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7 Οκτ 2013 (πριν από 4 χρόνια και 9 μήνες)

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PHYS132 Week 4 (DC circuits II)

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Review

Current always flows “downhill” from high electric potential to low electric potential. As
charges flow “downhill”, they lose electric potential.

According to the definition of resistance,
the resistance of an element equals the potential
differenc
e across it divided by the current through it (
R

=

V/I
)
.

Written another way, the potential on the “uphill” side of a resistor is greater than the
potential on the “downhill” side of a resistor by an amount equal to
IR
.

18.5 Kirchhoff’s Rules

Energy is conserved in a circuit. Since the charges do not ga
in kinetic energy, the energy per
charge provided by the battery must be dissipated by each charge on its travels through the
circuit. Mathematically, this is written as Σ∆
V
=0 [this is known as
Kirchhoff’s loop rule
].

Charge is conserved in a circuit. F
or every charge that enters a region, an equal charge must
exit it, so that the total charge of the region remains neutral. Mathematically, this is written
as Σ
I
in

Σ
I
out

= 0 [this is known as
Kirchhoff’s junction rule
].

Definition of Series and Parallel

The current is the same through all of elements placed in
series
. This means that there can
be no other path for the current

all the current that goes through one element must likewise
go through the next element in series.

Elements placed in
parallel

all share the same starting and ending points. This means that
they must have the same potential difference across them.

18.6 Series and Parallel Circuits (skip “Capacitors in Series” and “Capacitors in Parallel”)

When placed in series, the total resista
nce is equal to the sum of the individual resistances.
This means that the total resistance is
larger

than any of the individual resistances that make
up the series. To show this:

o

From

V=IR
, the potential difference across each resistor is
IR
i

so the to
tal potential
difference across all of them is ∆
V
total

=
IR
1

+
IR
2

+ … +
IR
n

o

From

V=IR
, the total potential difference across the entire series must equal the
current through them times the total “equivalent” resistance: ∆
V
total

=
IR
total

o

Combining the la
st two relationships and dividing through by
I
, we have

R
total

=
R
1

+
R
2

+ … +
R
n

When placed in parallel, the total resistance is equal to the
inverse

of the
sum of the
individual
inverted

resistances.

This means that t
he total resistance is
smaller

th
an any of the
individual resistances that make up the series
. To show this:

o

The
potential difference
is the same
across
all of the resistors in
parallel
.

o

From

V=IR
, the
current through
each resistor is

V/
R
i

so the total
current through
all
of them is
I
total

=

V/
R
1

+

V/
R
2

+ … +

V/
R
n

o

From

V=IR
, the total
current through
the entire series must equal the
potential
difference across them divided by the
total “e
quivalent” resistance:
I
total

=

V/
R
total

o

Combining the last two relationships and dividing through by

V
, we have

1/
R
total

=
1/
R
1

+
1/
R
2

+ … +
1/
R
n

Depending on which way one progresses, each individual emf can be positive or negative

o

When placed in ser
ies, the total emf is equal to the sum of the individual emfs.

o

When two sources of the same emf are placed in parallel, the total emf is the same as
the individual emfs

PHYS132 Week 4 (DC circuits II)

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18.7 Circuit Analysis using Ki
r
chhoff’s Rules

A common problem to test your understand
ing of circuit concepts is to provide a circuit with
multiple batteries and resistors in various configurations, and ask for several “unknowns”
(i.e., resistances, emfs and/or currents).

If all resistances are provided, first identify any resistors that sh
are the same current (in
series) or share the same end points (in parallel). Then replace those resistors with a single
resistor of equivalent resistance. Continue until all resistors have been replaced with a single
resistor. The resistance of that res
istor represents the total resistance of the circuit.

For circuits with multiple loops, Kirchhoff’s Loop Rule can be applied to each loop and
Kirchhoff’s Current Rule can be applied to each “junction”. This will give several
relationships, which can be co
mbined to solve for the unknowns.

The basic three steps are as
follows:

o

First, identify what your known values and unknown values are.

Use letters to represent the unknown values for the time being.

Draw arrows for all current directions that are given.
If any current directions
are not given, draw an arrow for that current anyway. If you happen to be
wrong, you’ll end up with a negative current (the current value will be correct;
the negative just indicates that you chose the wrong direction).

Identify
the high and low potential sides of each element (resistors and
batteries).

o

Second, apply the junction rule (conservation
of current) to

find the relationship
between the currents in various parts of the

circuit.

If there are no junctions (single
loop cir
cuit), you can skip this step. If there are two junctions (double loop circuit),
apply the junction rule to only one of the two junctions (applying it to the other
junction will give the same relationship).

o

Third
, apply

V
=
IR

(
from
definition of resistance) to

every element in the circuit. If
you don't know the value of

V
,
I

or
R
, you simply use the variable abbreviation

o

Fourth
, you apply the loop rule (conservation of energy) around

eac
h loop that is
present using the

V

s you found in step 1.

Again, if you don't know the value of any
of them, you simply

use the variable abbreviation instead of the variable value.

If
there is only one junction, there is only one loop and you’ll end up
with just one
equation. If there are two junctions, there are three loops. Apply it to only two of
those (the third is a combination of those two).

o

Last,
solve the equations for the unknown values. If you end up with three equations
(two
-
junction circui
t), you can have up to three unknown values (but you may need
combine the equations algebraically to find the unknown values).

Homework

(remember to provide rationale for conceptual questions)
:

Conceptual Questions

18.21, 18.23 (assume ideal battery)

Multiple Choice 18.3, 18.4

Problems
18.40, 18.41, 18.52, 18.54, 18.116