Transcript Series and Parallel Circuits

Series and Parallel Circuits
Characteristics of Series Circuits
 Only one path (electron has no choices,
must go through all components)
 If one goes out, they all go out
 Voltage adds VT=V1+V2+V3+….
 Current is constant IT=I1=I2=I3=….
 Resistance adds RT=R1+R2+R3+….
 Power adds PT=P1+P2+P3+….
Characteristics of Parallel
Circuits
 More than one path
 Voltage is constant VT=V1=V2=V3=….
 Current adds IT=I1+I2+I3+….
 Resistance reciprocals add to give the reciprocal
1
1
1
1
 

 ....
RT R1 R2 R3
 Power adds PT = P1 + P2 + P3 + ……
Solving Problems
 Set up chart
 Fill in given information.
 Fill in what is constant in that type of circuit
 Look for any place where you either have at
least 2 in a column or all but one in a row.
 Use Ohm’s law to complete columns and
characteristics of that type of circuit to
complete row.
Sample Problem
 Four resistors, 3 , 5 , 7  and 9, are
arranged in series across a 48-V battery.
 (a) Draw the circuit diagram.
Cont’d
You
have(b)
at but

Find
the
equivalent
resistance and (c)
You
have
all
one
in thethe
Since
it
is
a
series
circuit,
Record
Given
Values
You
Calculate
now
have
all
the
at
individual
least
2
in
all
leastcheck
2 in row,
a withsince
Double
the
total
Power
resistance
itthe
is apower….
current
is
constant
the
current
in
circuit.
(d)
Calculate
the
remaining
powers
using
columns,
P=VI
so
use
column,
so
adds
to give
the you
totaladd
andtoP get
= VI
series
circuit,
V=IR
voltage
drop across each resistor. (e) What
V=IR
the
total resistance
power is dissipated in each resistor?
VT = 48 V1 = 6
V2 = 10 V3 = 14 V4 = 18
IT = 2
I1 = 2
I2 = 2
I3 = 2
I4 = 2
RT = 24
R1 = 3
R2 = 5
R3 = 7
R4 = 9
PT = 96
P1 = 12
P2 = 20
P3 = 28
P4 = 36
Sample Problem
 Connect the same circuit up with the
resistors in parallel. (a) Draw the circuit
diagram.
Cont’d

(b)have
isallthe
voltage
across
each
You
but
one
in
the voltage
resistance
row,
so the
You
have
atparallel
least
2 circuit,
in
each
column
Since
Now
calculate
itWhat
is aall
of
powers
usingdrop
Pis=constant
VI……
Don’t
reciprocals
add(c)
to
give
the is
reciprocal
of thepowers
total add to
resistor?
What
the
equivalent
so
V=
Show
all
given
information
forget
toIR
double
check
since
the
individual
give
the total
resistance?
(d) What current flows through
each resistor? (e) Calculate the power
dissipated by each resistor.
VT = 48 V1 = 48 V2 = 48 V3 = 48 V4 = 48
IT = 37.7 I1 = 16
I2 = 9.6
I3 = 6.9
I4 = 5.3
RT = 1.27 R1 = 3
R2 = 5
R3 = 7
R4 = 9
PT =1814 P1 = 768 P2 = 461 P3 = 329 P4 = 256
Internal Resistance of a Battery
 All batteries have internal resistance
because some of the energy must be used to
drive the current through the battery itself.
 EMF (electromotive force) also known as
the open circuit reading is the maximum
amount of energy a battery could produce
 The VT (terminal voltage) also known as the
closed circuit reading is always less than the
EMF due to the internal resistance. It is the
actual amount used up in the external
circuit.
VT = EMF - Iri
IRE = EMF - Iri
 VT = Terminal Voltage
in Volts
 EMF = Electromotive
force in Volts
 I = Current in Amps
 ri = Internal resistance
of the battery in Ohms
 RE = total resistance of
external circuit
Cells in Series
 Positive terminal connected to negative
terminal of the next battery
 EMF adds
 Current is constant
 Internal resistance adds
 Gives you high voltage for short periods of
time
Cells in Parallel
 Positive terminal connected to the positive
terminal of the next battery
 EMF is constant
 Current adds
 Reciprocals of internal resistances add to
give the reciprocal of the total
 Provides energy for a long time
Sample Problem
 Three dry cells each have an EMF of 1.5 V and an
internal resistance of 0.1 . What is the EMF if
these cells are connected in series?
4.5 V
 What is the internal resistance of the battery?
0.3 Ohm
 What is the line current if this battery is connected
to a 10 Ohm resistor?
0.437 Amps
 What is the terminal voltage of the battery?
4.37 V
Sample Problem
 A battery gives an open-circuit reading
(EMF) of 3.00 V. The voltmeter is
disconnected and the battery is then
connected in series with an ammeter and an
external load of 11.5 . The ammeter
reading is 0.250 A. Calculate the internal
resistance of the battery.
0.5 Ohms
Complex Circuits
 Also known as combination circuits or
networks because parts of the circuit are
connected in series and parts in parallel
 Create chart… keep simplifying until you
have a simple circuit.
 Solve using rules for series and parallel
Sample Problem
 For each circuit below, determine the readings
on each meter.
V1 =
12 V
R1 =
10 Ω
R2 =
12 Ω
R3=
6Ω
R4 =
4Ω
You can now play the voltage game. Pick a path that has only
You
have
at
2 inasa going
column
unknown
onleast
it such
R2 then back
Theone
current
leaving
the
battery
has nothrough
choiceRbut
to go
1 and
You
have
at
least
2 ininformation
column with
so useR2V=IR
Since
R3V=IR
and
R4 are
ina parallel
, the
Record
the
given
so
use
to theRbattery.
total through
voltage drop
on R and
R must be
through
theThe
current
that resistor
is the
1. So
voltage
drop across each one must be the1 same 2
equal
the terminal voltage of the battery. So the voltage drop
same
as Ito
T
across R2 is 2 V
VT=
V1=
12
IT=
V2=
10
I 1=
1
RT=
1
R1Rcombines
with the for
2, R3, R4 combine
equivalent
an equivalent
resistance of R2,
R3,resistance
R4 for a combined
of 2 Ohms
resistance
since they
of 12
areOhms
in
since
parallel
that is in series
I3=
0.17
R2=
10
2
2
I2=
R1=
12
V3=
12
2
I4=
0.33
R3=
0.5
R4=
6
2
12
V4=
4
Kirchhoff’s Laws
 1st Law - Total current into a junction is
equal to the total current leaving the
junction…. Also known as the law of
conservation of charge
Sample Problem
 Find I4
I1 = 2 A
1 k
I2 = 3 A
1 k
I3 = 4 A
1 k
I4 = ?1A
1 k
Kirchhoff’s Laws
 2nd Law - The algebraic sum of the changes
in potential energy occurring in any closed
loop is zero due to the law of conservation
of energy.
Sign Conventions for 2nd Law
 Crossing a resistor with current then -IR
 Crossing a resistor against current then +IR
 Crossing a battery with current +V
 Crossing a battery against current -V
Sample Problem
20 Ω
60 Ω
A
30 V
Left Clockwise
loop starting at A
50 V
I1
I3
I2
5Ω
30 Ω
20I1 +5I2 +30I1 -30 =0
You
Set
You
itnext
next
equal
cross
cross
to
a20
the
battery
5you
Ohm
against
areresistor
backcurrent
atagainst
your
You
next
cross
the
30
Ohm
resistor
against
You
cross
thezero,
Ohm
resistor
against
so
starting
current
-30 so
position.
soso+30I
+5I
21 The total change in
current
current,
+20I
1
potential around any closed loop is zero.
Solving for Currents using K.
Laws
 You must have one first law equation and 2
second law equations.
 Rearrange equation into proper format.
 D(I1) + E(I2) + F(I3) = G
 Set up matrix A (3 x 3) and B (3x1)
 A-1B
Sample Problem
20 Ω
30 V
60 Ω
Bottom
junction
equation
50 V
I1
I3
I2
5Ω
I1 + I 3 = I2
30 Ω
Left Clockwise
Loop Equation: Right Clockwise Loop Equation:
20I1 +5I2 + 30I1 – 30 =0
50 - 5I2 -60I3 = 0
50I1 + 5I2 + 0I3 = 30
Rearrange all equations to fit the
DI1 + EI2 + FI3 = G format
0I1 – 5I2 -60I3 = -50
1I1 – 1I2 + 1I3 = 0
Create a 3 x 3 matrix named A
using the left side of each
equation
Create a 3 x 1 matrix
named B using the right
side of each equation
Matrix A
50
5
0
-5 -60
1
-1
Matrix B
30
-50
0
0
1
Using the matrices A and B, perform the
function A-1B. The output on your calculator
screen should be
0.47887
1.21126
0.73239
The top number is the value of I1, the
second line is the value of I2, and the
last line is the value of I3