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Series Parallel Circuits

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Introduction to Combination Circuits

Advantages of Series Circuits A series circuit may be used to connect small voltages

to obtain high voltages. High voltages may be reduced by connecting

resistances in series. Series circuits provide a means for reducing and

controlling the current by connecting resistances in series.

Series circuits are used where different voltage drops and a constant current are needed.

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Since the current is constant in a series circuit we are forced to use only those devices which require the same current.

If any part of a series circuit should burn out it would cause an open circuit and put the entire circuit out of operation.

Series circuits are used where different voltage drops and a constant current are needed.

Disadvantage of Series Circuits

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Disadvantages of Parallel Circuits

If a break should occur in any branch of a parallel circuit, it would not affect the other branch circuits.

Any device may be operated independently of any other device.

In a parallel circuit, more branches (devices) may be added at any time.

Advantages of Parallel Circuits

As more devices are added more current is drawn till eventually the fuse blows.

Parallel circuits are used where a constant voltage and a large current are required.

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If we combine series circuits with parallel circuits, we produce a combination circuit.

A combination circuit makes it possible to obtain the different voltages of a series circuit and the different currents of a parallel circuit.

Simple combination circuits are of two types.1. A parallel-series circuit in which one or more groups of

resistances in series are connected in parallel.2. A series-parallel circuit in which one or more groups of

resistances in parallel are connected in series.

Combination Circuits

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General Method for Solving Combination Circuits

1. A group of resistances is a simple combination of two or more resistances which are arranged in either simple series or simple parallel circuits. Identify these groups.

2. Every group must be removed from the circuit as a unit and replaced by a single resistor which offers the identical resistance. This equivalent resistance is the total resistance of the group.

Continued

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3. Redraw the circuit, using the equivalent resistance in place of each group.

4. Solve the resulting simple circuit for all missing values.

5. Go back to the original circuit to find the voltage, current, and resistance for each resistance in the circuit.

General Method for Solving Combination Circuits

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Solving Parallel-Series CircuitsExampleTwo groups of resistors are Group A (that has 10-, and 50 resistors connected in series) and Group B (that has two 30 resistors connected in series). Find all the missing values connected in parallel across a 120 V source of voltage, current, and resistance.

Continued

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1. Identify groups A and B as series circuits.

2. Find the equivalent resistance of each group.

3. Since the resistors of groups A and B are in series.

RA = R1 + R2

RA = 10 + 50 = 60 Ans.

RB = R3 + R4

RB = 30 + 30 = 60 Ans.

Solution

Continued

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4. Redraw the circuit, using these 60 resistors in place of the series groups.

5. Solve the new parallel circuit.

Find the voltage for each group.

VT = VA = VB = 120 V Ans.

Find the current in each group:

Solution

VA = IA x RA

120 = IA x 60

IA = 120/60 = 2 A Ans.

VB = IB x RB

120 = IB x 60

IB = 120/60 = 2 A Ans.

Continued

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6. Go back to the original circuit to find the voltage and current for each resistor. Find the current in each resistor.

IA = I1 = I2 = 2 A Ans.

IB = I3 = I4 = 2 A Ans.

Find the total current IT.

IT = IA + IB

IT = 2 + 2 = 4 A Ans.

Find the total resistance RT.

VT = IT x RT

120 = 4 x RT

RT = 120/4 = 30 Ans.

Continued

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V2 = I2 x R2

V2 = 2 x 50

V2 = 100 V

Find the voltage drop across each resistor.

V1 = I1 x R1

V1 = 2 x 10

V1 = 20 V

V3 = I3 x R3

V3 = 2 x 30

V3 = 60 V

V4 = I4 x R4

V4 = 2 x 30

V4 = 60 V

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Solving Series-Parallel Circuits

A resistor R1 = 10 is in series with two parallel resistors R2 = 40 and R3 = 60 . The three resistors are connected across a voltage source of 34 V. Find all the values of voltage, current, and resistance.

Example

Continued

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4. Solve the new series circuit. Find the total resistance RT.

RT = R1 + R4 RT = 10 + 24 = 34 Ans.

1. Find the resistance of the parallel group A (R2 and R3).2. Since R2 and R3 are in parallel,

RA = (R2 x R3) (R2 + R3)

RA = (40 x 60) (40 + 60) = 24 Ans.3. Redraw the circuit using this 24 resistor RA in

place of the parallel combination.

Continued

Solution

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Find the voltage in each part of the series circuit.

Find the total current IT.

VT = IT x RT

34 = IT x 34

IT = 34/34 = 1 A Ans. Find the current in each part of the series circuit.

IT = I1 = IA = 1 A Ans.

VA = IA x RA

VA = 1 x 24 = 24 V Ans. V1 = I1 x R1

V1 = 1 x 10 = 10 V Ans.

Continued

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Find the current in each resistor of group A.

5. Go back to the parallel group A to find V, I, and R for each resistance in the group. Since VA represents the total voltage of the parallel group A.

VA = V2 = V3 = 24 V

6. Check: IT = I2 + I3

IT = 0.6 + 0.4

1 = 1

V2 = I2 x R2

24 = I2 x 40

I2 = 24/40 = 0.6 A

V3 = I3 x R3

24 = I3 x 60

I3 = 24/60 = 0.4 A

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End of Lesson