Drawing a Circuit Drawing a Circuit • When illustrating an electrical When illustrating an electrical circuit for everyone to understand, circuit for everyone to understand, you must draw a standardized you must draw a standardized picture called a schematic. picture called a schematic. • A A schematic schematic is simply a picture of is simply a picture of an electrical circuit that uses an electrical circuit that uses standardized symbols to represent standardized symbols to represent the different parts of an the different parts of an electrical circuit. electrical circuit.
Transcript
Drawing a CircuitDrawing a Circuit
• When illustrating an electrical circuit for When illustrating an electrical circuit for everyone to understand, you must draw everyone to understand, you must draw a standardized picture called a a standardized picture called a schematic.schematic.
• A A schematic schematic is simply a picture of an is simply a picture of an electrical circuit that uses standardized electrical circuit that uses standardized symbols to represent the different parts symbols to represent the different parts of an electrical circuit.of an electrical circuit.
Basic Schematic SymbolsBasic Schematic Symbols
Battery/ Voltage source
Resistor
Switch
Fuse
Series CircuitSeries Circuit
• Because a series Because a series circuit has only one circuit has only one loop, loop, the amount of the amount of current that flows current that flows through one resistor through one resistor must be the same must be the same amount of current amount of current that flows for all the that flows for all the other resistorsother resistors as as well.well.
.5 A
.5 A
.5 A
.5 A
Series CircuitSeries Circuit
• Because no one Because no one resistor is connected resistor is connected directly to both directly to both terminals of the terminals of the power source. No power source. No one resistor receives one resistor receives all of the voltage, all of the voltage, the the voltage is divided out voltage is divided out amount each of the amount each of the resistorsresistors
100 V
20 V
50 V
30 V
Kirchhoff’s Voltage RuleKirchhoff’s Voltage Rule
• We know from the conservation from energy that all the We know from the conservation from energy that all the energy that the power source provides must equal the total energy that the power source provides must equal the total work that each resistor does.work that each resistor does.
• So then we can say the the energy that each charge has So then we can say the the energy that each charge has must to be used by all the resistors.must to be used by all the resistors.
• Since voltage is the energy of one charge, we can say that 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 the voltage of the power supply must be used up by all the resistors.resistors.
• Kirchhoff’s Voltage lawKirchhoff’s Voltage law– VVpower supplypower supply = Sum of all Voltage drops = Sum of all Voltage drops – VVinin = V = VR1R1 + V + VR2R2 + V + VR3R3 + V + VR4R4 …. ….
Simplifying a Series CircuitSimplifying a Series Circuit• When trying to analyze a series circuit, normally the first step is to reduce the circuit of several resistors to 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 sourcean equivalent circuit of only 1 resistor and 1 power source• For any simplified circuits we will have a/anFor any simplified circuits we will have a/an
– Equivalent Voltage (VEquivalent Voltage (Veqeq)) - How much voltage is truly being supplied to the circuit - How much voltage is truly being supplied to the circuit– Equivalent Current (IEquivalent Current (Ieqeq)) - How much current the power source has entering the circuit - How much current the power source has entering the circuit– Equivalent resistance (REquivalent resistance (Reqeq)) - How much resistance the circuit actually has - How much resistance the circuit actually has– Equivalent power (PEquivalent power (Peqeq)) - The total power of the circuit - The total power of the circuit
Original Circuit Equivalent Circuit
We can simply a series circuit by using 3 basic principles
But we will still need to find the equivalent resistance. For that we will need to use Ohm’s Law: V=IRV=IR
Kirrcoff’s Voltage law: VVEqEq = V = VR1R1 + V + VR2R2 + V + VR3R3
Because there is 1 pathway for current in a series circuit the current for all resistors is the same. IIEqEq = I = IR1R1 = I = IR2R2 = I = IR3R3
Vin
R3
R2
R1
Lets start witha series circuitwith 3 resistors
Vin
R3
R2
R1
Lets start witha series circuitwith 3 resistors
Because: VVeqeq = V = VR1R1 + V + VR2R2 + V + VR3R3
And since: V = IRWe can say: IIeqeqRReqeq = I = IR1R1RR11 + I + IR2R2RR22 + I + IR3R3RR33
Because All I’s are the same we call each one I
We can say: IRIReqeq = IR = IR11 + IR + IR22 + IR + IR33
The current is a common multiple on both sides Now we have an equation for the equivalent Resistance: RReqeq = R = R11 + R + R22 + R + R33
Equations for a Series Equations for a Series CircuitCircuit
• Equivalent VoltageEquivalent Voltage– VVeqeq = V = VR1R1 + V + VR2R2 + V + VR3R3….….
• Equivalent CurrentEquivalent Current– IIEqEq = I = IR1R1 = I = IR2R2 = I = IR3R3….….
• Equivalent ResistanceEquivalent Resistance– RReqeq = R = R11 + R + R22 + R + R33……
For the circuit shown what is the equivalent resistance?How much current leaves the battery?
300 V
10
10
10
For a series Req = R1 + R2 + R3
Req = 10 + 10 + 10
Req = 30
Using Ohms Law: 300 V = I(30 )
I= 10 Amps
Solve a Series CircuitSolve a Series Circuit
Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor.
+
-
12 V
Solve a Series CircuitSolve a Series Circuit
Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor.
+
-
12 V
3 A
Solve a Series CircuitSolve a Series Circuit
Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor.
+
-
12 V
3 A 3 A
3 A
3 A
Solve a Series CircuitSolve a Series CircuitFind the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor.
-
+
12 V
3 A 3 A
3 A
3 A
+
-
15 V
45 V+ -
Solve a Series CircuitSolve a Series CircuitFind the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor.
+
-
12 V
3 A 3 A
3 A
3 A
+
-
15 V 45 W
45 V+ -
V battery = 12V + 45V + 15V = 72 V
Parallel CircuitParallel Circuit• A parallel circuit is a A parallel circuit is a
circuit were each circuit were each resistor has its own resistor has its own loop, being loop, being connected directly to connected directly to the power source the power source itself. itself. Because of Because of this each resistor this each resistor receives the full receives the full voltage from the voltage from the power supply.power supply.
+ 20 V -
+ 20 V -
+ 20 V -
+ 20 V -
Parallel CircuitParallel Circuit
• Because each Because each resistor has its own resistor has its own connection to both connection to both terminals of the terminals of the power source power source each each resistor operates resistor operates independently of all independently of all the othersthe others, thus , thus each each can have their own can have their own currentcurrent, and in one is , and in one is off the others can off the others can still be on.still be on.
10 A
10 A10 A
6 A6 A 4 A
1 A 5 A5 A
5 A
Kirchhoff’s Current RuleKirchhoff’s Current Rule• We know from the conservation of charges that the total We know from the conservation of charges that the total
charge of a system can not change.charge of a system can not change.• Kirchhoff’s Current ruleKirchhoff’s Current rule states when a group of charges states when a group of charges
enter an intersection of wire, the total number of enter an intersection of wire, the total number of charges that leave the intersection must equal, the total charges that leave the intersection must equal, the total number of charges that entered the intersection..number of charges that entered the intersection..
Iin I2
I1 Iin = I1 + I2
We can simply a parallel circuit by using 3 basic principles
But we still need to find the equivalent resistance. For that we will still need to use Ohm’s Law: V=IR
Kirchhoff’s Current law: IIEqEq = I = IR1R1 + I + IR2R2 + I + IR3R3
Because each resistors is connected directly to the power source. VVEqEq = V = VR1R1 = V = VR2R2 = V = VR3R3
Lets simplifya parallel circuitwith 3 resistors
Finding Req
of a parallel circuitwith 3 resistors
Because: IIeqeq = I = IR1R1 + I + IR2R2 + I + IR3R3
Two Circuits, but What are Two Circuits, but What are They Good for?They Good for?
• Series Circuit is great for controlSeries Circuit is great for control– Switches are connect in series to turn things on and offSwitches are connect in series to turn things on and off– Fuses are connected in series as to turn the circuit offFuses are connected in series as to turn the circuit off– ““Dimmers”/Rheostats are used in series to regulate the Dimmers”/Rheostats are used in series to regulate the
current flow and voltage of another objectcurrent flow and voltage of another object
• Parallel is great when you want things to run Parallel is great when you want things to run independentlyindependently– Power stripsPower strips– Power outlets of a housePower outlets of a house– The Different electronics of a carThe Different electronics of a car
Sample problemSample problem
For the circuit shown what is the equivalent resistance?How much current leaves the battery?
300 V
10
10
10
For a parallel 1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/10 + 1/10 + 1/10
1/Req = 3/10
Using Ohms Law: 300 V = I(3.33)
I= 90.09 Amps
Req = 10/3
Using the PVIR BoxUsing 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 orparallel circuit
Solve a Parralell CircuitSolve a Parralell CircuitFind the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor.
20
10
5
2 A
Solve a Parralell CircuitSolve a Parralell CircuitFind the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor.
20
10
5
2 A- 20 V +
Solve a Parralell CircuitSolve a Parralell CircuitFind the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor.
20
10
5
2 A- 20 V +
4 A
1 A
Solve a Parralell CircuitSolve a Parralell CircuitFind the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor.
20
10
5
2 A- 20 V +
4 A
1 A
7 A
Solve a Parralell CircuitSolve a Parralell CircuitFind the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor.
