Download DC CIRCUITS UNIT 3 POWER POINT HONORS

SERIES AND PARALLEL
CIRCUITS
UNIT 3: DIRECT CURRENT CIRCUITS
SERIES CIRCUITS
• A series circuit is formed in a
circuit like this one, where
current flows through one path
that travels through one device
after another (the two light
bulbs).
• If you measure the current at
any point in this circuit, it will be
the same.
RESISTORS IN SERIES
• When two or more resistors are connected endto-end, they are said to be in series.
• The current is the same in resistors in series
because any charge that flows through one
resistor flows through the other.
• The sum of the potential differences (voltage
drops) across the resistors is equal to the total
potential difference across the combination (the
battery or power supply voltage).
RESISTORS IN SERIES
• Potentials add
• ΔV = IR1 + IR2
= I (R1+R2)
• Consequence of
Conservation of Energy
• The equivalent resistance
R1+R2 has the same effect on
the circuit as the original
combination of resistors
EQUIVALENT RESISTANCE – SERIES
• Req = R1 + R2 + R3 + …
• The equivalent resistance of a series combination of
resistors is the algebraic sum of the individual
resistances and is always greater than any of the
individual resistances.
• Note that the equivalent resistance is always
LARGER than any of the individual resistances when
in series.
EQUIVALENT RESISTANCE – SERIES
AN EXAMPLE
• Four resistors are replaced with their equivalent resistance.
• The current 𝐼 =
𝑉
𝑅𝑒𝑞
=
6.0 𝑉
18.0 Ω
= 0.33𝐴
QUICK QUIZ 1
WHEN A PIECE OF WIRE IS USED
TO CONNECT POINTS B AND C
IN THIS FIGURE, THE BRIGHTNESS
OF BULB R1
(A) INCREASES,
(B) DECREASES, OR
(C) STAYS THE SAME.
THE BRIGHTNESS OF BULB R2
(A) INCREASES,
(B) DECREASES, OR
(C) STAYS THE SAME.
QUICK QUIZ 1 ANSWER
R1 BECOMES BRIGHTER. CONNECTING A WIRE FROM B TO
C PROVIDES A NEARLY ZERO RESISTANCE PATH FROM B
TO C AND DECREASES THE TOTAL RESISTANCE OF THE
CIRCUIT FROM R1 + R2 TO JUST R1.
IGNORING INTERNAL RESISTANCE, THE POTENTIAL
DIFFERENCE MAINTAINED BY THE BATTERY IS
UNCHANGED WHILE THE RESISTANCE OF THE CIRCUIT
HAS DECREASED. THE CURRENT PASSING THROUGH BULB
R1 INCREASES, CAUSING THIS BULB TO GLOW BRIGHTER.
BULB R2 GOES OUT BECAUSE ESSENTIALLY ALL OF THE
CURRENT NOW PASSES THROUGH THE WIRE
CONNECTING B AND C AND BYPASSES THE FILAMENT OF
BULB R2.
QUICK QUIZ 2
 WITH THE SWITCH IN THIS CIRCUIT (FIGURE A) CLOSED, NO CURRENT
EXISTS IN R2 BECAUSE THE CURRENT HAS AN ALTERNATE ZERORESISTANCE PATH THROUGH THE SWITCH. CURRENT DOES EXIST IN
R1 AND THIS CURRENT IS MEASURED WITH THE AMMETER AT THE
RIGHT SIDE OF THE CIRCUIT. IF THE SWITCH IS OPENED (FIGURE B),
CURRENT EXISTS IN R2. AFTER THE SWITCH IS OPENED, THE READING
ON THE AMMETER (A) INCREASES, (B) DECREASES, (C) DOES NOT
CHANGE.
QUICK QUIZ 2 ANSWER
(B). WHEN THE SWITCH IS OPENED,
RESISTORS R1 AND R2 ARE IN SERIES, SO
THAT THE TOTAL CIRCUIT RESISTANCE IS
LARGER THAN WHEN THE SWITCH WAS
CLOSED. AS A RESULT, THE CURRENT
DECREASES.
VOLTAGE DIVIDER CIRCUIT
For this circuit, I = V/Req so
I = V / (RA + RB)
The voltage drop VB, across
resistor RB is given by
VB = IRB
𝑉𝐵 =
𝑉
𝑅𝐴 +𝑅𝐵
𝑅𝐵 or
𝑅𝐵
𝑉𝐵 =
𝑅𝐴 + 𝑅𝐵
𝑉
RESISTORS IN PARALLEL
• For resistors in parallel, the
potential
difference across each resistor
(from a to b) is the same because
each is connected directly across the battery terminals.
• The current, I, that enters a point must be equal to the
total current leaving that point.
• I = I1 + I2
• The currents I1 and I2 are generally not the same.
• This is a consequence of conservation of charge.
EQUIVALENT RESISTANCE –
PARALLEL, EXAMPLES
• Equivalent resistance replaces the two original
resistances
• Household circuits are wired so the electrical
devices are connected in parallel
• Circuit breakers may be used in series with other
circuit elements for safety purposes
EQUIVALENT RESISTANCE –
PARALLEL CIRCUITS
• Equivalent Resistance
1
1
1
1




R eq R1 R 2 R 3
• The inverse of the equivalent
resistance of two or more
resistors connected in parallel is
the algebraic sum of the
inverses of the individual
resistance.
For parallel R’s, the equivalent Req is always less
than the smallest resistor in the group.
CURRENT DIVIDER CIRCUIT
• How does the current split
itself between the R1 and
R2 branches?
• Since the voltage drop
across every branch is 10V,
• 𝑰𝟏 = 𝑹𝑽 =
𝟏
• 𝑰𝟐 = 𝑹𝑽
𝟐
=
𝟏𝟎
𝟏
𝟏𝟎
𝟐
=1Ω
=2Ω
= 𝟏𝟎 𝑨
= 10 V
=𝟓𝑨
• Add total 𝑰 = 𝑰𝟏 + 𝑰𝟐 =
𝟏𝟎 + 𝟓 = 𝟏𝟓𝑨
• So do you see how 2/3 of the total current goes
down the 1Ω resistor and 1/3 of the total current
goes down the 2 Ω resistor?
PROBLEM-SOLVING STRATEGY, 1
• When two or more unequal resistors are connected in
series, they carry the same current, but the potential
differences across them are not the same.
• The resistors add directly to give the equivalent resistance
of the series combination Req = R1 + R2
• The equivalent resistance is always larger than any of the
individual resistances.
PROBLEM-SOLVING STRATEGY, 2
• When two or more unequal resistors are connected in
parallel, the potential differences (voltage drop) across
them are the same. The currents through them are not
the same.
• The equivalent resistance of a parallel combination is found
through reciprocal addition
1
1
1
1




R eq R1 R 2 R 3
• The equivalent resistance is always less than the smallest
individual resistor in the combination
PROBLEM-SOLVING STRATEGY, 3
• A complicated circuit consisting of several resistors and
batteries can often be reduced to a simple equivalent
circuit with only one resistor
• Replace any groups of resistors in series or in parallel
using steps 1 or 2.
• Sketch the new circuit after each change has been made.
• Continue to replace any series or parallel combinations .
• Continue until one equivalent resistance is found.
PROBLEM-SOLVING STRATEGY, 4
• If the current or the potential difference across a
resistor in the complicated circuit is to be
identified, start with the final circuit found in step
3 and gradually work back through the circuits
• Use ΔV = I R and the procedures in steps 1 and 2
QUICK QUIZ 3
WITH THE SWITCH IN THIS CIRCUIT (FIGURE A) OPEN, THERE IS NO
CURRENT IN R2. THERE IS CURRENT IN R1 AND THIS CURRENT IS
MEASURED WITH THE AMMETER AT THE RIGHT SIDE OF THE CIRCUIT.
IF THE SWITCH IS CLOSED (FIGURE B), THERE IS CURRENT IN R2.
WHEN THE SWITCH IS CLOSED, THE READING ON THE AMMETER (A)
INCREASES, (B) DECREASES, OR (C) REMAINS THE SAME.
QUICK QUIZ 3 ANSWER
(A). WHEN THE SWITCH IS CLOSED, RESISTORS R1
AND R2 ARE IN PARALLEL, SO THAT THE TOTAL
CIRCUIT RESISTANCE IS SMALLER THAN WHEN THE
SWITCH WAS OPEN. AS A RESULT, THE TOTAL
CURRENT INCREASES.
EQUIVALENT
RESISTANCE –
COMPLEX
CIRCUIT
Parallel part:
1
1 1
3
= + =
𝑅𝑒𝑞
6 3
6
so Req = 6/3 = 2 Ω
KIRCHHOFF’S LAWS
• There are ways in which resistors can be
connected so that the circuits formed cannot be
reduced to a single equivalent resistor
• Two rules, called Kirchhoff’s Rules can be used
instead
STATEMENT OF KIRCHHOFF’S LAWS
• Junction Rule - KCL
• The sum of the currents entering any junction must equal the sum
of the currents leaving that junction
• A statement of Conservation of Charge
• Loop Rule - KVL
• The sum of the potential differences across all the elements
around any closed circuit loop must be zero
• A statement of Conservation of Energy
MORE ABOUT THE JUNCTION
RULE
• I1 = I2 + I3
• From Conservation of
Charge
• Diagram b shows a
mechanical analog
SETTING UP KIRCHOFF’S RULES
• Assign symbols and directions to the currents in all
branches of the circuit
• If a direction is chosen incorrectly, the resulting
answer will be negative, but the magnitude will
be correct
• When applying the loop rule, choose a direction for
traversing the loop
• Record voltage drops and rises as they occur
MORE ABOUT THE LOOP RULE
• Traveling around the loop
from a to b
• In a, the resistor is traversed
in the direction of the current,
the potential across the
resistor is –IR
• In b, the resistor is traversed in
the direction opposite of the
current, the potential across
the resistor is +IR
LOOP RULE, FINAL
• In c, the source of emf is
traversed in the direction of
the emf (from – to +), the
change in the electric
potential is +ε
• In d, the source of emf is
traversed in the direction
opposite of the emf (from +
to -), the change in the
electric potential is -ε
JUNCTION EQUATIONS FROM
KIRCHHOFF’S RULES
• Use the junction rule as often as needed so long as,
each time you write an equation, you include in it a
current that has not been used in a previous junction
rule equation
• In general, the number of times the junction rule
can be used is one fewer than the number of
junction points in the circuit
LOOP EQUATIONS FROM
KIRCHHOFF’S RULES
• The loop rule can be used as often as needed so
long as a new circuit element (resistor or battery)
or a new current appears in each new equation
• You need as many independent equations as you
have unknowns
PROBLEM-SOLVING STRATEGY –
KIRCHHOFF’S RULES
• Draw the circuit diagram and assign labels and symbols
•
•
•
to all known and unknown quantities. Assign directions to
the currents.
Apply the junction rule to any junction in the circuit
Apply the loop rule to as many loops as are needed to
solve for the unknowns
Solve the equations simultaneously for the unknown
quantities.
• For example 2 equations and 2 unknowns
or 3 equations and 3 unknowns
EXAMPLE 1 BY
STANDARD
METHOD
I1
I3
I2
I1
I3
Left loop: 18 - 5I1 - 5I2 - 1.5 I1 = 0
Whole loop: 18 - 5I1 - 5 I3 - 5I3 -1.5I1= 0
Junction: I1 = I2 + I3
Three equations, three unknowns, solve.
You should get I1 = 1.83A, I2 = 1.22A, I3 = 0.61A
EXAMPLE 1MESH METHOD
I1
I2
Loop 1: 18 - 5I1 - 5(I1-I2) - 1.5 I1 = 0
Loop 2: 0 - 5I2 - 5 I2 - 5(I2-I1) = 0
By using the mesh method, you have one
less equation to solve.
Two equations, two unknowns, solve.
You should get I1 = 1.83A, I2 = 0.61
EXAMPLE 2 –
STANDARD
METHOD
Try this one
I1
I2
I1
I3
Full loop: 12 – 1(I1) – 3(I1) – 8(I3) = 0
Left loop: 4 – 1(I2) – 5(I2) – 8(I3) = 0
Junction: I1 + I2 = I3
3 equations, 3 unknowns, solve for I1, I2, I3
I1 = 1.31A, I2 = -.46A, I3 = 0.85A
EXAMPLE 2
MESH METHOD
Try this one
I1
I2
Loop 1: 4 – 1(I1-I2) – 5(I1-I2) – 8(I1) = 0
Loop 2: 12 – 1(I2) – 3(I2) – 5(I2-I1) –
1(I2-I1) – 4 = 0
2 equations, 2 unknowns, solve for I1, I2
I1 = 0.85, I2 = 1.31, I2 – I1 = 0.46A
EXAMPLE 3 –
MESH METHOD
I1
Try this one
I2
Loop 1: 20 – 30I1 – 5 (I1-I2) – 10 = 0
Loop 2: 10 - 5(I2-I1) – 20(I2) = 0
2 equations, 2 unknowns, solve for I1, I2
I1 = 0.353A , I2 = 0.47A
CAPACITORS
A parallel plate capacitor
consists of two metal plates, one
carrying charge +q and
the other carrying charge –q.
It is the capacity to store charge
that resulted in the name
capacitor.
It is common to fill the region
between the plates with an
electrically insulating
substance called a dielectric.
THE RELATION BETWEEN CHARGE AND
VOLTAGE FOR A CAPACITOR
• The magnitude of the charge on each
plate of the capacitor is directly
proportional to the magnitude of the
potential difference between the plates.
q = CV
• The capacitance C is the proportionality
constant.
• SI Unit of Capacitance:
• coulomb/volt = farad (F)
• Typical measurements are in microfarads
(μF) or picofarads (pF).
• The base units for Farad are A2s4kg-1m-2
• The Farad is named after Michael
Faraday
• The electrons are moving
from the left plate around the
loop to the right plate.
• The electrons are not moving
across the dielectric, that’s just
the E field lines.
CAPACITORS
•
•
•
•
•
THE DIELECTRIC CONSTANT, κ
The insulating material between the plates
is called a dielectric.
If a dielectric is inserted between the
plates of a capacitor, the capacitance can
increase markedly.
κ is a number greater than 1.0 & unitless
Eo and E are , respectively, the magnitudes
of the E field between the plates without
and with a dielectric.
The dielectric has a lower electric field as
shown in (c), allowing for more charge on
the plates.
𝑬
Dielectric constant
𝜿=
𝒐
𝑬
CAPACITORS
Silicon dioxide
3.8
CAPACITANCE FOR PARALLEL PLATE
CAPACITORS
𝑞
𝑉 𝐸𝑜
𝐸𝑜 =
𝐸= =
𝜖𝑜 𝐴
𝑑
𝜅
ϵo = permittivity of free space
= 8.85E-12 C2/N m2
Parallel plate capacitor
filled with a dielectric
𝜅𝜖𝑜 𝐴
𝑞=
𝑉
𝑑
and since q = CV
𝜅𝜖𝑜 𝐴
𝐶=
𝑑
SAMPLE PROBLEM
• Two parallel plates both have an area of 0.015
m2 and are placed 2.0 x 10-3 m apart. Calculate
the capacitance of this arrangement.
•𝑪 =
𝜺𝒐 𝑨
𝒅
= 𝟖. 𝟗𝑬 − 𝟏𝟐 𝒙
𝟎.𝟎𝟏𝟓
𝟐𝑬−𝟑
= 𝟔. 𝟕𝑬 − 𝟏𝟏 𝑭
ENERGY STORAGE IN A CAPACITOR
1
1
1 2
𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑞𝑉 = 𝐶𝑉 𝑉 = 𝐶𝑉
2
2
2
Volume  Ad
V =Ed
1 𝜅𝜖0 𝐴
𝐸𝑛𝑒𝑟𝑔𝑦 = (
)(𝐸𝑑)2
2 𝑑
𝐸𝑛𝑒𝑟𝑔𝑦 1
𝐸𝑛𝑒𝑟𝑔𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
= 𝜅𝜖𝑜 𝐸 2
𝑣𝑜𝑙𝑢𝑚𝑒 2
ENERGY SAMPLE PROBLEM
• A capacitor of value 100 mF stores an energy of
250 J. Calculate the potential difference across
the capacitor.
• 𝐸𝑛𝑒𝑟𝑔𝑦 =
•𝑉 =
1
𝐶𝑉 2
2
2 𝑥 𝐸𝑛𝑒𝑟𝑔𝑦
𝐶
=
500
0.1
= 71 𝑉
CAPACITORS IN PARALLEL
Adding up
capacitors in
parallel is opposite
of what happens
with resistors:
Parallel : just add
them up!
q  q1  q2  C1V  C2V  C1  C2 V
Parallel capacitors
CP  C1  C2  C3  
CAPACITORS IN SERIES
1
q
q
1 

V  V1  V2  
 q  
C1 C2
 C1 C2 
Series capacitors
1
1
1
1




CS C1 C2 C3
Adding up
capacitors in
series is
opposite of
what
happens
with
resistors:
Series :
inverse like
resistors in
parallel
CAPACITOR
CHARGING
Capacitor charging
𝑞 = 𝑞𝑜 [1 − 𝑒 −𝑡/𝑅𝐶 ]
time constant
  RC
When t = RC, then cap is
63.2% charged up.
When t= 3RC cap 95%
charged up.
.632qo
t=RC
CAPACITORS
• The RC time constant, also called tau,
• τ= RC
• τ is the time constant (in seconds) of an RC
circuit. It is equal to the product of the circuit
resistance (in ohms) and the circuit capacitance
(in farads). Why?
• R=V/I= Volt sec/coul
• C = Q/V = Coul/volt
• So RC = seconds
CAPACITOR CHARGING
𝒒 = 𝒒𝒐 𝟏 − 𝒆−𝒕/𝑹𝑪
𝒒
= 𝟏 − 𝒆−𝒕/𝑹𝑪
𝒒𝒐
𝒒
𝟏−
= 𝒆−𝒕/𝑹𝑪
𝒒𝒐
𝒒
𝒍𝒏 𝟏 −
= 𝒍𝒏 𝒆−𝒕/𝑹𝑪
𝒒𝒐
𝒕
𝒒
−
= 𝒍𝒏(𝟏 − )
𝑹𝑪
𝒒𝒐
𝒒
𝒕 = −𝑹𝑪[𝒍𝒏 𝟏 −
]
𝒒𝒐
CAPACITOR CHARGING EXAMPLE
• Suppose that RC = 5.00 s and that the capacitor has
acquired a charge q to one-fourth of its final
equilibrium value qo so that q = 0.250 q0. The time
required for the capacitor to acquire the charge is
• 𝒕 = −𝑹𝑪[𝒍𝒏
𝟏−
𝒒
𝒒𝒐
• 𝒕 = −𝟓. 𝟎𝟎 𝒍𝒏 𝟏 −
𝟏. 𝟒𝟒 𝒔𝒆𝒄
]
𝟎.𝟐𝟓𝟎𝒒𝒐
𝒒𝒐
= −𝟓. 𝟎𝟎 𝒍𝒏 . 𝟕𝟓𝟎 =
CAPACITOR
DISCHARGING
Capacitor discharging
q  qo e
t RC
time constant
  RC
Io = Vo /R
0.368Io
t = RC
KIRCHOFF’S LAWS WITH CAPACITORS
If we were to apply Kirchoff’s Voltage Law
(KVL) to this circuit with the switch closed, we
would get
Vo – q/C – IR = 0
KIRCHOFF’S LAWS WITH CAPACITORS
If the capacitor was
originally uncharged and we
closed the switch, then
because qo is zero,
Vo – IR = 0
so Io = Vo/R but only at t=0!
Once the capacitor has been charged to its
maximum value Q then current >> zero & Vcap=Vo
Q = CVo because I=0 in equation on last slide
(Vo – q/C – IR = 0)
ELECTRICAL SAFETY
• Electric shock can result in fatal burns
• Electric shock can cause the muscles of vital
organs (such as the heart) to malfunction
• The degree of damage depends on
• the magnitude of the current
• the length of time it acts
• the part of the body through which it passes
EFFECTS OF VARIOUS CURRENTS
• 5 mA or less
• can cause a sensation of shock
• generally little or no damage
• 10 mA
• hand muscles contract
• may be unable to let go a of live wire
• 100 mA
• if passes through the body for 1 second or less,
can be fatal
GROUND WIRE
• Electrical equipment
manufacturers use
electrical cords that
have a third wire,
called a ground
• Prevents shocks
GROUND FAULT INTERRUPTS (GFI)
• Special power outlets
• Used in hazardous areas
• Designed to protect people from electrical shock
• Senses currents (of about 5 mA or greater) leaking to
ground
• Shuts off the current when above this level
THREE TYPES OF SAFETY DEVICES
• There are three types of safety devices that are
used in household electric circuits to prevent
electric overcurrent conditions:
• Fuses
• Circuit Breakers
• Ground Fault Interrupters