Electricity Magnetism Lecture 10: Kirchhoff’s Rules Lecture 10 - Kirchhoff's Rule… · If the batteries are ideal andVA = 1.5 V A)VAB = 0.0 V B)VAB = 0.5 V C)VAB = 1.5V D)VAB =

Electricity & MagnetismLecture 10: Kirchhoff’s Rules

Today’s Concept:

Kirchhoff’s Rules

Electricity & Magne7sm Lecture 10, Slide 1

With the switch closed, the voltage measured

A)VleE < Vright

B)VleE = Vright

C)VleE > Vright

(c) Now test your prediction. Connect the circuit in Figure 22-7. Use the voltmeter to measure the voltage across the battery and then use it to measure the voltage across the bulb.

Voltage across the Battery

Voltage across the bulb

What do you conclude about the voltage across the battery and the voltage across the bulb?

Now let's measure voltage and current in your circuit at the same time. To do this, connect a voltmeter and an ammeter so that you are measuring the voltage across the battery and the current entering the bulb at the same time. (See Figure 22-8.)

V

+-

A

+-

V

+ –

+

–

A

Figure 22-8: Meters connected to measure the voltage across the battery and the current through it. (The positive terminal of the battery is at the bottom.)

Activity 22-5: Current and Voltage Measurements(a) Measure the voltage across the battery when the switch is closed and the

light is lit. Enter the value in the table below in Activity 22-6d

Page 22-12 Workshop Physics II Activity Guide SFU

© 1990-93 Dept. of Physics and Astronomy, Dickinson College Supported by FIPSE (U.S. Dept. of Ed.) and NSF. Modified at SFU by N. Alberding & S. Johnson, 2007,2014.

(b) Measure the current through the circuit when the switch is closed and the light is lit. Enter the value in the table below in Activity 22-6d

Now suppose you connect a second bulb, as shown in Figure 22-9.

A

+-

+-

A+ -

V

+-

V

+-

V+

–

+

A

-

+ –A

Figure 22-9: Two bulbs connected in series with a voltmeter and an ammeter.

Activity 22-6: Current and Voltage Measurements with Two BulbsThe predictions below should be completed before class.(a) How do you think the voltage across the battery will compare to that with only one bulb? (More, less or the same within measurement error?)

(b) What do you think will happen to the brightness of the first bulb when you add a second bulb? Explain.

(c) What will happen to the current drawn from the battery? Explain.

(d) Connect a second bulb as shown, and test your predictions. Measure the voltage across both the bulbs and the current entering both bulbs with the switch closed and record in the table.

Measurements 22-5 and 22-6

1 bulb 2 bulbs

voltage

current

Workshop Physics II: Unit 22 – Batteries, Bulbs, & Current Flow Page 22–13Authors: Priscilla Laws, John Luetzelschwab, David Sokoloff, & Ron Thornton

© 1990-93 Dept. of Physics and Astronomy, Dickinson College Supported by FIPSE (U.S. Dept. of Ed.) and NSF. Modified at SFU by N. Alberding & S. Johnson, 2007, 2014.

With the switch closed, the current measured

A)IleE < IrightB)IleE = IrightC)IleE > Iright

(c) Now test your prediction. Connect the circuit in Figure 22-7. Use the voltmeter to measure the voltage across the battery and then use it to measure the voltage across the bulb.

Voltage across the Battery

Voltage across the bulb

What do you conclude about the voltage across the battery and the voltage across the bulb?

Now let's measure voltage and current in your circuit at the same time. To do this, connect a voltmeter and an ammeter so that you are measuring the voltage across the battery and the current entering the bulb at the same time. (See Figure 22-8.)

V

+-

A

+-

V

+ –

+

–

A

Figure 22-8: Meters connected to measure the voltage across the battery and the current through it. (The positive terminal of the battery is at the bottom.)

Activity 22-5: Current and Voltage Measurements(a) Measure the voltage across the battery when the switch is closed and the

light is lit. Enter the value in the table below in Activity 22-6d


© 1990-93 Dept. of Physics and Astronomy, Dickinson College Supported by FIPSE (U.S. Dept. of Ed.) and NSF. Modified at SFU by N. Alberding & S. Johnson, 2007,2014.

(b) Measure the current through the circuit when the switch is closed and the light is lit. Enter the value in the table below in Activity 22-6d

Now suppose you connect a second bulb, as shown in Figure 22-9.

A

+-

+-

A+ -

V

+-

V

+-

V+

–

+

A

-

+ –A

Figure 22-9: Two bulbs connected in series with a voltmeter and an ammeter.

Activity 22-6: Current and Voltage Measurements with Two BulbsThe predictions below should be completed before class.(a) How do you think the voltage across the battery will compare to that with only one bulb? (More, less or the same within measurement error?)

(b) What do you think will happen to the brightness of the first bulb when you add a second bulb? Explain.

(c) What will happen to the current drawn from the battery? Explain.

(d) Connect a second bulb as shown, and test your predictions. Measure the voltage across both the bulbs and the current entering both bulbs with the switch closed and record in the table.

Measurements 22-5 and 22-6

1 bulb 2 bulbs

voltage

current

Workshop Physics II: Unit 22 – Batteries, Bulbs, & Current Flow Page 22–13Authors: Priscilla Laws, John Luetzelschwab, David Sokoloff, & Ron Thornton

© 1990-93 Dept. of Physics and Astronomy, Dickinson College Supported by FIPSE (U.S. Dept. of Ed.) and NSF. Modified at SFU by N. Alberding & S. Johnson, 2007, 2014.

If the batteries are ideal andVA = 1.5 VA)VAB = 0.0 VB)VAB = 0.5 VC)VAB = 1.5VD)VAB = 3.0 VE)something elseF)

V

V

VV

V

Figure 23-2: Voltmeters connected to measure the potential difference across (a) a single battery, (b) a single battery and two batteries connected in series, and (c) a single battery and two batteries connected in parallel.

Activity 23-2: Combinations of Batteries(a) Predict the voltage for each combination of batteries in Fig 23-2. Write

you prediction beside the meter symbols.(b) Measure the voltages you predicted and write them below the predicted

values on the figure.

Using a MultimeterA digital multimeter (DMM) is a device that can be used to measure either current, voltage or resistance depending on how it is set up. We have already used one to measure voltage. The following activity will give you some practice in using it as an ohmmeter. You will need:! ! • A digital multimeter! ! • A D-cell alkaline battery w/ holder ! ! • A SPST switch! ! • 4 alligator clip wires! ! • 1 resistor, 10 Ω

VΩCOMMAA

Ω

V A

MA

Figure 23-6: Diagram of a typical digital multimeter that can be used to measure resistances, currents, and voltages


© 1990-93 Dept. of Physics and Astronomy, Dickinson College Supported by FIPSE (U.S. Dept. of Ed.) and NSF. Modified at SFU by S. Johnson, N. Alberding, 2014.

V

V

VV

V






VΩCOMMAA

Ω

V A

MA




VAVAVAB

V

V

VV

V






VΩCOMMAA

Ω

V A

MA




VAV

V

VV

V






VΩCOMMAA

Ω

V A

MA




VAVAB

If the batteries are ideal andVA = 1.5 VA)VAB = 0.0 VB)VAB = 0.5 VC)VAB = 1.5VD)VAB = 3.0 VE)something elseF)

Comments

"Please explain Kirchhoff in human language."Not very keen on circuits..."I like circuits :P"The Blue Wire"The whole concept of the joined 2 parallel curcuits"Direction of current flow through complicated resistor set-up - 'gains and drops'.

water, pipes, pumps, tanks ...

will talk about these

Current through is same.

Voltage drop across is IRi

Resistors in series:

Voltage drop across is same.

Current through is V/Ri

Resistors in parallel:

Solved Circuits

V

R1 R2

R4

R3V

R1234I1234=

Last Time


THE ANSWER: Kirchhoff’s Rules

I1234

New Circuit


Kirchhoff’s Voltage Rule

Kirchhoff's Voltage Rule states that the sum of the voltage changes caused by any elements (like wires, baYeries, and resistors) around a circuit must be zero.

WHY?The poten@al difference between a point and itself is zero!


Kirchhoff's Current Rule states that the sum of all currents entering any given point in a circuit must equal the sum of all currents leaving the same point.

WHY? Electric Charge is Conserved

Kirchhoff’s Current Rule


Kirchhoff’s Laws

1) Label all currents Choose any direc7on

2) Label +/− for all elements Current goes + ⇒ − (for resistors)

3) Choose loop and direc@onMust start on wire, not element.

4) Write down voltage drops First sign you hit is sign to use.

R4

I1

I3I2 I4

+

+

+ +

+

−

−

−

−

−

+

+

+

−

−

−

R1

E1

R2

R3E2

E3

R5

A

B

5) Write down node equa@on Iin = Iout

I5

We’ll do calcula@on first todayIt’s actually the easiest thing to do!


CheckPoint: Gains and Drops


In the following circuit, consider the loop abc. The direc7on of the current through each resistor is indicated by black arrows.

If we are to write Kirchoff's voltage equa7on for this loop in the clockwise direc7on star7ng from point a, what is the correct order of voltage gains/drops that we will encounter for resistors R1, R2 and R3?

A. drop, drop, dropB. gain, gain, gainC. drop, gain, gainD. gain, drop, dropE. drop, drop, gain

With the current VOLTAGE DROP

DROP

Against the current VOLTAGE GAIN

GAIN

GAIN

2V

1V

1V

Conceptual Analysis: – Circuit behavior described by Kirchhoff’s Rules:

• KVR: Σ Vdrops = 0 • KCR: Σ Iin = Σ Iout

Strategic Analysis– Write down Loop Equa7ons (KVR)– Write down Node Equa7ons (KCR)– Solve

I2

Calculation

In this circuit, assume Vi and Ri are known.

What is I2 ?


+ −

+ −

+ −

This is easy for baYeries

V1R1

R2


What is I2 ?

R3

V2

V3

I1

I3

I2

Label and pick direc7ons for each current

Label the + and − side of each element

− +

+ −

− +

For resistors, the “upstream” side is +

Now write down loop and node equa7ons

Calculation


How many equa7ons do we need to write down in order to solve for I2?

A) 1 B) 2 C) 3 D) 4 E) 5

Why?– We have 3 unknowns: I1, I2, and I3

– We need 3 independent equa7ons to solve for these unknowns

V1R1

R2

R3

V2

V3

+ −

+ −

+ −− +

+ −

− +

I1

I3

I2


What is I2 ?

Calculation


Which of the following equa7ons is NOT correct? A) I2 = I1 + I3 B) − V1 + I1R1 − I3R3 + V3 = 0C) − V3 + I3R3 + I2R2 + V2 = 0D) − V2 − I2R2 + I1R1 + V1 = 0

Why?– (D) is an aYempt to write down KVR for the top loop– Start at nega7ve terminal of V2 and go clockwise

Vgain (−V2) then Vgain (−I2R2) then Vgain (−I1R1) then Vdrop (+V1)

V1R1

R2

R3

V2

V3

+ −

+ −

+ −− +

+ −

− +

I1

I3

I2


What is I2 ?

Calculation


A) Any 3 will do B) 1, 2, and 4 C) 2, 3, and 4

We have the following 4 equa7ons:

1. I2 = I1 + I3 2. − V1 + I1R1 − I3R3 + V3 = 0 3. − V3 + I3R3 + I2R2 + V2 = 0 4. − V2 − I2R2 − I1R1 + V1 = 0Why?

– We need 3 INDEPENDENT equa7ons– Equa7ons 2, 3, and 4 are NOT INDEPENDENT

Eqn 2 + Eqn 3 = − Eqn 4 – We must choose Equa7on 1 and any two of the remaining ( 2, 3, and 4)

We need 3 equa7ons: Which 3 should we use?

V1R1

R2

R3

V2

V3

I1

I3

I2


What is I2 ?

Calculation


V1R1

R2

R3

V2

V3

I1

I3

I2

We have 3 equa7ons and 3 unknowns.I2 = I1 + I3

V1 + I1R1 − I3R3 + V3 = 0V2 − I2R2 − I1R1 + V1 = 0

The solu7on will get very messy!Simplify: assume V2 = V3 = V V1 = 2V R1 = R3 = R R2 = 2R

2VR

2R

R

V

V

I1

I3

I2

Calculation


What is I2 ?


In this circuit, assume V and R are known. What is I2 ?

With this simplifica7on, you can verify:I2 = ( 1/5) V/RI1 = ( 3/5) V/RI3 = (−2/5) V/R

We have 3 equa7ons and 3 unknowns.I2 = I1 + I3

−2V + I1R − I3R + V = 0 (outside)−V − I2(2R) − I1R + 2V = 0 (top)

2VR

2R

R

V

V

I1

I3

I2

current direc7on

Calculation: Simplify


We know:I2 = ( 1/5) V/RI1 = ( 3/5) V/RI3 = (−2/5) V/R

a b

Suppose we short R3: What happens to Vab (voltage across R2?)

A) Vab remains the same

B) Vab changes sign C) Vab increasesD) Vab goes to zero

Why? Redraw:

2VR

2R V

V

I1

I3

I2a b

c

d

2VR

2R

R

V

V

I1

I3

I2

Vab + V − V = 0BoYom Loop Equa7on:

Follow Up

Vab = 0


V R R

a b

Is there a current flowing between a and b ?

A) YesB) No

a & b have the same poten7al No current flows between a & b

Current flows from baYery and splits at aSome current flows down

Some current flows rightElectricity & Magne7sm Lecture 10, Slide 16

Clicker Question

CheckPoint: Circuits w/ Resistors and a Battery 1


Consider the circuit shown below. Which of the following statements best describes the current flowing in the blue wire connec7ng points a and b?

A. Posi7ve current flows from a to bB. Posi7ve current flows from b to aC. No current flows between a and b

I1R − I2 (2R) = 0

I4R − I3 (2R) = 0

I = I1 − I3

I + I2 = I4

I2 = ½ I1

I4 = 2 I3

I1 − I3 + ½ I1 = 2I3 I1 = 2I3 I = +I3

II1

I2

I3I4

What is the same? Current flowing in and out of the baTery.

What is different? Current flowing from a to b.

2R3

2R3

Prelecture CheckPoint


2RI1/3R

2/3I

V

R 2R

a b

I2/3I

V/2

I

1/3

0

2/3I

2/3I

2/3I

1/3I1/3I

1/3I

2/3I1/3I


CheckPoint: Circuits w/ Resistors and a Battery 2


Consider the circuit shown below. In which case is the current flowing in the blue wire connec7ng points a and b bigger?

IA IB

Current will flow from leE to right in both cases.

Case A Case B They are the sameA B C

In both cases, Vac = V/2

c c

IA = IR − I2R

= IR − 2I4R IB = IR − I4R

I2R = 2I4R

V0

r

R VL

r

V0

+

VLR

Usually can’t supply too much current to the load without voltage “sagging”

Model for Real Battery: Internal Resistance


Using Breadboards (protoboards)

Original Breadboards

Circuit Technique

58 CHAPTER 6. INTRODUCTORY ELECTRONICS NOTES: PRACTICE

Figure 6.1: Bad and Good breadboarding technique.

• Try to build your circuit so that it looks like its circuit diagram:

– Let signal flow in from the left, exit on the right (in this case, the “signal” is justV ; the “output” is just I, read on the ammeter);

– Place ground on a horizontal breadboard bus strip below your circuit. When youreach circuits that include negative supply, place that on a bus strip below theground bus.

– Use colour coding to help you follow your own wiring: use black for ground, redfor the positive supply. Such colour coding helps a little now, a lot later, whenyou begin to lay out more complicated digital circuits.

Figure 6.2 shows bad and good examples of breadboard layouts. Figure 6.3 showsthe layout of a typical breadboard. Typically, one places components in the middlegroups with vertical interconnects and power lines and grounds in the horizontalinterconnects at top and bottom.

Figure 6.2: Bad and good breadboard layouts of a simple circuit

Good and Bad component layout








Figure 6.2: Bad and good breadboard layouts of a simple circuitConnections among pins in the breadboard.

Use horizontal rows for voltage busses: +5V, ±12V, gnd.

Use vertical rows for connecting components

together.








Figure 6.2: Bad and good breadboard layouts of a simple circuit

+5V bus

gnd bus

to +5V ofpower supply

to gnd ofpower supply

to scope

connection

Documents

Electricity Magnetism Lecture 10: Kirchhoff’s Rules Lecture 10 - Kirchhoff's Rule… · If the batteries are ideal andVA = 1.5 V A)VAB = 0.0 V B)VAB = 0.5 V C)VAB = 1.5V D)VAB =