Download AP Physics 2 Electrical Circuits 2015-16

Drawing a Circuit
• When illustrating an electrical circuit
for everyone to understand, you must
draw a standardized picture called a
schematic.
• A schematic is simply a picture of an
electrical circuit that uses
standardized symbols to represent
the different parts of an electrical
circuit.
Basic Schematic
Symbols
Battery/ Voltage source
Resistor
Switch
Fuse
Series Circuit
• Because a series
circuit has only one
loop, the amount of
current that flows
through one
resistor must be
the same amount of
current that flows
for all the other
resistors as well.
.5 A
.5 A
.5 A
.5 A
Series Circuit
• Because no one
resistor is
connected directly
to both terminals of
the power source.
No one resistor
receives all of the
voltage, the voltage
is divided out
amount each of the
resistors
50 V
30 V
20 V
100 V
Kirchhoff’s Voltage Rule
• We know from the conservation from energy that
all the energy that the power source provides must
equal the total work that each resistor does.
• So then we can say the the energy that each charge
has must to be used by all the resistors.
• Since voltage is the energy of one charge, we can
say that the voltage of the power supply must be
used up by all the resistors.
• Kirchhoff’s Voltage law
– Vpower supply = Sum of all Voltage drops
– Vin = VR1 + VR2 + VR3 + VR4 ….
Simplifying a Series Circuit
• When trying to analyze a series circuit, normally the first
step is to reduce the circuit of several resistors to an
equivalent circuit of only 1 resistor and 1 power source
• For any simplified circuits we will have a/an
– Equivalent Voltage (Veq) - How much voltage is truly
being supplied to the circuit
– Equivalent Current (Ieq) - How much current the power
source has entering the circuit
– Equivalent resistance (Req) - How much resistance the
circuit actually has
– Equivalent power (Peq) - The total power of the circuit
Original Circuit
Equivalent Circuit
Lets start with
a series circuit
with 3 resistors
R2
R3
R1
Vin
We can simply a series circuit by using 3 basic principles
Kirrcoff’s Voltage law:
VEq = VR1 + VR2 + VR3
Because there is 1 pathway for current in a series circuit
the current for all resistors is the same.
IEq = IR1 = IR2 = IR3
But we will still need to find the equivalent resistance.
For that we will need to use Ohm’s Law: V=IR
Lets start with
a series circuit
with 3 resistors
Because:
And since:
We can say:
R2
R3
R1
Vin
Veq = VR1 + VR2 + VR3
V = IR
IeqReq = IR1R1 + IR2R2 + IR3R3
Because All I’s are the same we call each one I
We can say:
IReq = IR1 + IR2 + IR3
The current is a common multiple on both sides
Now we have an equation for the equivalent
Resistance: Req = R1 + R2 + R3
Equations for a Series
Circuit
• Equivalent Voltage
– Veq = VR1 + VR2 + VR3….
• Equivalent Current
– IEq = IR1 = IR2 = IR3….
• Equivalent Resistance
– Req = R1 + R2 + R3…
• Equivalent power
– PowEq = PowR1 +PowR2 + PowR3 …
Sample problem
For the circuit shown what is the equivalent resistance?
How much current leaves the battery?
For a series Req = R1 + R2 + R3
Req = 10 W + 10 W + 10 W
Req = 30 W
Using Ohms Law:
300 V = I(30 W)
I= 10 Amps
10 W
10 W
10 W
300 V
Solve a Series Circuit
Find the voltage of the battery, the current leaving
the battery, and the power of the 5 ohm resistor.
15 W
+
12 V
-
4W
5W
Solve a Series Circuit
Find the voltage of the battery, the current leaving
the battery, and the power of the 5 ohm resistor.
15 W
+
12 V
-
4W
3A
5W
Solve a Series Circuit
Find the voltage of the battery, the current leaving
the battery, and the power of the 5 ohm resistor.
3A
15 W
+
12 V
-
4W
3A
5W
3A
3A
Solve a Series Circuit
Find the voltage of the battery, the current leaving
the battery, and the power of the 5 ohm resistor.
+ 45 V 3A
15 W
12 V
+
4W
3A
+
15 V
5W
3A
3A
-
Solve a Series Circuit
Find the voltage of the battery, the current leaving
the battery, and the power of the 5 ohm resistor.
+ 45 V 3A
15 W
+
12 V
-
4W
3A
+
15 V
5W
3A
3A
45 W
-
V battery = 12V + 45V + 15V = 72 V
Parallel Circuit
• A parallel circuit is
a circuit were each
resistor has its own
loop, being
connected directly
to the power
source itself.
Because of this
each resistor
receives the full
voltage from the
power supply.
+ 20 V -
+ 20 V -
+ 20 V -
+ 20 V -
Parallel Circuit
• Because each
resistor has its own
connection to both
terminals of the
power source each
resistor operates
independently of all
the others, thus
each can have their
own current, and in
one is off the
others can still be
on.
5A
5A
6A
1A
5A
4A
6A
10 A
10 A
10 A
Kirchhoff’s Current Rule
• We know from the conservation of charges that the
total charge of a system can not change.
• Kirchhoff’s Current rule states when a group of
charges enter an intersection of wire, the total
number of charges that leave the intersection must
equal, the total number of charges that entered the
intersection..
Iin
I2
I1
Iin = I1 + I2
Lets simplify
a parallel circuit
with 3 resistors
We can simply a parallel circuit by using 3 basic
principles
Kirchhoff’s Current law:
IEq = IR1 + IR2 + IR3
Because each resistors is connected directly to the
power source.
VEq = VR1 = VR2 = VR3
But we still need to find the equivalent resistance.
For that we will still need to use Ohm’s Law:
V=IR
Finding Req
of a parallel circuit
with 3 resistors
Because:
And since:
Ieq = IR1 + IR2 + IR3
V = IR or even better I = V/R
We have: (Veq/Req) = (V1/R1) + (V2/R2) + (V3/Re3)
Because All V’s are the same we call each one V
We have: (V/Req) = (V/Req) + (V/Req) + (V/Req) (V/Req)
The Voltage is a common multiple on both sides
Now we have an equation for the equivalent
Resistance: (1/Req)= (1/R1) + (1/R2)+ (1/R3)
Equations for a Parallel
Circuit
• Equivalent Voltage
– Veq = VR1 = VR2 = VR3….
• Equivalent Current
– IEq + IR1 + IR2 + IR3….
• Equivalent Resistance
– 1/Req = 1/R1 + 1/R2 + 1/R3…
• Equivalent power
– PowEq = PowR1 +PowR2 + PowR3 …
Two Circuits, but What
are They Good for?
• Series Circuit is great for control
– Switches are connect in series to turn things on
and off
– Fuses are connected in series as to turn the circuit
off
– “Dimmers”/Rheostats are used in series to regulate
the current flow and voltage of another object
• Parallel is great when you want things to run
independently
– Power strips
– Power outlets of a house
– The Different electronics of a car
Sample problem
For the circuit shown what is the equivalent resistance?
How much current leaves the battery?
10 W
For a parallel 1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/10 W + 1/10 W + 1/10 W
1/Req = 3/10 W
Req = 10/3 W = 3.33 W
Using Ohms Law:
300 V = I(3.33W)
I= 90.09 Amps
10 W
10 W
300 V
Using the PVIR Box
The box is just an easy way to organize
the information provided
Because V = IR, and P = IV we can use
The box to solve any given series or
parallel circuit
Solve a Parralell Circuit
Find the voltage of the battery, the current leaving
the battery, and the power of the 5 ohm resistor.
2A
10 W
5W
20W
Solve a Parralell Circuit
Find the voltage of the battery, the current leaving
the battery, and the power of the 5 ohm resistor.
-
20 V +
2A
10 W
5W
20W
Solve a Parralell Circuit
Find the voltage of the battery, the current leaving
the battery, and the power of the 5 ohm resistor.
-
20 V +
2A
10 W
4A
5W
1A
20W
Solve a Parralell Circuit
Find the voltage of the battery, the current leaving
the battery, and the power of the 5 ohm resistor.
-
20 V +
2A
10 W
4A
5W
1A
20W
7A
Solve a Parralell Circuit
Find the voltage of the battery, the current leaving
the battery, and the power of the 5 ohm resistor.
-
20 V +
2A
10 W
4A
5W
1A
20W
7A
P = IV
P = (4A)(20V)
P = 80 Watts