Phys 122 Lecture 14G. Rybka

Kirchoff Loop Laws …Rule 1: “What goes in, must go out “

Rule 2: “What goes up, must come down”

I1 I3

I2

Business…• Exam next Thursday !• Material is everything through TODAY.

– No RC circuits that we will cover Monday• Practice exam and Eqn. posted on home page after lecture• Try the SmartPhysics HOMEWORK this week on today’s lessons. Very good practice for multi-­loop problems and a very pedagogical example

Can we solve this from simple series and parallel combinations?

Answer: no

Kirchoff Voltage Law: KVL"When any closed circuit loop is traversed, the algebraic sum of the changes in potential must equal zero."

• This is just a restatement of what you already know: that thepotential difference is independent of path!

∑ =loop

nV 0

We will follow the ECE conventions: (engineering)voltage drops enter with a + sign andvoltage gains enter with a -­ sign in this equation.

ε1R1 ε2

R2ILet’s go clockwise around circuit:

ε2R1 R2ε1 I

− ε1 + IR1 + IR2 + ε2 = 0

KVL:

ABCDE

DROP

GAIN

With the current VOLTAGE DROPAgainst the current VOLTAGE GAIN

CheckPointThis resistor labyrinth stuff is crazy!

Examine this loop

a

d

b

ec

fε1

R1

I

R2 R3

R4

I

ε2

Þ IR IR IR IR1 2 2 3 4 1 0+ + + + − =ε ε

Þ IR R R R

=−

+ + +ε ε1 2

1 2 3 4

Vnloop

=∑ 0KVL:

Let’s go around the loop counterclockwise from a to f

Clicker• Consider the circuit shown.

– The switch is initially open and the current flowing through the bottom resistor is I0.

– After the switch is closed, the current flowing through the bottom resistor is I1.

– What is the relation between I0 and I1?

(a) I1 < I0 (b) I1 = I0 (c) I1 > I0

• Write a loop law for original loop:

-12V + I1R = 0

I1 = 12V/R

• Write a loop law for the new loop:

-12V -12V + I0R + I0R = 0

I0 = 12V/R

R

12 V

12 V

R

12 VIa

b

…or, alternative reasoning• Consider the circuit shown.

– The switch is initially open and the current flowing through the bottom resistor is I0.

– After the switch is closed, the current flowing through the bottom resistor is I1.

– What is the relation between I0 and I1?

(a) I1 < I0 (b) I1 = I0 (c) I1 > I0

• The key here is to determine the potential (Va-­Vb) before the switch is closed.

• From symmetry, (Va-­Vb) = +12V.

• Therefore, when the switch is closed, NO additional current will flow!

• Therefore, the current before the switch is closed is equal to the current after the switch is closed.

R

12 V

12 V

R

12 VIa

b

Kirchoff's Current Law: KCL

Conservation of Charge (at a junction)

"At any junction point in a circuit where the current can divide (also called a node), the sum of the currents into the node must equal the sum of the currents out of the node."

∑= outin II

I1 I3

I2

Wire connects a to b;; DV through R and 2R to a/b is V/2. (symmetry from below);; Therefore: I1R = I2 2R = V/2;; I1 = V/2R

(twice as much current goes down the left)

From a/b to bottom of circuit, ΔV/2 as well;; I3 2R = I4R = V/2;; I3 = V/4R(twice as much current down right side)

Junction Rule at a: I1 = Iab + I3 à V/(2R) = Iab + V/(4R) à Iab is positive

II1 I2

I3 I4

CheckPoint

Which of the following statements best describes the current flowing in the blue wire connecting points a and b?

What is the same? Current flowing in and out of the battery. Equivalent resistance is the same.

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

No current would flow

RI2R

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

Current 1/3 I will flow

IA IB

Current will flow from left to right in both cases.

In both cases, Vac = V/2

c c

IA = IR − I2R= IR − 2I4R

IB = IR − I4R

CheckPoint 7

I2R = 2I4R

Have a look at this if you missed it. Some reasoning below …

Analysis of a circuit

Outside loop:

Top loop:

Junction: ε1

I1ε2

ε3

R

R

RI2

I3

I R I R1 3 3 1 0+ + − =ε ε

I R I R1 2 2 1 0+ + − =ε ε

I I I1 2 3= +

IR1

1 2 323

=− −ε ε ε

IR2

1 3 223

=+ −ε ε ε

IR3

1 2 323

=+ −ε ε ε

No getting around it3 equations/3 unknowns typically

Clicker

Consider the circuit shown:

– What is the relation between Va -­Vdand Va -­Vc ?

(a) (Va -Vd) < (Va -Vc)(b) (Va -Vd) = (Va -Vc)(c) (Va -Vd) > (Va -Vc)

12 VI1 I2

ab

d c

50Ω

20Ω 80Ω

• Remember: potential is independent of path

Going from a to d or c is like going to the same place, electrically

à (Va -Vd) = (Va -Vc)

Clicker

• Consider the circuit shown:

–What is the relation between I1 and I2? 12 V

I1 I2

ab

d c

50Ω

20Ω 80Ω

(a) I1 < I2 (b) I1 = I2 (c) I1 > I2

• Note that: Vb -Vd = Vb -Vc• Therefore, I I1 220 80( ) ( )Ω Ω= I I1 24=