1 SERIES CIRCUITS Benchmark Companies Inc PO Box 473768 Aurora CO 80047.

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SERIES CIRCUITS

Benchmark Companies IncPO Box 473768Aurora CO 80047

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Series CircuitA series circuit is defined as having all components connected in series (head

to toe) with each other

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Voltage, Current, and Resistance in Series Circuits

In this section we will discuss the properties of Ohms Laws and the relationship between each.

Current (I) AmpsVoltage (V) VoltsResistance (Ω) Ohms

V=IR

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The total current in a series circuit is equal to the current in any other part of the circuit.

I1 = current in the first partI2 = current in the second partI3 = current in the third part, etc.

Formula: IT = I1 = I2 = I3 = etc.

Total Current in a Series Circuit

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The total current in a series circuit is equal to the current in any other part of the circuit.

Total Current in a Series Circuit

Therefore IT=IR1=IR2=IR3= etc.

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where VT = total voltage

V1 = voltage across first part

V2 = voltage across second part

V3 = voltage across third part, etc.

The total voltage in a series circuit is equal to the sum of the voltage drops across all the parts of the circuit.

Formula: VT = V1 + V2 + V3 + etc.

Total Voltage in a Series Circuit

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Therefore: VT = VR1 + VR2 + VR3.


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Therefore: VT = 10V + 10V + 20V= 40V.


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where RT = total resistance

R1 = resistance of first part

R2 = resistance of second part

R3 = resistance of third part, etc.

The total resistance of a series circuit is equal to the sum of the resistances of all the parts of the circuit.

Formula: RT = R1 + R2 + R3 + etc.

Total Resistance in a Series Circuit

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The total resistance of a series circuit is equal to the sum of the resistances of all the parts of the circuit.

Therefore: RT = 10Ω + 10Ω + 20Ω = 40Ω

Total Resistance in a Series Circuit

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Series Equivalent Circuit

1 1 2 2 3 3

1 2 3

1 2 3

1 2 3

1 2 3

V I R V I R V I R

R R R R

V V V V

V I R I R I R

V I R R R

V I R

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Example 4-1

A 6-V 20- filament and a 12-V 40- filament are connected in series with a 20- limiting resistor dropping 6 V across it and 0.3 A through it.

Find:

(a) the total voltage

(b) the total current

(c) the total resistanceContinued

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(c) Find the total resistance. 1. Write the formula. RT = R1 + R2 + R3

2. Substitute numbers. RT = 20 + 40 + 20 = 80

Solution: using the equations derived previously

(a) Find the total voltage.1. Write the formula. VT = V1 + V2 + V3

2. Substitute numbers. VT = 6 + 12 + 6 = 24 V

(b) Find the total current.1. Write the formula. IT = I1 = I2 = I3

2. Substitute numbers. IT = I1 = =I2 =I3 = 0.3 A

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Using Ohm’s Law in Series Circuits

Ohm’s law may be used for the individual parts of a series circuit. When it is used on a particular part of a circuit, great care must be taken to use only the voltage, current, and resistance of that particular part. This may be easily remembered by using the correct subscripts when writing the Ohm’s-law formula for a particular part.

For the first part: V1 = I1 x R1

For the second part: V2 = I2 x R2

For the third part: V3 = I3 x R3

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Using Ohm’s Law for Total Values in Series Circuits

A resistor, doorbell, and buzzer are connected in series across a voltage source. The resistor has a resistance R1

of 45 , the doorbell resistance R2 is 60 , and the buzzer resistance R3 is 50 . I3 is 0.2 A. Find the value of VT.

Example

Finding the Total Voltage

Continued

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3. Find the total voltage.

VT = IT x RT

VT = 0.2 x 155 = 31 V Ans.

1. Find the total current.

IT = I1 = I2 = I3 = 0.2 A Ans.

2. Find the total resistance.

RT = R1 + R2 + R3

RT = 45 + 60 + 50 = 155 Ans.

Solution

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Control of Current in a Series Circuit

Example 4-27

1. Write the formula. VT = VX + VL

2. Substitute numbers. 24 = VX + 123. Transpose the 12. 24 - 12 = VX 4. Subtract. 12 = VX = 12 V

What resistance must be added in series with a lamp rated at 12 V, 0.3 A order to operate it from a 24 V source?

Solution

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Series Circuit

Current is constant

Why? Only one path for the

current to take Kirchhoff’s Current Law

Voltages through circuit equals zero Kirchhoff’s Voltage Law

1 2 3

1 2 3

V V V V

R R R R

V I R

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Introducing

Kirchhoff’s Voltage Law

Net Voltage for a circuit = 0

inV VoltageAcrossEachResistor

Sum of all voltage drops and voltage rises in a circuit (a closed loop) equals zero

V

V1 V2

1 2

1 2 0

V V V

V V V

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Example

One 10Ω resistor connected to the 6V power source (batteries). Add another 10Ω resistor to the circuit in series to the first resistor.

Q: What is the equivalent resistance, R? What will happen to the value of the current through each resistor? What will happen to the value of the voltage across each resistor?

1 2

1 2

...

...

V I R

R R R

V V V

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Example: What is happening in theory

1 2

1 1

2 2

1 2

Initially,

=6V, 10

6V0.6A

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Add the second resistor:

6V, 10 10 20

6V0.3A The current through the circuit is halved!

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0.3A 10 3V

0.3A 10 3V

6V T

V R

VV I R I

R

V R R R

VI

R

V I R

V I R

V V V

he voltage across each resistor is halved!

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Example: The actual data

In reality, the data we get is not the same as what we get in theory.

Why?

Because when we calculate numbers in theory, we are dealing with an ideal system. In reality there are sources of error in every aspect, which make our numbers imperfect.

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1. Write the formulas for finding IT , VT , and RT , in series circuit.

Exercise

2. A 5 V 17 and a 10 V 40 filaments are in series with a 17 limiting resistor using 6 V and 0.3 A.

Find: (a) VT; (b) IT; and (c) RT.

(21 V; 74 ; 0.3 A)

IT = I1 = I2 = I3 = etc.VT = V1 + V2 + V3 + etc.RT = R1 + R2 + R3 + etc.

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3. A resistance R1 = 25 , a doorbell R2 =

50 , and a buzzer R3 =38 , are

connected in series.The current through the buzzer is I3 = 0.4 A. Find VT.

Exercise

4. Three resistances are connected in series. V1 = 12.6 V, I1 = 1.1 A, R1 = 11 ,

V2 = 24 V, R2 = 21 , V3 = 48 V, R3 = 42 .

Find VT , IT , RT. (84.6 V, 1.14 A, 74 )

(45.2 V)

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End of Lesson

1 SERIES CIRCUITS Benchmark Companies Inc PO Box 473768 Aurora CO 80047.

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