
from parallel
capacitors
• Since these two capacitors are in series both
capacitors have 36mC of charge on them.
• There is a voltage drop across the 3mF capacitor
which we find by solving V = Q/C
• V = 36mC/3mF = 12 volts
Finally
C = 4mF
36mC
C = 3mF
C = 2mF
18 V
• This leaves 6 volts to drop across the capacitors in
parallel. Since 6 volts drops across both capacitors
we find the total charge on each by using Q = VC
• Q4mF=(4mF)(6V) = 24mC
• Q2mF=(2mF)(6V) = 12mC
Sum up to
36mC
Example 2
Finding the total charge on each
capacitor
10F
18F
10F
8F
9F
We use…
15F
Q = VC
Finding the total charge on each
capacitor
32V
10F
18F
10F
8F
9F
15F
Ceq = 30F so the total charge
Qtotal = VCeq=(32)(30) = 960C
On each capacitor
The same voltage drop across each branch
32V
10F
18F
10F
8F
9F
15F
For capacitors in series, we use the equivalent
capacitance of those capacitors
On each capacitor
The same voltage drop across each branch
WE USE: Q=VC
32V
10F
18F
10F
8F
9F
15F
Ceq =6F
Q18 = 32 x 6 = 192 C
Q9 = 32 x 6 = 192 C
Q8= 8 x 32 = 256C
Ceq =6F
Q10 = 32 x 6 = 192 C
Q15 = 32 x 6 = 192 C
Q10= 10 x 32 = 320C
On each capacitor
The same voltage drop across each branch
32V
10F
18F
10F
8F
9F
256C
15F
192C
320C
192C
If we add up all the charge:
192 + 256 + 192 + 320 = 960C
Multi-loop Circuits and Kirchoff's Rules
• Before talking about
what a multi-loop
circuit is, it is helpful
to define two terms,
– Junction
– Branch
Junctions
A junction is a point where at least three
circuit paths meet.
A branch is a path connecting two junctions.
• circuits where there are more than one battery in
different branches are known as Multi-Loop Circuits
• To analyze the current in each branch we use
Kirchoff’s Rules
What we can do with these Rules
• We can find the current in all branches of a
multi-loop circuit
• We can find the EMF (Voltage) of a battery
• We can find the value of a resistor
Kirchoff’s Rules
• Kirchoff's first rule
– Junction rule: The sum of the currents coming in
to a junction is equal to the sum leaving the
junction. (Conservation of charge)
• Kirchoff's second rule
– Loop Rule: The sum of all the potential
differences around a complete loop is equal to
zero. (Conservation of energy)
Analyzing a Circuit
• Label the current and the current direction in each
branch.
– Direction you choose does not matter. Simply pick a
direction.
– If you guess wrong, you¹ll get a negative value. Your value
will still be correct but the direction guess will be reversed
• Use Kirchoff's first rule to write down current
equations for each junction that gives you a different
equation.
What we do
R2
I1
R4
R3
V1
R1
I2
V2
• At points a and b we have junctions
• Rule 1 at “a”
I2+I3=I1
• Rule 1 at “b” will yield the same equation
V3
I3
Analyzing a Circuit
• Use Kirchoff's second rule to write down loop
equations for as many loops as it takes to include
each branch at least once.
• To write down a loop equation, you choose a
starting point, and then walk around the loop in one
direction until you get back to the starting point. As
you cross batteries and resistors, write down each
voltage change. Add these voltage gains and losses
up and set them equal to zero.
Analyzing a Circuit
• When you cross a battery from the - side to the +
side, that's a positive change. Going the other way
gives you a drop in potential, so that's a negative
negative change
change.
-
+
Positive change
• Crossing a resistor is also a change in potentials.
With the current is a negative change and against the
current is a positive change
What we do
R2
I1
R4
R3
V1
R1
I2
V3
V2
• Loop 1: +I1R1+V1+I1R2+I2R3-V2 = 0
• Loop 2: +V2 – I2R3+I3R4-V3 = 0
I3
Finally
• We know have three equations with three
unknowns. We can use linear algebra to solve
for I1, I2, I3
+I1R1+V1+I1R2+I2R3 = 0
+V2 – I2R3+I3R4-V3 = 0
I2+I3=I1
Example
1W
3W
2W
4V
12V
4W
1V
Example
10 W
5W
10 W
10V
15V
4W
2V
HW 2 Solutions
• Number 1
– Req = 12.85714 Ohms
– Itotal = 1.5556 A
R8: 12.44 Voltage Drop 1.556 Amps
R1: 1.111 Voltage Drop 1.111 Amps
R3: 3.333 Voltage Drop 1.111 Amps
R10: 4.4449 Voltage Drop 0.444 Amps
R2: 3.111 Voltage Drop 1.556 Amps
HW 2 Solutions
• Number 2
Req = 22 Ohms Itotal = 3 Amps
R10: 15 Volt Drop 1.5 Amps
R5: 15 Volt Drop 3 Amps
R6: 6 Volt Drop 1 Amps
R3: 6 Volt Drop 2 Amps
R4: 12 Volt drop 3 Amps
R15: 18 Volt Drop 1.2 Amps
R10: 18 Volt Drop 1.8 Amps
HW 3 Solutions
1.
2.
3.
4.
5.
Ceq=(1/50 + 1/80,000)-1 =4.999 x 10-5 F
Ceq= 90nF + .03mF = 1.2 x 10-7 F
Ceq= (4+2+1)mF = 7mF
Ceq=(1/(4+2)+1/3)-1mF = 2 mF
Ceq= (1/5mF+1/12mF + 1/(4mF + 6mF + 8mF))1
= 2.951 mF
6. Ceq=54mF + (1/72mF + 1/87mF)-1= 9.34 x 10-5 F
HW 3
7) Qt=.04426 C
Q1 =.04426 C
Q2 =.04426 C
Q3 = .00984 C
Q4 = .01475 C
Q5 = .01967
8) Qt=.00934 C
Q54 =. 0054 C
Q72 =.00394 C
Q87 = .00394 C
RC LAB
Is an electric circuit composed of
resistors and capacitors driven by a
voltage or current source.
RC Circuit Equations
• RC circuits have a charging/discharging
constants.
– Luckily they are the same
– Simply the product of the resistor and capacitor
values
Phet
Determining the Charge, Voltage, and
current in the circuit
• Kirchoff’s Rules allows us to solve the circuit to
understand how these values
• Add a little calculus into it and you get the
equations to describe the circuit