Circuits - Teach Yourself

8/3/2019 Circuits - Teach Yourself

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Electric Circuits, DCCircuits are classified by the type of path that the electricity follows as it goes around the circuit.There are two types of circuits - series circuits and parallel circuits. In a series circuit there is onlyone path and all the circuit components are in a line, connected by the conductor, so that all theelectrons flow through each component. A parallel circuit offers different paths for the electrons.

Series Circuits: Heres a simple series circuit withthree resistors; R 1, R 2, and R 3. When the electrons leavethe battery they all go through the first resistor theyencounter. Then all of them go through the next one andthe next one. Then they all go back into the battery.

The current is the same at every point in the circuit.

Each resistor has a potential difference across it.

The voltage, current, and resistance in a series circuit behave according to the following rules.

Rules For Resistance in Series:

1. The current in every part of the circuit is the same.

2. The total resistance in the circuit is equal to the sum of all the resistances.

3. The voltage provided by the voltage source is equal to the sum of all the voltage dropsacross each of the resistors

Mathematically:

1 2 n R R R R= + + +

I = I 1 = I 2 = = I n

1 2 nV V V V = + + +

On the AP Physic Test, the equation for series resistance is given as:

S

i

i R R=

This simply says that the total resistance is the sum of the individual resistances. No equations for voltage and current are given. Youll need to remember that thecurrent is the same at every point in the circuit and the voltage drops add up to thetotal voltage provided by the voltage source.

V

R1 R 2

R3

I I

I I


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Find (a) the total resistance and (b) the current in the circuit to the right.

(a) To find the total resistance just add up the three resistances.

1 2 3S

i

i R R R R R R= = + +

25 35 32 92 R = + + =

(b) Find I: V = IR ,V

I R

=

180.196 0.20

92

V I A A= = =

.

Calculate (a) the current and (b) the total voltage in thiscircuit.

(a) Find current through R 1 :

V IR=

12.00.27

45.0

V V I A

R

= = =

This is the total current.

(b) We can then use Ohms law to find the voltage. First we find the total resistance, then useV=IR

1 2 R R R= + R = 45.0 + 75.0 = 120.0

V = IR V = 0.267 A (120.0 A)

32V V =

2 5 3 5

3 2 1 8 V

45 75

12 V


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Parallel Circuits: Parallel circuits offer the current more than one path.

The current, I , comes to the junction between R 1 and R 2 and splits. Some of the current goesthrough R 1 and some goes through R 2. If you add up the current in each branch, I 1 and I 2, their sumwould equal I .

Each branch of the parallel circuit sees the same electric potential, V .

Imagine a 12 V car battery. Two vertical rods are driven into the soft lead electrodes of the battery.The potential difference across the electrodes and now the rods is 12 V. You clip the leads of alight bulb onto the vertical rods and the voltage drop across the bulb (which acts like a resistor) is12 V. The metal rod and the wire leads for the bulb have essentially no resistance and

there is no potential difference from one spot to the other in a short length of conductor, remember we said the resistance is virtually zero. Only the bulb has a potential difference. Next, we add asecond bulb, but nothing happens to the voltage drop. Each of the bulbs sees a voltage difference of 12 V. Each side of the two bulbs are connected together and have the same potential. We couldeven add a third bulb. In fact we could add as many bulb as we like. Each would see a voltage dropof 12 V.

Rules for Parallel Circuits:

1. The voltage drop is the same across each branch of parallel circuit.

R

R

1

2

V

I I 1

I 2

I 1

I

P o w e

r

P o w e

r

P o w e

r

1 2 V1 2 V1 2 V

12 V 12 V 12 V


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2. The total current is the sum of the currents through each branch.

3. The total resistance is less than the resistance of any one branch. The reciprocal of the totalresistance is equal to the sum of the reciprocals of the resistance of each branch.

The equation to find the total resistance: 1 1

P i i

R R=

Which basically means:1 2

1 1 1 1

n R R R R= + + +

To find the total current:

I = I 1 + I 2 + + I n

And, of course, the voltage is the same for all branches in the parallel circuit.

1 2 nV V V V = = = =

Ohm's law applies separately to each branch.

In the following circuit: find (a) the total resistance, (b) total current, and (c) the current througheach branch.

(a) 1 2

1 1 1 1 1

85 35 R R R= + = +

25 R =

(b) 45

1.825

V V V IR I A

R= = = =

(c) To find the current through each branch, we use Ohms law for each branch:

11

450.53

85

V V I A

R

= = =

2

2

451.3

35

V V I A

R

= = =

Combination Circuits: Sometimes we have a circuit that has components in series with oneanother and components that are in parallel. These are called combination circuits. To solve

problems in these circuits its best to simplify things by finding the equivalent circuit. Basicallyyou take all the resistances and, by adding the series ones and solving the parallel ones, you end upwith one equivalent resistor. You basically are finding the total resistance for the entire circuit.Which is really just what you do when you find the total resistance of a parallel circuit.

3 5

4 5 V

8 5

R

R

1

2


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Equivalent Circuits: The idea here is to take a complicated circuit that has series and parallel components and work through it finding and adding various resistances until you end upwith a single equivalent resistance that has the same value of resistance as the entire circuit.

Heres a simple example - this circuit has two resistors in parallel.

Using the parallel resistance equation, we can calculate the total resistance, which is the equivalentresistor.

Here are some other examples:

What is (a) the total resistance and (b)the total current?

(a) The first resistor is in series with theother two resistors that are in parallel. Sowe have to add the first resistorsresistance to the resistance for the parallel

225.0

250.0

12.0 V

5 5

2 4 V

8 5

R R1 2

R 3

3 5


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part of the circuit.

1 23 R R R= + 23 2 3

1 1 1 1 1

85 55 R R R= + = +

12 33.4 33 R = =

(b) Use Ohms law to find the current since we know the voltage and the total resistance.

240.73

33

V V V IR I A

R= = = =

Heres another lovely circuit. Find the totalvoltage.

We know the current that goes through R 3 well callit I 3. We also know the values of all the resistors. Wecan calculate the total resistance. But how do we findthe current?

123 4 R R R= +

123 1 2 3

1 1 1 1 1

11 17 35 R R R R= + = +

+ +

123 15.6 R = 15.6 12 27.6 R = + =

We can find the voltage drop across R 3 using Ohms law.

( )3 3 3 0.55 35 19.2V IR V I R A V = = = =

We also know that the same voltage drop will exist on the other leg of the parallel segment of thecircuit. We know that the equivalent resistor must drop this same voltage and we know the valuefor this resistor, good old R 123 . So using Ohms law we can find the current through the parallelsegment. Since it is in series with R 4, this current must be the total current.

123

123

19.21.2

15.6

V V V V IR I A

R R= = = = =

Now we can use Ohms law to find the voltage supplied by the battery. We have found the totalresistance and the total current.

( )1.2 27.6 33V IR A V = = =

3 5

1 1

R

R

1

2R3

1 5

R4

1 7


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You have three 75 W bulbs that you will use in a household circuit so figure 117 V. Whattype of circuit (series or parallel) would give you the greatest brightness (assume that the bulbsthat operate with the greatest power arethe brightest).

Lets look at a series circuit first.

We know that one bulb will operate at 75 W.Using this, we can figure out the resistance of the bulb.

Power is: P IV = Ohms law is: V IR=

We can solve Ohms law for I :V

V IR I R

= =

We plug this into the equation for power and solve for R , this will give us the bulbs resistance.

( )22 2 117

18375

V V V V P V R

R R P W

= = = = =

The resistance doesnt change its pretty much constant, no matter how much potential differencewe apply. Now we can find the power developed by each bulb in the series circuit.

Find the total resistance and then the total current. This is the same current for each bulb, so we canuse it to find the power consumed by a bulb.

( )1 2 3 3 183 549 R R R R

= + + =

=

1200.219

549

V V V IR I A

R= = = =

The voltage drop for a bulb is: ( )0.219 183 40.1V IR A V = = =

The power is: ( )0.219 40.1 8.8 P IV A V W = = =

So in a series circuit with three other bulbs, the 75 W bulbis really an 8.8 W bulb. So its not so bright.

Now lets look at the same three bulbs in a parallel circuit.The voltage drop for each bulb, since they are in branches,is 117 V. Therefore each bulb produces 75 W and is as

bright as it was designed to be.

So the bulbs are brightest in the parallel circuit.

1 2 0 V

120V


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Houses are wired so that all the lights and the things you plug in will be in parallel. This way eachitem has a voltage drop of 120 V.

What happens to the current when you add a resistor R to (a) a series circuit and (b) a parallelcircuit? The series circuit has two identical resistors R . The parallel circuit also has two of the

same resistors R in parallel.

Series Circuit: In a series circuit, the total resistance is the sum of the resistances. When you add athird resistor you end up changing the resistance from 2R to 3R. The current is equal to:

V I

R= When the resistance is increased, the current gets smaller . It goes from:

2 3

V V I I

R R= =

Parallel Circuit: Lets look at the total resistance for a parallel circuit.

1 1 1 2

2T

T

R R

R R R R= + = =

The current is:2 2

2

V V V I I

R R R= = =

When we add a third resistor:1 1 1 1 3

3T T

R R

R R R R R= + + = =

So the current becomes:3V

I R

= The current gets bigger with each extra resistor.

V

V


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Measuring Current, Voltage, and Resistance: Measuring these values is done withan appropriate meter. Specifically, the voltmeter , ammeter (current meter), and Ohmmeter (resistance meter).These are simple to use- youve already used them combined together into a multimeter.

Measuring Voltage: The voltmeter has a huge internal resistance. Place it in parallel withthe thing you want to read the voltage of.

Heres a simple circuit with a single load (in this case, a resistor).

The voltmeter is placed in parallel with the load our single resistor.

The voltmeter, because its in parallel with the resistor has the same potential difference across it asdoes the resistor. It has a huge resistance millions of Ohms worth, so the current that goes throughit is very, very small and the equivalent resistance of the parallel circuit weve formed (which iswhat we are measuring) is essentially the resistance of the single resistor.

Let R 1 be 35.0 and R m (the resistance of the meter) be 5.50 M . The voltage is 12.0 V. Let uscalculate the equivalent resistance of the parallel circuit and the current through each of the

branches (this would be I 1 and I m ).

So the equivalent resistance:

RV

RV

V

R

V

1

R m


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61

1 1 1 1 1 1 1

35.0 5.50 10 P mi i

R R R R R x= = + = +

61 1 1 1 1 10.02857 0.1818 10 0.02857 0.000001818 0.028572 x R

= + = + =

135.0

10.028572

R = =

So the equivalent resistance is simply the resistance of the lone resistor!

We know the voltage drop is the same for each resistor, so we can find the current.

The current before the meter was placed into the circuit was:

12.0 0.34335.0

V V V IR I A R

= = = =

You can see that the current is still the same after adding the meter (since the equivalent resistanthasnt essentially changed).

The current through the 35.0 resistor is 0.343 A. The current through the meter is:

6

6

12.02.182 10

5.5 10

V V I x A

R x

= = =

This value is so small that we can ignore it it is insignificant.

So when we add a voltmeter, we dont change anything in the circuit. The voltage is the same, thecurrent is the same, and the resistance is the same. The reading for the voltage drop is a correct one.

Measuring Current: To measure current we put the ammeter in series with the circuit. Weknow that the current in a series circuit is the same at all points in the circuit. So we can accuratelymeasure the current. The resistance of the ammeter is very, very small so that the voltage drop itcauses is so small that it too is insignificant.

Measuring Resistance: To measure the resistance of a component or of a circuit you haveto remove the items of interest from the circuit. Then you hook in your Ohmmeter. You place it in

parallel with the components of interest. The Ohmmeter makes its own little circuit. It provides asmall current at a given volume, measures the current, and gives you the resistance for the circuit.


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RC Circuit

Capacitors:

Symbol for capacitor

Parallel Combination:

1 2eqC C C = +

Total charge is:

1 2Q Q Q= +

V is the same for all capacitors in parallel

Find the equivalent capacitance for this circuit, thecharge on the 1.85 F capacitor, and the totalcharge in all the capacitors.

1 2 3eqC C C C = + +

2.35 1.85 3.15 7.35eqC F F F F = + + = Q CV =

( )2 1.85 12.0 2.22Q F V C = =

( )7.35 12.0 88.2Q F V C = =

Series Combination:

1 2

1 1 1 1

eq nC C C C = + + +

V across capacitors may be different

Q for each is the same.

q1

q2

c1

c2

V

V=12.0 V

2.35 F

1.85 F

3.15 F


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Find the equivalent capacitance for thiscircuit, the total charge in all the capacitors,and the charge on the 2.85 F capacitor.

1 2 3eqC C C C = + + .

1 2 3

1 1 1 1

eqC C C C = + +

1 1 1 1

3.35 2.85 5.25eqC F F F = + +

1.19eqC F =

Q CV =

( )1.19 12.0 14.3Q F V C = =

2 14.3Q C =

RC Circuits:

Takes time to charge a resistor

Charges to maximum equilibrium value

Q C =

Max voltage across capacitor Once capacitor is charged, current is zero.

At t = 0, current flows

V=12.0 V

3.35 F

2.85 F

5.25 F

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Circuits - Teach Yourself