Electrical Circuits
~Moving Charge Put to Use
The Circuit• All circuits, no matter how simple or complex, have one
thing in common, they form a complete loop.• As mentioned before, circuits should have various circuit
elements in the loop.
Series Circuit
• Have you ever driven down a 1 lane road? • You can keep moving until…• If there is an accident all traffic stops, there is no other
road to follow.
• A series circuit is similar to a one lane road, current can flow in only one path.
• Even if you add a 2nd resistor in series, there is still just 1 path.
Series Circuit
R1
R2
Series Circuit• One path means all components have the same
current• The Voltage provided by the source must equal the
Voltage drop across the resistor(s)
source resistorV V
RVI
Series Circuit• One path means all components have the
same current• What is the voltage drop across R1?
1 2source R RV V V
R1
V
R2
I 1 1RV IR
2 2RV IR
Series Circuit• How do we find Req?
1 2series R RV V V
R2
R1
V I
e 1 2Rs q s sI I R I R
e 1 2R q R R Divide both sides by Is
ReqV
The Series Circuit (cont.)• Every series configuration can be reduced to a single
value for resistance known as the equivalent resistance, or Req.
• The formula for Req is as follows for series:
• This can be used as a step to solve for the current in the circuit or the voltage across each resistor.
R1
R2
Req
1 2eqR R R
I
Sample Problem (Series)• A circuit is configured in series as shown below.
– What is the equivalent resistance (Req)?
– What is the current through the circuit?
(Hint: Use Ohm’s Law.)
10
20
30
6V
1 2 3eqR R R R 10 20 30eqR 60eqR
eq eqeq eq
eq eq
V VR I
I R
6
60eq
VI
0.1eqI A
Ieq = 0.1A
606V
Sample Problem (Series) (cont.)• We still have one question to ask. What are the voltages
across each resistor?
– For the 10 Resistor:
– For the 20 Resistor:
– For the 30 Resistor:
• What do you notice about the
voltage sum?
10
20
30
6V
Ieq = 0.1A
VR V IR
I
V IR 0.1 10V A
0.1 20V A
0.1 30V A
1V V
2V V
3V VV IR
V IR
Voltages across resistors in series add to make up the total voltage.
1 2 3 6V V V V 6 eqV V
Series Circuit Summary• Current is constant throughout the entire circuit.
• Resistances add to give Req.
• Voltages across each resistor add to give Veq.
1 2eqI I I
1 2eqR R R
1 2eqV V V
Devices that Make Use of the Series Configuration
• Although not practical in every application, the series connection is crucial as a part of most electrical apparatuses.– Switches
• Necessary to open and close entire circuits.
– Dials/Dimmers• A type of switch containing a variable resistor
(potentiometer).
– Breakers/Fuses• Special switches designed to shut off if current is too
high, thus preventing fires.
– Ammeters• Since current is constant in series, these current-
measuring devices must be connected in that configuration as well.
The Parallel Circuit (cont.)
• Parallel circuits are similar to rivers with branches in them.
• The current from the river divides into multiple paths.• After the paths, the water recombines into the same
amount of flowing water.1 2eqI I I
IeqIeqI1
I2
• A parallel circuit is similar to a river that branches, current can flow in multiple paths.
• Once the paths end, the total flow remains the same
Parallel Circuit
R2R1
The Parallel Circuit• Notice that the circuit branches out to each resistor,
allowing multiple paths for current to flow.
• If there are exactly two clear paths from the ends of one resistor to the ends of the other resistor.
R1 R2
Branch
X
BranchX
A break in one of the branches of a parallel circuit will not disable current flow in the remainder of the circuit.
Parallel Circuit• How do we find Req a parallel circuit?
1 2parallel R RI I I
R2
V
e 1 2Rp p p
q
V V V
R R
e 1 2
1 1 1
R q R R
Divide both sides by Vp
ReqVR1 R2
V
Use Ohm’s law
The Parallel Circuit (cont.)• Notice how every resistor has a direct connection to the
DC source. This allows the voltages to be equal across all resistors connected this way.
• An equivalent resistance (Req) can also be found for parallel configurations. It is as follows:
R1 R2
1 2eqV V V
Req
1 2
1 1 1
eqR R R
Sample Problem (Parallel)• A circuit is configured in parallel as shown below.
– What is the equivalent resistance of the circuit?
30 6V
1 2 3
1 1 1 1
eqR R R R
1 1 1 1
30 30 60eqR
11 1 130 30 60
eqR
12eqR
126V
Sample Problem (Parallel)• What is the current in the entire circuit?
• What is the current across each resistor?
30 6V
eq eqeq eq
eq eq
V VR I
I R 6
12eq
VI
0.5eqI A
VI
R
6
30
VI
6
60
VI
0.2I A 0.1I A
The 30 Resistors The 60 Resistor
Parallel Circuit Summary• There are several facts that you must always keep in
mind when solving parallel problems.– Voltage is constant throughout the entire parallel circuit.
– The Inverses of the Resistances add to give the inverse of Req.
– Current through each resistor adds to give Ieq.
– Make use of Ohm’s Law.V VR I V IR
I R
1 2eqV V V
1 2eqI I I
1 2
1 1 1
eqR R R
Devices that Make Use of the Parallel Configuration
• Although not practical or safe in every application, the parallel circuit finds definite use in some electrical apparatuses.– Electrical Outlets
• Constant voltage is a must for appliances.
– Light Strands• Prevents all bulbs from going out when a single
one burns out.
– Voltmeters• Since voltage is constant in parallel, these
meters must be connected in this way.
Lights demo• DC source with 3 lights in series
• DC source with 3 lights in parallel
• DC source with 2 lights in series 1 parallel
• DC source with 1 lights in series 2 parallel
Combination Circuits• Some circuits have series/parallel combinations • These can be reduced using equivalent resistance
formulas.• Now let’s solve a problem involving this circuit.
R1R2
R3 R4
SeriesParallel
Sample Problem (Combo)
What is the equivalent resistance (Req) of the circuit?– First, we must identify the various combinations present.
Series
Parallel
1 2eqR R R 10 30eqR
40eqR 1 2
1 1 1
eqR R R
1 1 1
20 20eqR
10eqR
Series Parallel
1040
301020 20
25V
Sample Problem (Combo)• The simplified circuit only shows the equivalent
resistances. Is the circuit now fully simplified?• Now, we must identify the final configuration.
Series
Parallel
1040
301020 20
25V
4010
25V1 2eqR R R 40 10eqR 50eqR
Series
50
Sample Problem (Combo)• The circuit is further simplified below. Can it be
simplified again?• Now, the circuit is completely simplified.• What is the current in the entire circuit?
4010
25V
Series
50
5025V
eq eqeq eq
eq eq
V VR I
I R 25
50eq
VI
0.5eqI A
Combination Circuits
• Parallel Paths: Must make a complete loop through two resistors with out touching any other component.
• Series Paths: Must form a path through multiple resistors with out crossing an intersection.
Kichoff's current laws
1
• Choose a direction and label the current in each branch• Identity the number of unknowns as develop as many equation
• Label the polarity of each Vr for all resistors.
• Apply Kirchoff’s junction rule (sum of current in equals sum of current out.• Apply Kirchoff’s loop rule. The sum of all voltages around a loop must equal zero• Solve the simultaneous equations
40V
30 1 35V
10
10
I1
I2
I3
Kichoff's current laws
140V
30 1 35V
10
10
25V3010
30
15V
• Make a terminal voltage slide
25V3010
30
15V
24μf
12μf
36μf
220V
114
117
220V
114
117
140