Chapter 18Direct Current Circuits

Read and take notes on

pages 548-558 in Conceptual Physics

TextOr lesson 4 B Two Types of Connections

on Physics Classroom

Bright Storm on Electric Circuits

(Start to minute 1:50)

Read and take notes on

pages 594-597 in College Physics Text

Electric Circuits

• Electric circuits control the flow of electricity and the energy associated with it.

• Circuits are used in many applications.• Kirchhoff’s Rules will simplify the analysis of

simple circuits.• Some circuits will be in steady state.– The currents are constant in magnitude and direction.

• In circuits containing resistors and capacitors, the current varies in time.

Introduction

Khan Academy on Circuits and Ohm’s Law

Sources of emf

• The source that maintains the current in a closed circuit is called a source of emf.– Any devices that increase the potential energy of charges

circulating in circuits are sources of emf.– Examples include batteries and generators

• SI units are Volts– The emf is the work done per unit charge.

Section 18.1

emf and Internal Resistance

• A real battery has some internal resistance.

• Therefore, the terminal voltage is not equal to the emf.

Section 18.1

More About Internal Resistance• The schematic shows the

internal resistance, r• The terminal voltage is ΔV =

Vb-Va

• ΔV = ε – Ir• For the entire circuit,

ε = IR + Ir

Section 18.1

Internal Resistance and emf, Cont.

• ε is equal to the terminal voltage when the current is zero.– Also called the open-circuit voltage

• R is called the load resistance.• The current depends on both the resistance

external to the battery and the internal resistance.

Section 18.1

Internal Resistance and emf, Final

• When R >> r, r can be ignored.– Generally assumed in problems

• Power relationship– I e = I2 R + I2 r•When R >> r, most of the power delivered

by the battery is transferred to the load resistor.

Section 18.1

Batteries and emf

• The current in a circuit depends on the resistance of the battery.– The battery cannot be considered a source of

constant current.• The terminal voltage of battery cannot be

considered constant since the internal resistance may change.

• The battery is a source of constant emf.

Section 18.1

Bright Storm on Resistors in Series

Resistors in Series• When two or more resistors are connected end-to-

end, they are said to be in series.• The current is the same in all resistors because any

charge that flows through one resistor flows through the other.

• The sum of the potential differences across the resistors is equal to the total potential difference across the combination.

Khan Academy on Resistors in Series Circuits (A must watch, explains much more than how to add resistors)

Resistors in Series, Cont.

Section 18.2

• Potentials add– ΔV = IR1 + IR2 = I (R1+R2)– Consequence of Conservation of Energy

• The equivalent resistance has the effect on the circuit as the original combination of resistors.

Equivalent Resistance – Series

• Req = R1 + R2 + R3 + …• The equivalent resistance of a series

combination of resistors is the algebraic sum of the individual resistances and is always greater than any of the individual resistors.

• If one element in the series circuit fails, the circuit would no longer be complete and none of the elements would work.

Section 18.2

Equivalent Resistance – SeriesAn Example

Section 18.3

• Four resistors are replaced with their equivalent resistance.

EXAMPLE 18.1 Four Resistors in Series

(a) Four resistors connected in series. (b) The equivalent resistance of the circuit in (a). Goal Analyze several resistors connected in series. Problem Four resistors are arranged as

shown in figure a. Find (a) the equivalent resistance of the circuit and (b) the current in the circuit if the emf of the battery is 6.0 V. Strategy Because the resistors are connected in series, summing their resistances gives the equivalent resistance. Ohm's law can then be used to find the current.

SOLUTION

(a) Find the equivalent resistance of the circuit. Sum the resistances to find the equivalent resistance. Req = R1 + R2 + R3 + R4

Req = 2.0 Ω + 4.0 Ω + 5.0 Ω + 7.0 Ω

Req = 18.0 Ω

(b) Find the current in the circuit.

Apply Ohm's law to the equivalent resistor in figure b, solving for the current.

I = ΔV

= 6.0 V

= 1/3 A Req 18.0 Ω

LEARN MORE

Remarks A common misconception is that the current is "used up" and steadily declines as it progresses through a series of resistors. That would be a violation of conservation of charge. Question The internal resistance of the battery is neglected in this example. How would it affect the solution for the final current, if taken into account? (Select all that apply.)

It would not affect the current. It would increase the equivalent

resistance of the circuit. It would not affect the equivalent resistance of the

circuit. It would decrease the current. It would decrease the equivalent

resistance of the circuit. It would increase the current.

Bright Storm on Resistors in Parallel

(Start to minute 3:30)

Read and take notes on

pages 598-599 in College Physics Text

Resistors in Parallel• The potential difference across each resistor is the

same because each is connected directly across the battery terminals.

• The current, I, that enters a point must be equal to the total current leaving that point.– I = I1 + I2

– The currents are generally not the same.– Consequence of Conservation of Charge

Section 18.3

Equivalent Resistance – ParallelAn Example

Section 18.3

• Equivalent resistance replaces the two original resistances.• Household circuits are wired so the electrical devices are

connected in parallel.– Circuit breakers may be used in series with other circuit elements for

safety purposes.

Khan Academy on Resistors in Parallel

(Another must watch!)

Equivalent Resistance – Parallel• 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.

Section 18.3

EXAMPLE 18.2 Three Resistors in Parallel

Three resistors connected in parallel. The voltage across each resistor is 18 V. Goal Analyze a circuit having resistors connected in parallel. Problem Three resistors are connected in parallel as in the figure. A potential difference of 18 V is maintained between points a and b. (a) Find the current in each resistor. (b) Calculate the power delivered to each

resistor and the total power. (c) Find the equivalent resistance of the circuit. (d) Find the total power delivered to the equivalent resistance. Strategy To get the current in each resistor we can use Ohm's law and the fact that the voltage drops across parallel resistors are all the same. The rest of the problem just requires substitution into the equation for power delivered to a resistor, = I2R, and the reciprocal-sum law for parallel resistors.

SOLUTION

(a) Find the current in each resistor. Apply Ohm's law, solved for the current I delivered by the battery to find the current in each resistor.

I1 = ΔV

= 18 V

= 6.0 A R1 3.0 Ω

I2 = ΔV

= 18 V

= 3.0 A R2 6.0 Ω

I3 = ΔV

= 18 V

= 2.0 A R3 9.0 Ω

(b) Calculate the power delivered to each resistor and the total power.

Apply = I2R to each resistor, substituting the results from part (a).

3 Ω: 1 = I12R1 = (6.0 A)2(3.0 Ω) = 110 W

6 Ω: 2 = I22R2 = (3.0 A)2(6.0 Ω) = 54 W

9 Ω: 3 = I32R3 = (2.0 A)2(9.0 Ω) = 36 W

Sum to get the total power.

tot = 110 W + 54 W + 36 W = 2.0 102 W

(c) Find the equivalent resistance of the circuit.

Apply the reciprocal-sum rule.

1 =

1 +

1 +

1 Req R1 R2 R3 1

= 1

+ 1

+ 1

= 11

Req 3.0 Ω 6.0 Ω 9.0 Ω 18 Ω

Req = 18

Ω = 1.6 Ω 11

(d) Compute the power dissipated by the equivalent resistance.

Use the alternate power equation.

= (ΔV)2

= (18 V)2

= 2.0 102 W Req 1.6 Ω

LEARN MORE

Remarks There's something important to notice in part (a): the smallest 3.0 Ω resistor carries the largest current, whereas the other, larger resistors of 6.0 Ω and 9.0 Ω carry smaller currents. The largest current is always found in the path of least resistance. In part (b) the power could also be found with = (ΔV )2/R. Note that 1 = 108 W, but is rounded to 110 W because there are only two significant figures. Finally, notice that the total power dissipated in the equivalent resistor is the same as the sum of the power dissipated in the individual resistors, as it should be. Question If a fourth resistor were added in parallel to the other three, how would the equivalent resistance change?

It would be larger. It would be smaller. More information is needed to determine the effect.

Bright Storm on Solving Complex Resistor

Problems (It is long but VERY effective!!!)

If you didn’t read and take notes on

pages 548-558 in Conceptual Physics Text then read

and take notes on the following link

lesson 4 E Combination Circuitson Physics Classroom

Read and take notes on

pages 600-602 in College Physics Text

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 + …

– Draw the simplified circuit diagram.

Section 18.3

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:

– Remember to invert the answer after summing the reciprocals.

– Draw the simplified circuit diagram.

Section 18.3

Problem-Solving Strategy, 3• Repeat the first two steps as necessary.– 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.

Section 18.3

Problem-Solving Strategy, 4

• Use Ohm’s Law.– Use ΔV = I R to determine the current in the

equivalent resistor.– Start with the final circuit found in step 3 and

gradually work back through the circuits, applying the useful facts from steps 1 and 2 to find the current in the other resistors.

Section 18.3

Khan Academy Example Problem (Great Practice Problem)

Example• Complex circuit

reduction– Combine the resistors in

series and parallel.• Redraw the circuit

with the equivalents of each set.

– Combine the resulting resistors in series.

– Determine the final equivalent resistance.

Section 18.3

EXAMPLE 18.3 Equivalent Resistance

Figure (a) The four resistors shown above can be reduced in steps to an equivalent 14 Ω resistor. Goal Solve a problem involving both series and parallel resistors. Problem Four resistors are connected as shown in figure a. (a) Find the equivalent resistance between points a and c. (b) What is the current in each resistor if a 42-V battery is connected between a and c? Strategy Reduce the circuit in steps, as shown in figures b and c, using the sum rule for resistors in series and the reciprocal-sum rule for resistors in parallel.

Finding the currents is a matter of applying Ohm's law while working backwards through the diagrams.

SOLUTION

(a) Find the equivalent resistance of the circuit. The 8.0-Ω and 4.0-Ω resistors are in series, so use the sum rule to find the equivalent resistance between a and b: Req = R1 + R2 = 8.0 Ω + 4.0 Ω = 12.0 Ω

The 6.0-Ω and 3.0-Ω resistors are in parallel, so use the reciprocal-sum rule to find the equivalent resistance between b and c (don't forget to invert!): 1

= 1

+ 1

= 1

+ 1

= 1

Req R1 R2 6.0 Ω 3.0 Ω 2.0 Ω Req = 2.0 Ω

In the new diagram, figure b, there are now two resistors in series. Combine them with the sum rule to find the equivalent resistance of the circuit. Req = R1 + R2 = 12.0 Ω + 2.0 Ω = 14.0 Ω

(b) Find the current in each resistor if a 42-V battery is connected between points a and c.

Find the current in the equivalent resistor in figure c, which is the total current. Resistors in series all carry the same current, so the value is the current in the 12-Ω resistor in figure b and also in the 8.0-Ω and 4.0-Ω resistors in figure a:

I = ΔVac =

42 V = 3.0 A

Req 14 Ω Apply the junction rule to point b:

(1) I = I1 + I2

The 6.0-Ω and 3.0-Ω resistors are in parallel, so the voltage drops across them are the same:

ΔV6Ω = ΔV3Ω → (6.0 Ω)I1 = (3.0 Ω)I2 → 2.0 I1 = I2

Substitute this result into Equation (1), with I = 3.0 A.

3.0 A = I1 + 2I1 = 3I1 → I1 = 1.0 A I2 = 2.0 A

LEARN MORE

Remarks As a final check, note that ΔVbc = (6.0 Ω)I1 = (3.0 Ω)I2 = 6.0 V and ΔVab = (12 Ω)I1 = 36 V; therefore, ΔVac + ΔVab + ΔVbc = 42 V, as expected. Question Which of the original resistors dissipates energy at the greatest rate?

the 4.0 Ω resistor the 6.0 Ω resistor the 8.0 Ω resistor the 3.0 Ω resistor

Gustav Kirchhoff

• 1824 – 1887• Invented spectroscopy

with Robert Bunsen• Formulated rules about

radiation

Section 18.4

Read and take notes on

pages 603-604 in College Physics Text

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.

Section 18.4

Statement of Kirchhoff’s Rules

• Junction Rule– The sum of the currents entering any junction

must equal the sum of the currents leaving that junction.• A statement of Conservation of Charge

• 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

Section 18.4

More About the Junction Rule

• I1 = I2 + I3

• From Conservation of Charge

• Diagram b shows a mechanical analog.

Section 18.4

Loop Rule

• A statement of Conservation of Energy• To apply Kirchhoff’s Rules,– Assign symbols and directions to the currents in all

branches of the circuit.• If the direction of a current is incorrect, the answer will

be negative, but have the correct magnitude.– Choose a direction to transverse the loops.• Record voltage rises and drops.

Section 18.4

More About the Loop Rule• 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.

Section 18.4

Loop Rule, Final• 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

Section 18.4

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.

Section 18.4

Loop Equations from Kirchhoff’s Rules

• 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.

Section 18.4

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.

Section 18.4

EXAMPLE 18.4 Applying Kirchhoff's Rules

Goal Use Kirchhoff's rules to find currents in a circuit with three currents and one battery. Problem Find the currents in the circuit shown in the figure by using Kirchhoff's rules. Strategy There are three unknown currents in this circuit, so we must obtain three independent equations, which then can be solved by substitution. We can find

the equations with one application of the junction rule and two applications of the loop rule. We choose junction c. (Junction d gives the same equation.) For the loops, we choose the bottom loop and the top loop, both shown by blue arrows, which indicate the direction we are going to traverse the circuit mathematically (not necessarily the direction of the current). The third loop gives an equation that can be obtained by a linear combination of the other two, so it provides no additional information and isn't used.

SOLUTION

Apply the junction rule to point c. I1 is directed into the junction, I2 and I3 are directed out of the junction. I1 = I2 + I3

Select the bottom loop and traverse it clockwise starting at point a, generating an equation with the loop rule. ΣΔV = ΔVbat + ΔV4.0 Ω + ΔV9.0 Ω = 0

6.0 V - (4.0 Ω)I1 - (9.0 Ω)I3 = 0

Select the top loop and traverse it clockwise from point c. Notice the gain across the 9.0 Ω resistor because it is traversed against the direction of the current! ΣΔV = ΔV5.0 Ω + ΔV9.0 Ω = 0

-(5.0 Ω)I2 + (9.0 Ω)I3 = 0

Rewrite the three equations, rearranging terms and dropping units for the moment, for convenience. (1) I1 = I2 + I3 (2) 4.0I1 + 9.0I3 = 6.0 (3) -5.0I2 + 9.0I3 = 0 Solve Equation (3) for I2 and substitute into Equation (1). I2 = 1.8I3

I1 = I2 + I3 = 1.8I3 + I3 = 2.8I3

Substitute the latter expression into Equation (2) and solve for I3. 4.0 (2.8I3) + 9.0I3 = 6.0 → I3= 0.30 A

Substitute I3 back into Equation (3) to get I2. -5.0I2 + 9.0(0.30 A) = 0 → I2 = 0.54 A

Substitute I3 into Equation (2) to get I1. 4.0I1 + 9.0(0.30 A) = 6.0 → I1 = 0.83 A

LEARN MORE

Remarks Substituting these values back into the original equations verifies that they are correct, with any small discrepancies due to rounding. The problem can also be solved by first combining resistors. Question How would the answers change if the assumed directions of the currents used in solving the problem were all reversed in the figure above, but you followed the circuit paths in the same way to write down the equations based on Kirchhoff's rules? (Select all that apply.)

The sign of each current-times-resistance term would reverse. The sign

used for the voltage across the battery would reverse. The sign of the

resistances would reverse. The signs of the currents after solving would reverse.

EXAMPLE 18.5 Another Application of Kirchhoff's Rules

Goal Find the currents in a circuit with three currents and two batteries when some current directions are chosen wrongly. Problem Find I1, I2, and I3 in figure a. Strategy Use Kirchhoff's two rules,

the junction rule once and the loop rule twice, to develop three equations for the three unknown currents. Solve the equations simultaneously.

SOLUTION

Apply Kirchhoff's junction rule to junction c. Because of the chosen current directions, I1 and I2 are directed into the junction and I3 is directed out of the junction. (1) I3 = I1 + I2 Apply Kirchhoff's loop rule to the loops abcda and befcb. (Loop aefda gives no new information.) In loop befcb, a positive sign is obtained when the 6.0 Ω resistor is traversed because the direction of the path is opposite the direction of the current I1. (2) Loop abcda: 10 V - (6.0 Ω)I1 - (2.0 Ω)I3 = 0 (3) Loop befcb: -(4.0 Ω)I2 - 14 V + (6.0 Ω)I1 - 10 V = 0 Using Equation (1), eliminate I3 from Equation (2) (ignore units for the moment): 10 - 6I1 - 2.0 (I1 + I2) = 0

(4) 10 = 8.0I1 + 2.0I2 Divide each term in Equation (3) by 2 and rearrange the equation so that the currents are on the right side: (5) -12 = -3.0I1 + 2.0I2 Subtracting Equation (5) from Equation (4) eliminates I2 and gives I1: 22 = 11I1 → I1 = 2.0 A

Substituting this value of I1 into Equation (5) gives I2: 2.0I2 = 3.0I1 - 12 = 3.0(2.0) - 12 = -6.0 A

I2 = -3.0 A

Finally, substitute the values found for I1 and I2 into Equation (1) to obtain I3: I3 = I1 + I2 = 2.0 A - 3.0 A = -1.0 A

LEARN MORE

Remarks The fact that I2 and I3 are both negative indicates that the wrong directions were chosen for these currents. Nonetheless, the magnitudes are correct. Choosing the right directions of the currents at the outset is unimportant because the equations are linear, and wrong choices result only in a minus sign in the answer. Question Is it possible for the current through a battery to be directed from the positive terminal toward the negative terminal? (Select all that apply.)

Yes. It happens inside a rechargeable flashlight battery or a car battery

during recharge. Yes. It can happen when the voltage applied to a battery

is greater than its emf. No. The charge carriers inside a battery is never reversible.

Household Circuits• The utility company

distributes electric power to individual houses with a pair of wires.

• Electrical devices in the house are connected in parallel with those wires.

• The potential difference between the wires is about 120V.

Section 18.6

Household Circuits, Cont.• A meter and a circuit breaker are connected in series

with the wire entering the house.• Wires and circuit breakers are selected to meet the

demands of the circuit.• If the current exceeds the rating of the circuit

breaker, the breaker acts as a switch and opens the circuit.

• Household circuits actually use alternating current and voltage.

Section 18.6

Circuit Breaker Details• Current passes through

a bimetallic strip.– The top bends to the left

when excessive current heats it.

– Bar drops enough to open the circuit

• Many circuit breakers use electromagnets instead.

Section 18.6

240-V Connections

• Heavy-duty appliances may require 240 V to operate.

• The power company provides another wire at 120 V below ground potential.

Section 18.6

Electrical Safety

• Electric shock can result in fatal burns.• Electric shock can cause the muscles of vital

organs (such as the heart) to malfunction.• The degree of damage depends on– The magnitude of the current– The length of time it acts– The part of the body through which it passes

Section 18.7

Effects of Various Currents

• 5 mA or less– Can cause a sensation of shock– Generally little or no damage

• 10 mA– Hand muscles contract– May be unable to let go of a live wire

• 100 mA – If passes through the body for just a few seconds, can be

fatal

Section 18.7

Ground Wire

• Electrical equipment manufacturers use electrical cords that have a third wire, called a case ground.

• Prevents shocks

Section 18.7

Ground Fault Interrupts (GFI)

• Special power outlets• Used in hazardous areas• Designed to protect people from electrical

shock• Senses currents (of about 5 mA or greater)

leaking to ground• Shuts off the current when above this level

Section 18.7

Electrical Signals in Neurons• Specialized cells in the body, called neurons, form

a complex network that receives, processes, and transmits information from one part of the body to another.

• Three classes of neurons– Sensory neurons

• Receive stimuli from sensory organs that monitor the external and internal environment of the body

– Motor neurons• Carry messages that control the muscle cells

– Interneurons• Transmit information from one neuron to another

Section 18.8

Diagram of a Neuron

Section 18.8

Link to Homework Questions on Web

AssignUnit 3 C Circuits

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AssignUnit 3 C Circuits

Grading Rubric for Unit 3C Circuits Name: ______________________ Conceptual notes Lesson 4 B Two Types of Connections Physics Classroom --------------------------____ Lesson 4 E Combination Circuits Physics Classroom --------------------------------____ Conceptual Physics Text Pgs 548-558 ----------------------------------------------------____ Advanced notes from text book:

Pgs 594-597 --------------------------------------------------------------------------------------------_____ Pgs 598-599 --------------------------------------------------------------------------------------------_____ Pgs 600-602 --------------------------------------------------------------------------------------------_____ Pgs 603-604 --------------------------------------------------------------------------------------------_____

Example Problems and Questions: 18.1 a,b--------------------------------------------------------------------------------------------------_____ 18.2 a-d--------------------------------------------------------------------------------------------------_____ 18.3 a,b--------------------------------------------------------------------------------------------------_____ 18.4 three currents------------------------------------------------------------------------------------------_____ 18.5 three currents------------------------------------------------------------------------------------------_____ Web Assign 1, 2, 3, 4, 5, 6, 7, 8 -------------------------------------------------------------------------------------____