## Physics_documet_with_circurit_diagram_tables_and_equations.pdf-Chapter 28 Direct Current cshowing page 1-5 out of 7

##### Page 1

Chapter 28 Direct Current circuits

28.1 Electromotive Force (emf)

Sources of emf: A battery or any other device that provides

electrical energy.

“charge pump”

r

:

internal resistance

Ir

: potential decrease by r

IR = V

: potential on R, or

terminal voltage of the battery

V = - Ir, or

= IR +Ir

•

is the open-circuit voltage. (if I = 0, V =

= V

max

)

•

If r is small (a good battery), V

.

•

If r is large (a bad battery), V <<

.

•

Power dissipation

I

=

I

2

R

+

I

2

r

+

_

Resistor

Example

: A battery has an emf of 12 V, and internal

resistance of 0.05

.

Its terminals are connected to a load

resistance of 3

.

(a)Find the current.

I

=

R

+

r

=

12

V

3.05

=

3.93

A

(b) Find the terminal voltage:

V

=

Ir

=

12

V

(3.93

A

)(0.05

)

=

11.8

V

Find the equivalent resistance

6

8

4

3

12

2

14

##### Page 2

28.2 Resistors in Series and in Parallel

In Series

:

The current is the same through each resistor in series

connection:

I = I

1

= I

2

The total voltage = sum of voltages of each resistor:

V

=

V

1

+

V

2

=

IR

1

+

IR

2

=

I

(

R

1

+

R

2

)

=

IR

eq

Generally

,

R

eq

=

R

1

+

R

2

+

R

3

+

...

R1

I

In Parallel:

The voltage (potential difference) across each resistor in a

parallel circuit is the same:

V = V

1

= V

2

The total current = sum of currents passing through each

resistor in a parallel circuit

I

=

I

1

+

I

2

=

V

R

1

+

V

R

2

=

V

1

R

1

+

1

R

2

=

v

R

eq

Generally,

1

R

eq

=

1

R

1

+

1

R

2

+

1

R

3

+

...

##### Page 4

28.3 Kirchhoff’s Rules

(1)

Junction

Rule

: "The sum of the current entering any

junction must equal the sum the currents leaving that

junction."

I

in

= I

out

I

1

= I

2

+ I

3

“

Conservation of Charge

”

(2)

Loop Rule

: "The sum of potential differences across all

the elements around any closed circuit loop must be

zero."

--> “

Conservation of Energy

”

Strategy to solve complex circuits:

•

Label all quantities and assign directions of the currents.

•

Apply the junction rule (Kirchhoff’s

1

st

rule).

•

Apply Kirchhoff’s loop rule to loops. Pay attention to signs.

•

Solve the equations.

I

V

=

V

b

V

a

=

IR

b

a

I

V

=

V

b

V

a

=+

IR

b

a

V

=

V

b

V

a

=+

b

a

+

_

+

_

V

=

V

b

V

a

=

b

a

I

2

I

1

I

3

Example

: Find the current

I

1

, I

2

, and

I

3

(1)

Assign directions of currents.

Just make a reasonable guess.

(2)

Junction rule:

I

1

+ I

2

= I

3

(3) Identify loops (3):

Apply Kirchhoff’s 2

nd

rule 3-1= 2 times.

Lower loop:

10 V – (6

)

I

1

- (2

)

I

3

= 0

Upper loop: -14 V – 10 V + (6

)

I

1

- (4

)

I

2

= 0

(4) 3 unknowns, and 3 independent equations.

I

1

= 2 A,

I

2

= -3 A, and

I

3

= -1 A

Negative value indicates that the actual direction of current is

opposite to the labeled direction.

4

6

2

14V

10V

I

3

I

2

I

1

+

+

_

_

##### Page 5

Water in the bucket

Charge in the capacitor

Water flow in the pipe

Electric current

q

t

C

0.632

I

t

I

0

0.368I

0

I

0

=

R

R

S

I

+q

-q

t

flow

rate

0.368F

0

Water in bucket

t

W

max

0.632

Bucket

pipe

Reservoir