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1 Text section 28.3 Kirchhoff’s circuit rules Practice: Chapter 28, problems 17, 19, 25, 26, 43 Junction Rule : total current in = total current out at each junction (from conservation of charge). Loop Rule : Sum of emfs and potential differences around any closed loop is zero (from conservation of energy).

Text section 28.3 Kirchhoff’s circuit rulesluke/1e03/lecture17.… ·  · 2010-02-081 Text section 28.3 Kirchhoff’s circuit rules Practice: Chapter 28, problems 17, 19, 25, 26,

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Text section 28.3

Kirchhoff’s circuit rules

Practice: Chapter 28, problems 17, 19, 25, 26, 43

Junction Rule: total current in = total current out at each junction (from conservation of charge).

Loop Rule: Sum of emfs and potential differences around any closed loop is zero (from conservation of energy).

2

Junction Rule: conservation of charge.

I1 I2

I3 I1 = I2 + I3

or

I1’ I2’ (= -I2)

I3’ (= -I3)

I1’ + I2’ + I3’ = 0

-Q +Q

C

- +

R

R I

I

changes going from left to right ΔV = -IR

ΔV = +IR

ΔV =

ΔV = Q/C

Loop Rule: conservation of energy. Follow a test charge q around a loop:

around any loop in circuit.

3

18 Ω

6 Ω 3 Ω 9 V

3 V I1

I2

I3 a

b c

d

Find the current through each battery.

18 Ω

6 Ω 3 Ω 9 V

3 V I1

I2

I3 a

b c

d

The junction rule will give:

A)  I1 + I2 + I3 = 0 B)  -I1 + I2 + I3 = 0 C)  I1 - I2 + I3 = 0 D)  I1 + I2 - I3 = 0 E) none of these

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18 Ω

6 Ω 3 Ω 9 V

3 V I1

I2

I3 = - (I1 + I2) a

b c

d

Find the current through each battery.

18 Ω

6 Ω 3 Ω 9 V

3 V I1

I2

I3 a

b c

d

The junction rule will give:

A)  I1 + I2 + I3 = 0 B)  -I1 + I2 + I3 = 0 C)  I1 - I2 + I3 = 0 D)  I1 + I2 - I3 = 0 E) none of these

5

18 Ω

6 Ω 3 Ω 9 V

3 V I1

I2

I3 a

b c

d

The loop rule applied to loop abcda will give:

A)  9A - 18I1 -3I3 = 0 B)  9A + 18I1 -3I3 = 0 C)  9A + 18I1 +3I3 = 0 D)  9A - 18I1 +3I3 = 0 E) none of these

18 Ω

6 Ω 3 Ω 9 V

3 V I1

I2

I3 a

b c

d

The loop rule applied to loop abda will give:

A)  12A - 18I1 + 6I2 = 0 B)  12A - 18I1 - 6I2 = 0 C)  6A - 18I1 - 6I2 = 0 D)  6A + 18I1 + 6I2 = 0 E)  6A - 18I1 + 6I2 = 0

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18 Ω

6 Ω 3 Ω 9 V

3 V I1

I2

I3 = - (I1 + I2) a

b c

d

Find the current through each battery.

Answers: I1 = 400 mA, I2 = 200 mA

18 Ω

6 Ω 3 Ω 9 V

3 V I1

I2

I3 = - (I1 + I2)

a

b c

d

Solution: Loop abcda:

1

2 Loop dbcd:

7× - : 60 I2 = 12A I2 = 200 mA 1 2

3 × - : 60 I1 = 24A I1 = 400 mA 1 2

Junction b: I1 + I2 + I3 = 0 so I3 = -(I1 + I2)

.....

.....

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10 Ω

20 Ω 40 Ω 6 V

3 V I1

I2

I3 = - (I1 + I2) a

b c

d

Find the current through each battery.

10 Ω

20 Ω 40 Ω 6 V

3 V I1

I2

I3 = - (I1 + I2) a

b c

d

Find the current through each battery.

Answers: I1 = 171 mA, I2 = -64.3 mA

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10 Ω

20 Ω 40 Ω 6 V

3 V I1

I2

I3 = - (I1 + I2)

a

b c

d Solution:

Loop abcda: 1

2 Loop dbcd:

4x - 5x : -140 I2 = 9 I2 = -64 mA 1 2

3x - 2x : 70 I1 = 12 I1 = 171 mA 1 2

Junction b: I1 + I2 + I3 = 0 so I3 = -(I1 + I2)

.....

.....

200 Ω

600 Ω

200Ω

300Ω

12 V a b

c

d

What is Vab (i.e., Va - Vb ) when the switch is open?

Exercise for fun: Find the current through the switch when it is closed.

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200 Ω

600 Ω

200Ω

300Ω

a b

c

d

The loop rule requires that Vab (i.e., Va - Vb )

obeys:

I1 I2

A)  Vab = (200Ω)I1 + (200Ω)I2 B)  Vab = (200Ω)I1 - (200Ω)I2 C)  Vab = -(200Ω)I1 + (200Ω)I2 D)  Vab = -(200Ω)I1 - (200Ω)I2

1 Ω

2 Ω

a b

c

d

1Ω Reff = ?

“Series and parallel“ rules don’t help in this case. You have to go back to the fundamentals—Kirchhoff’s Circuit Rules.

(Answer: Reff = 1.4Ω )

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1 Ω

2 Ω

a b

c

d

Reff = VTOTAL/ITOTAL

ITOTAL

VTOTAL I3

I2 I1

I5

I4

1)  Use Kirchhoff’s rules to write everything in terms of, one variable, e.g., I3.

2)  Divide VTOTAL/ITOTAL .

Show that:

I1 = I5 = 3I3

I2 = I4 = 2I3

ITOTAL= 5I3

VTOTAL= (7Ω) I3

1 Ω

2 Ω

a b

c

d

Reff = VTOTAL/ITOTAL

ITOTAL

VTOTAL I3

I2 I1

I5

I4

1)  Use Kirchhoff’s rules to write everything in terms of, e.g., I3. (Results: I1 = I5 = 3I3, I2 = I4 = 2I3; ITOTAL= 5I3; VTOTAL= (7Ω) I3.)

2)  Divide VTOTAL/ITOTAL .

Show that:

I1 = I5 = 3I3

I2 = I4 = 2I3

ITOTAL= 5I3

VTOTAL= (7Ω) I3

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Solution to part A:

250

250

100

400

12 V a b

c

d

I1

I1 + I3

I1 - I3

I2

I3

Solution to part b:

Loop adba:

1

Loop abca:

2

Then:

Solve…

12

R1

R2

V1

V2

V

and

(from ; and etc)

1 Ω

2 Ω

a b

c

d

5Ω Reff = ?

Notice that “series and parallel“ don’t help in this case. You have to go back to Kirchhoff’s Rules.

(Answer: Reff = ( 155/74 ) Ω )