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CHAPTER 19 DC Circuits

Units

• EMF and Terminal Voltage • Resistors in Series and in Parallel • Kirchhoff’s Rules • EMFs in Series and in Parallel; Charging a Battery • Circuits Containing Capacitors in Series and in Parallel • RC Circuits – Resistor and Capacitor in Series • Electric Hazards • Ammeters and Voltmeters

MEANING OF ELECTRIC SYMBOLS

ammeter voltmeter galvanometer

resistor

capacitor

cell

switch

electric connection

battery of cells in series

battery of cells in parallel

EMF and Terminal Voltage

Electric circuit needs battery or generator to produce current – these are called sources of emf. Battery is a nearly constant voltage source, but does have a small internal resistance, which reduces the actual voltage from the ideal emf:

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This resistance behaves as though it were in series with the emf.

Example 1: Battery with internal resistance A 65.0 resistor is connected to the terminals of a battery whose

emf is 12.0V and whose internal resistance is 0.50 . Calculate a) The current in the circuit.

b) The terminal voltage of the battery, abV .

c) the power dissipated in the resistor R and the battery’s internal resistance r.

Resistors in Series and in Parallel

A series connection has a single path from the battery, through each circuit element in turn, then back to the battery.

The current through each resistor is the same; the voltage depends on the resistance. The sum of the voltage drops across the resistors equals the battery voltage.

12.00.183A

65.0 0.5

VI

R r

12.0 (0.183A)(0.5 ) 11.9V V

2 2

2 2

(0.183A) (65.0 ) 2.18W

(0.183A) (0.5 ) 0.02W

R

r

P I R

P I r

3

From this we get the equivalent resistance (that single resistance that gives the same current in the circuit).

When several resistors are put in series the total or equivalent resistances is the sum of the separate resistances.

Equivalent Resistance – Series

Four resistors are replaced with their equivalent resistance

This sum applies to any number of resistances in series. Note that when you add more resistance to the circuit, the current through the circuit will decrease. For example, if a 12-V battery is connected to a 4 resistor, the current will be 3A. But if the 12-V battery is connected to three 4 resistors in series, the total resistance is 8 and the current through the entire circuit will be only 1 A.

Resistors in Series and in Parallel

A parallel connection splits the current; the voltage across each resistor is the same:

The wiring in houses and buildings is arranged so all electric devices are in parallel. With parallel wiring, if you disconnect one device, the current to the other devices is not interrupted.

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In a parallel circuit, the total current I that leaves the battery splits into three separate paths. Let

1 2 3, , I I and I be the currents through each of the resistors, 1 2 3, , R R and R respectively. The

current I flowing into junction A must equal the current flowing out of the junction. The total current is the sum of the currents across each resistor:

Equivalent Resistance – Parallel

Equivalent resistance replaces the two original resistances

• Equivalent Resistance

• The inverse of the equivalent resistance of two or more resistors connected in parallel is the algebraic sum of the inverses of the individual resistance

– The equivalent is always less than the smallest resistor in the group

Problem-Solving Strategy, 1

• Combine all resistors in series – They carry the same current – The potential differences across them are not the same – The resistors add directly to give the equivalent resistance of the series

combination: Req = R1 + R2 + …

1 2 3

1 1 1 1

eqR R R R

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Problem-Solving Strategy, 2

• Combine all resistors in parallel – The potential differences across them are the same – The currents through them are not the same – The equivalent resistance of a parallel combination is found through reciprocal

addition:

Problem-Solving Strategy, 3

• A complicated circuit consisting of several resistors and batteries can often be reduced to a simple circuit with only one resistor

– Replace any resistors in series or in parallel using steps 1 or 2. – Sketch the new circuit after these changes have been made – Continue to replace any series or parallel combinations – Continue until one equivalent resistance is found

Problem-Solving Strategy, 4

• If the current in or the potential difference across a resistor in the complicated circuit is to be identified, start with the final circuit found in step 3 and gradually work back through the circuits

– Use ΔV = I R and the procedures in steps 1 and 2

Equivalent Resistance – Complex Circuit

1 2 3

1 1 1 1

eqR R R R

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Example 2: Series and parallel resistors Two 100 resistors are connected in (a) series. What is the current through each resistor and what is the equivalent resistance of each circuit? (b) parallel, to a 24.0V battery. Any given charge (or electron) can flow through only one or the other of the two resistors. The total current I from the battery split to flow through each resistor, so I equals the sum of the separate currents through the two resistors.

Example 3: Circuit with series and parallel How much current is drawn from the battery shown to the right? The current I that flows out of the battery all passes through the 400 resistor, but then splits into I1 and I2 passing through the 500 and 700 resistors.

1 2

24.00.120A

100 100

V VI

R R

24.0200

0.120Aeq

V VR

I

1 2

24.0 24.0

100 100

V V V VI

R R

0.24A + 0.24A=0.48A

24.050

0.48Aeq

V VR

I

1 11 1 10.0020 0.0014

500 700PR

10.0034

1

1290

0.0034PR

400 290 690eqR

12.00.0174A 17 A

690eq

V VI m

R

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Example 4: Current in one branch What is the current through the 500 resistor?

Example 5: Analyzing a circuit A 9.0-V battery whose internal resistance r is 0.050 connected in the circuit. How much current is drawn from the battery? To find the current out of the battery first determine the equivalent resistance of the entire circuit, including the r.

This 2.7 is in series with the 6.0 resistor. The net resistance of the lower arm of the circuit is

This 4.8 is in series with the 5.0 resistor and the 0.50 internal resistance of the battery so the total equivalent resistance of the circuit is

The internal voltage of the battery is

V IR

(0.0174A)(400 ) 7.0abV V

2

1

5.01.0 10 A 10 A

500

VI x m

1

5.07 A

700

VI m

1

1 1 1 3

8.0 4.0 8.0eqR

1 2.7eqso R

2 6.0 2.7 8.7eqR

1

3

1 1 10.21

10.0 8.7eqR

3 1

1 4.8

0.21eqso R

4.8 5.0 0.50 10.3eqR

9.00.87A

10.3eq

VI

R

abV Ir

9.0 (0.87A)(0.50 ) 8.6V V

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To determine the current in the 6.0 resistor:

It must be the same as the current through the8.7 . Why? The

voltage across that will be 8.7 the emf of the battery minus the

voltage drops across r and the 5.0 resistor:

Example 6: Resistor ladder. Estimate the equivalent resistance of the ladder of equal 100 resistors. Start at the end where the three are in series.

Next we can see that this combination is in parallel with the next resistor to the left. The equivalent resistance of the resistors in the dashed box is

which numerically is

The resistor in the dashed box (c) are in series, and are equivalent to

which equals Now this 11 / 4R is in parallel with the next step on the ladder

The resistance in the dashed box (d) is equivalent to

so

This in turn is in series with two more, yielding the final equivalent resistance of

9.0 (0.87A)(0.50 5.0 )' 0.48A

8.7

VI

1 2 33R R R R 3(100 ) 300

1

1

1 1 1 4 3 so

3 3 4eq

eq

RR

R R R R

300 / 4 75

2 3 / 4 11 / 4R R R

1100 / 4 275

1 1 4 15

11 11eqR R R R

2

11

15eq

RR

11 41

15 15eq

RR R R R 100 so 273eqR R

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Gustav Kirchhoff

• 1824 – 1887 • Invented spectroscopy with Robert Bunsen • Formulated rules about radiation

Kirchhoff’s Rules

• There are ways in which resistors can be connected so that the circuits formed cannot be reduced to a single equivalent resistor

• Two rules, called Kirchhoff’s Rules can be used instead

For these circuits we use Kirchhoff’s rules. Junction rule: The sum of currents entering a junction equals the sum of the currents leaving it.

• I1 = I2 + I3 • From Conservation of Charge • Diagram b shows a mechanical analog

Loop Rule

– The sum of the potential differences across all the elements around any closed circuit loop must be zero

• A statement of Conservation of Energy

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• Traveling around the loop from a to b • In a, the resistor is transversed in the direction of the

current, the potential across the resistor is –IR • In b, the resistor is transversed in the direction opposite of

the current, the potential across the resistor is +IR

• In c, the source of emf is transversed in the direction of the emf (from – to +), the change in the electric potential is +ε

• In d, the source of emf is transversed in the direction opposite of the emf (from + to -), the change in the electric potential is -ε

Junction Equations from Kirchhoff’s Rules

• Use the junction rule as often as needed, so long as, each time you write an equation, you include in it a current that has not been used in a previous junction rule equation

– In general, the number of times the junction rule can be used is one fewer than the number of junction points in the circuit

• The loop rule can be used as often as needed so long as a new circuit element (resistor or battery) or a new current appears in each new equation

• You need as many independent equations as you have unknowns

Problem-Solving Strategy – Kirchhoff’s Rules

• Draw the circuit diagram and assign labels and symbols to all known and unknown quantities

• Assign directions to the currents. • Apply the junction rule to any junction in the circuit • Apply the loop rule to as many loops as are needed to solve for the unknowns • Solve the equations simultaneously for the unknown quantities • Check your answers

EMFs in Series and in Parallel; Charging a Battery

When two or more sources of emf, such as batteries, are arranged in series, the total voltage is the algebraic sum of their respective voltages. When a 20-V and a 12-V battery are connected oppositely the net voltage is 8-V. That is, a positive test charge moved from a to b gains in potential by 20-V, but when it passes from b to c it drops by 12-V. So the net change is 20V – 12 V = 8 V.

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You might think that connecting batteries in reverse like this would be wasteful. For most purposes that would be true. But such a reverse arrangement is precisely how a battery charger works. The 20-V source is charging up the 12-V battery. Because of its greater voltage, the 20-V source is

forcing charge back into the 12-V battery. Electrons are being forced into its negative terminal and removed from its positive terminal. EMF’s in parallel only make sense if the voltages are the same; this arrangement can produce more current than a single emf.

Example 8: Jump starting a car A good battery is used to jump start a car with a weak battery. The good battery has an emf of 12.5 V and an internal resistance of 0.020 . The

weak battery has an emf of 10.1 V and internal resistance of 0.10 . Each jumper cable is 3.0 m long and 0.50 cm in diameter, and can be attached as shown. Assume the starter motor can be represented as a

resistor 0.15SR .

Determine the current through the starter motor (a) if only the weak battery is connected to it. The current with only the weak battery and no jumper cables is an emf

connected to two resistances in series:

Determine the current through the starter motor (b) if the good battery is also connected. Apply Kirchhoff’s rules – first determine the resistance of the jumper cables using their dimensions and the resistivity for copper Find the resistance of the jumper cables that connect the good battery. Kirchhoff’s loop rule for the full outside gives:

(a) Since The loop rule for the lower loop, including the weak battery and the starter, gives

(b)

0.10 0.15 0.25

/I V R 10.1V / 0.25 40A

81.68 10x m

/JR L A 8 2 2(1.68 10 )(3.0 ) / (0.25 10 )x m m x m 0.0026

1 1 3

1 3

12.5 (2 ) 0

12.5 (0.025 ) (0.15 ) 0

J SV I R r I R

V I I

(2 ) (0.0052 0.020 ) 0.025RJ r

3 210.1 (0.15 ) (0.10 ) 0V I I

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The junction rule at point B gives

(c)

We have three equations in three unknowns: 1 3 2I I I and we substitute this:

Combining this last equation with (b) gives The other currents are The terminal voltage of the weak 10.1V battery is

Circuits Containing Capacitors in Series and in Parallel

Capacitors in parallel have the same voltage across each one Capacitors in parallel are shown to the right. In this case, the total capacitance is the sum:

Capacitors in series have the same charge: In this case, the reciprocals of the capacitances add to give the reciprocal of the equivalent capacitance:

1 2 3I I I

3 2 3

3 2

12.5 ( )(0.025 ) (0.15 ) 0

12.5 (0.175 ) (0.025 ) 0

V I I I

V I I

3 71I A

2 15 and 76I A I A

10.1 ( 5 )(0.10 ) 10.6BAV V A V

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CHAPTER 16 – 17 – 18 CURRENTS – RESISTANCE – CIRCUITS

CONCEPTS

1. The number of electrons that pass a certain point in a conductor in a given amount of time is defined as an electric current. 2. An electric current in a metallic solid consists of moving electrons. 3. Electrical conductivity in liquid solutions depends on the presence of free ions. 4. Conductivity in ionized gases depends on positive ions, negative ions, and free electrons. 5. The condition that must exist between two points in a conductor in order to maintain a flow of charge is a potential difference. 6. The ratio of the potential difference across a conductor to the current in the conductor is called resistance. 7. If the diameter of a wire were to increase, its electrical resistance would decrease. 8. If the length of a copper wire is reduced by half, the resistance of the wire will be halved. 9. The resistance of a metallic wire conductor is inversely proportional to its cross-sectional area. 10. If both the cross-sectional area and the length of a metallic conductor were doubled, the resistance of the conductor would be unchanged. 11. The graph that best represents how the resistance (R) of a series of copper wires of uniform length and temperature varies with cross-sectional area (A) is D.

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12. The graph that best represents the relationship between the resistance of a copper wire of uniform cross-sectional area and the wire’s length at constant temperature is C. 13. As the temperature of a metallic conductor increases, its resistance usually increases. 14. The graph to the right shows the relationship between current and potential difference for four resistors, A, B, C, and D. The resistor with the greatest resistance is A.

15. The slope of the line on the graph to the right represents resistance of a material.

16. The graph to the right shows how the voltage and current are related in a simple electric circuit. For any point on the line, the ratio of V to I represents resistance in ohms.

17. The graph o the right that best represents a material behaving according to Ohm’s law is D.

18. The graph on the right that best represents the relationship between current (I) and the potential difference (V) in a circuit in which resistance remains constant is B.

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19. The two resistor arrangements to the right that have equivalent resistance is B and C. 20. Diagram A shown below shows the correct current direction in a circuit segment.

21. The diagram on the right represents a segment of a circuit. The current in wire X may be 2 amperes. 22. The quantity that must be the same for each component in any series circuit is current. 23. As more resistors are added in series across a battery, the potential drop across each resistor decreases. 24. The unit of electric power is volt2/ohm. 25. As the resistance of a constant-voltage circuit is increased, the power developed in the circuit decreases. 26. As the resistance of a lamp operating at a constant voltage increases, the power dissipated by the lamp decreases. 27. Joule/second is a unit equivalent to a watt, the SI unit of power. 28. The diagram to the right represents a segment of a circuit. The current in ammeter A is 7 amperes. 29. The diagram to the right shows the current in a segment of a direct current circuit. The reading of ammeter A is 5 A. 30. An electric heater rated at 4,800 watts is operated on 120 volts. If the heater were replaced by one having a greater resistance, the amount of heat produced each second would decrease.