combined series and parallel circuits...series and parallel circuits 1. draw a diagram of the...

Combined Series and Parallel Circuits

Objectives: 1. Calculate the equivalent resistance, current, and voltage of series and parallel circuits. 2. Calculate the equivalent resistance of circuits combining series and parallel connections. 3. To understand the origins of both of Kirchhoff's rules and how to use them to solve a circuit problem. 4. Solve circuit problems.

Resistance and Current Series Circuit • Equivalent resistance is equal to the sum of all the resistance in the circuit.

• Circuit current is equal to the voltage source divided by the equivalent resistance.

1 2 3 ...

/

eq n

source eq

R R R R R

I V R

= + + +

=

Resistance and Current

1 2 ...

1 2 .....

1 1

1/ 1/ 1/ 1/

/

eq n

n

source

R R R RI I I II V R

= + +

= + +=

Parallel Circuit • The reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances. • The total current is the sum of all the currents. • The potential difference across each resistor is the same

Household Circuits

Why do the lights dim when the hair dryer goes on?

Small resistance from wiring

This is called a combination series and parallel circuit

Series and Parallel Circuits 1. Draw a diagram of the circuit 2. Find any resistors in parallel. They must have the same

potential difference across them. Calculate the single equivalent resistance of a resistor that can replace them.

3. Are any resistors (including the parallel equivalent resistor) in series? Resistors in series have one and only one current path through them. Calculate the new single equivalent resistance that can replace them. Draw a new schematic diagram using that resistor.

4. Repeat steps 2 and 3 until you can reduce the current to a single resistor. Find the total circuit current. Then go backwards to find the currents through and the voltages across individual resistors.

1. The sum of the currents entering any junction must equal the sum of the currents leaving that junction. (junction rule)

2. The sum of the potential differences across all the elements around any closed circuit loop must equal zero. (loop rule)

Kirchhoff’s Rules Gustav Kirchhoff - 1845

In this example you will notice that 8 Amps of current enter the junction and 3 and 5 Amps leave the junction. This makes a total of 8 Amps entering and 8 Amps leaving.

In this example you will notice 8 Amps and 1 Amp entering the junction and 9 Amps leaving. This makes a total of 9 Amps entering and 9 Amps leaving.

In this example you will notice 8 Amps and 1 Amp entering the junction while 7 Amps and 2 Amps leave. This makes a total of 9 Amps entering and 9 Amps leaving.

This is a simple circuit showing the potential differences across the source and the resistor. According to Kirchhoff's 2nd law the sum of the potential differences will be zero.

This diagram shows the potentials in the little circles and then shows the potential differences off to the side. Notice that the potential difference is actually the difference between one potential and another. Moving from a low potential to a high potential is considered a potential rise or positive potential difference. Moving from a high potential to a lower potential is considered a potential drop or negative potential difference.

This animation shows the same circuit as above but only looks at the potential differences as you go around the loop. Again, Kirchhoff's 2nd law says the sum of the potential differences has to be zero.

Now lets try some problems

Don’t wait to get totally lost. Ask your questions as they come to you.

#1

Series Circuit Rt = R1 + R2 + R3 + … Rt = 4 + 6 + 3 + 1 = 14 Ω I = V ÷ R I = 40 ÷ 14 = 2.86 amps

#2

Series Circuit Rt = R1 + R2 + R3 + … Rt = 5 + 4 + 12 = 21 Ω I = V ÷ R I = 10 ÷ 21 = 0.476 amps

#3

Series Circuit Rt = R1 + R2 + R3 + … Rt = 3 + 1 + 7 = 11 Ω I = V ÷ R I = 120 ÷ 11 = 10.9 amps

#4

Series Circuit Rt = R1 + R2 + R3 + … Rt = 5 + 1 + 6 + 3 + 4 + 1 = 20 Ω I = V ÷ R I = 9 ÷ 20 = 0.45 amps

#5

Series Circuit Rt = R1 + R2 + R3 + … Rt = 12 + 20 + 5 = 37 Ω I = V ÷ R I = 60 ÷ 37 = 1.62 amps

#6

Parallel Circuit 1/Rt = 1/R1 + 1/R2 + 1/R3 + … 1/Rt = 1/2 + 1/2 + 1/2 = 1.5 Rt = 0.667 Ω I = V ÷ R I = 6 ÷ 0.667 = 9.00 amps

#7

Parallel Circuit 1/Rt = 1/R1 + 1/R2 + 1/R3 + … 1/Rt = 1/6 + 1/8 + 1/4 = 0.542 Rt = 1.85 Ω I = V ÷ R I = 120 ÷ 1.85 = 64.9 amps

#8

Parallel Circuit 1/Rt = 1/R1 + 1/R2 + 1/R3 + … 1/Rt =1/2.5 + 1/6 + 1/1 = 1.57 Rt = 0.638 Ω I = V ÷ R I = 14 ÷ 0.638 = 21.9 amps

Now let’s put ‘em together Simplify diagram in steps to a single resistor Calculate total resistance and current for the whole circuit Then work backwards to find voltages and currents at individual resistors It takes time and care to do this right

DON’T TRY TO RUSH THROUGH IT!

#9

The 2 Ω and 3 Ω resistors are in series with one another

They combine to form a 5 Ω resistor

The three resistors are hooked up in parallel with each other.

1/Rt = 1/R1 + 1/R2 + 1/R3 + … 1/Rt =1/5 + 1/1 + 1/6 = 1.37 Rt = 0.730 Ω The circuit now looks like this I = V/R = 120/.730 = I = 164 amps

#10

Combine the 3 Ω and the 7 Ω resistors that are in series with one another to make a 10 Ω resistor

Combine the 1 Ω and the 2 Ω resistor that are in series with one another to make a 3 Ω resistor

Then re-draw the circuit It should look like this

The three resistors are hooked up in parallel with each other.

1/Rt = 1/R1 + 1/R2 + 1/R3 + … 1/Rt =1/10 + 1/4 + 1/3 = 0.683 Rt = 1.46 Ω The circuit now looks like this I = V/R = 40/1.46 = I = 27.4 amps

#11

Combine the resistors hooked up in series with one another and re-draw the circuit

It should look like this

The three resistors are hooked up in parallel with each other.

1/Rt = 1/R1 + 1/R2 + 1/R3 + … 1/Rt =1/4 + 1/4 + 1/10 = 0.600 Rt = 1.67 Ω The circuit now looks like this I = V/R = 220/1.67 = I = 132 amps

#12

What type of circuit is this? Find the total resistance Find the total current The ___________ is the same for all

devices in a series circuit Therefore I3 = To find V2 we need to know… V = IR so V2 = 1.17(4)

Series 12 Ω 1.17 amps Current 1.17 amps R2 and I2

4.68 V

#13

What type of circuit is this? Find the total resistance Find the total current The ___________ is the same for all

lines in a parallel circuit Therefore V3 = To find I3 we need to know… I = V/R so I3 = 120(2.5)

Parallel 1.11 Ω 108 amps Voltage 120 V V3 & R3

48 amps

#14

Both

R2 Ω

R3 Ω

110 V

R1 Ω R5 Ω

R4 Ω R6 Ω

Is this a series or a parallel circuit? Combine the resistors in series first. Then re-draw the circuit. It should look something like this

Now find the total resistance Do we need the total current? If

so, what is it? How is resistor #5 hooked up? Those two resistors combine to

form a resistor that is _____ Ω How is that 5 Ω resistor hooked up

to the circuit?

2.4 Ω No In series with #6 5 Ω In parallel with the

other combined resistors

More about Resistor #5

What is the same for all the lines in a parallel circuit?

What is the voltage across the 5 Ω resistor?

What is the current through that 5 Ω resistor?

What is the current through resistor #5? Why? V = IR V = 22(3)

Voltage 110 V 22 amps 22 amps Series w/ #6 66 V

Let’s look at Resistor #4

How is resistor #4 hooked up? Those two resistors combine to

form a resistor that is __ Ω How is that 11 Ω resistor

hooked up to the circuit? What is the same for all the

lines in a parallel circuit? What is the voltage across the

11 Ω resistor?

In series with #3 11 Ω Parallel with the other

combined resistors Voltage 110 V

More about Resistor #4

What is the current through that 11 Ω resistor?

What is the current through resistor #4?

Why?

10 amps 10 amps Series w/ #3

Now move on to the rest of the packet. We have the answers at all the problems.

We have the solutions to #17 - #22. #23 is extra credit (it is really hard)