Series Circuits Have only ONE LOOP or circuit for the current
to travel through.
Slide 4
Resistors in Series When two or more resistors are connected
end-to-end, they are said to be in series The current is the same
in all resistors because any charge that flows through one resistor
flows through the other The sum of the voltages across the
resistors is equal to the total voltages across the combination
Kirchoffs Voltage Law, the Conservation of Voltage
Slide 5
Resistors in Series Potentials add V = IR 1 + IR 2 = I (R 1 +R
2 ) Consequence of Conservation of Energy The equivalent resistance
has the effect on the circuit as the original combination of
resistors
Slide 6
Equivalent Resistance Series R e = R 1 + R 2 + R 3 + 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 resistors Batteries and even wires
contribute small amounts of resistance but we can ignore this for
now
Slide 7
Equivalent Resistance Series Four resistors are replaced with
their equivalent resistance An Example
Slide 8
Resistors in Parallel The voltage across each resistor 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 = I 1 + I 2 The currents are
generally not the same Consequence of Kirchoffs Second Law, the
Conservation of Charge
Slide 9
Equivalent Resistance Parallel 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
An Example
Slide 10
Equivalent Resistance Parallel Equivalent Resistance 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 The equivalent is always less than the
smallest resistor in the group
Slide 11
Example 1 In the depicted circuit, the voltage supplied by the
battery is 12V, and the resistors have values of R 1 = 10, R 2 = 5
and R 3 = 15. What is the current flowing through each branch?
Slide 12
Example 1 The current through each resistor can be found with
Ohms Law I 1 = 1.2A;I 2 = 2.4A;I 3 = 0.8A To check this, find the
current through the whole circuit by finding the total resistance,
then using Ohms Law again. R tot = 2.7I tot = 4.4A This matches the
sum of the individual currents
Slide 13
Example 2 If, V = 24V; R 1 = 2; R 2 = 3.3; R 3 = 7 and R 4 =
12.2 Find the current through the circuit and the current through
each resistor.
Slide 14
Example 2 Find the total resistance: R tot = o.973 Use this to
find the current through the circuit I = 24.7A The current through
each resistor is just the voltage divided by each individual
resistance: I 1 = 12A;I 2 = 7.3A; I 3 = 3.4AI 4 = 2A
Slide 15
Series Circuit A series of sources separated by resistors is
equivalent to a single source having the net voltage and a single
resistor having the combined resistance. R1R1 A B R2R2 R3R3 R4R4
V1V1 V2V2 V3V3 R = R 1 + R 2 + R 3 + R 4 V = V 1 - V 2 + V 3 The
same current passes through every resistor in a given branch,
regardless of the presence of sources in that branch, and the
resistors are in series even though they are not directly connected
to one another..
Slide 16
Parallel Circuit The current in each branch of a parallel
circuit depends inversely on the total resistance: the larger the
resistance, the less current flows through the branch If we know I
but not V R1R1 R2R2 V
Slide 17
20.2
Slide 18
Review of Circuit Rules For SERIES Circuits: There is ONE path
for the current Current is CONSTANT Voltage DROPS across each
resistance Resistors add simply Additional resistances DECREASE
current
Slide 19
Review of Circuit Rules For PARALLEL Circuits: There are
MULTIPLE paths for the current Current may NOT be CONSTANT Voltage
is CONSTANT to each branch Resistors add RECIPROCALLY Additional
resistance INCREASES current
Slide 20
Review of Circuit Rules Sometimes you will have BOTH series AND
parallel resistors in the SAME circuit!! You then need to SIMPLIFY
the circuit in your analysis.
Slide 21
Example 3 Consider this circuit. (a) If possible, simplify it
and determine an equivalent resistance between C and G. (b) What
current is provided by the source? (c) What is the voltage across
points G and E? Given: Nine resistors, R = 1.0 k each, and V = 12V
Find: R e, I, and V between G and E
Slide 22
Example 3 Solution To solve this one we'll need the equivalent
resistance of the circuit Procedure Redraw this circuit to make it
look more manageable Lift up the inside square, with the resistor
and source attached, and place it outside E-F-G-H see diagram on
right Branches A-B-C and A-D-C are in parallel, as are E-F-G and
E-H-G, and each has a resistance of 1.0 k + 1.0 k = 2.0 k
(resistors add in series)
Slide 23
Example 3 The equivalent circuit is shown to the right The
resistance of each square E-F-G-H and A-B- C-D reduces to and R =
1.0 k .
Slide 24
Example 3 The three 1.0-k resistors then are in series with the
source, (a) The equivalent resistance is 3.0 k . (b) Since V = IR
e,
Slide 25
Example 3 (c) A current of 4.0 mA leaves the battery and splits
at C Because the two branches C-D-A and C-B-A have the same
resistance, the current divides into two equal streams of 2.0 mA
each The voltage drop in going from C to A is given by V AC = IR =
(2.0 mA)(1.0 k + 1.0 k ) = 4.0 V In going from A to E, there is
another drop of V EA = IR = (4.0 mA)(1.0 k ) = 4.0 V. C is 12V
above G, A is 8.0 V above G, and E is 4.0V above G.
Slide 26
Example 4 Consider the following circuit: A battery supplying
12 volts leads to a resistor (R 1 = 1.3), then splits into three
branches. The first branch has R 2 = 4.5, the second branch has R 3
= , and the third branch contains R 4 = 5 AND R 5 = 2.2 in series.
Finally, the three branches reunite, and lead to R 6 = 7 before
reconnecting to the battery. Draw a diagram of this circuit Find
the total resistance Find the overall current in the circuit.
Slide 27
Problem-Solving Strategy, 1 Combine all resistors in series
They carry the same current The potential differences across them
are not the same The resistors add directly to give the equivalent
resistance of the series combination: R e = R 1 + R 2 +
Slide 28
Problem-Solving Strategy, 2 Combine all resistors in parallel
The potential differences 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:
Slide 29
Problem-Solving Strategy, 3 A complicated circuit consisting of
several resistors and batteries can often be reduced to a simple
circuit with only one resistor 1. Replace any resistors in series
or in parallel using steps 1 or 2. 2. Sketch the new circuit after
these changes have been made 3. Continue to replace any series or
parallel combinations 4. Continue until one equivalent resistance
is found
Slide 30
Problem-Solving Strategy, 4 If the current in 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 V = I R and the
procedures in steps 1 and 2
Slide 31
Equivalent Resistance Complex Circuit
Slide 32
Capacitors in Circuits A circuit is a collection of objects
usually containing a source of electrical energy (such as a
battery) connected to elements that convert electrical energy to
other forms A circuit diagram can be used to show the path of the
real circuit
Slide 33
Capacitors in Parallel When capacitors are first connected in
the circuit, electrons are transferred from the left plates through
the battery to the right plate, leaving the left plate positively
charged and the right plate negatively charged The flow of charges
ceases when the voltage across the capacitors equals that of the
battery The capacitors reach their maximum charge when the flow of
charge ceases
Slide 34
Capacitors in Parallel The total charge is equal to the sum of
the charges on the capacitors Q total = Q 1 + Q 2 The potential
difference across the capacitors is the same And each is equal to
the voltage of the battery
Slide 35
More About Capacitors in Parallel The capacitors can be
replaced with one capacitor with a capacitance of C eq The
equivalent capacitor must have exactly the same external effect on
the circuit as the original capacitors Capacitors in parallel all
have the same voltage differences as does the equivalent
capacitance
Slide 36
Capacitors in Parallel The equivalent capacitance of several
capacitors in parallel is the sum of all the individual capacitors.
C = C 1 + C 2 + The equivalent capacitance of a parallel
combination of capacitors is greater than any of the individual
capacitors
Slide 37
Example 5 The figure below shows two capacitors attached to a
12-V battery. Determine the equivalent capacitance and the charge
it would carry. What is the charge on each of the capacitors in the
figure? Given: C 1 = 20 F, C 2 = 30 F, and V = 12 V Find: C, Q, Q
1, and Q 2 +++ -- - 20 F30 F 12 V
Slide 38
Example 5 Solution: Capacitors are in parallel and the
potential across each capacitor is 12 V Q = CV = (50 x 10 -6 F)(12
V) = 6.0 x 10 -4 C
Slide 39
Capacitors in Series When a battery is connected to the
circuit, electrons are transferred from the left plate of C 1 to
the right plate of C 2 through the battery As this negative charge
accumulates on the right plate of C 2, an equivalent amount of
negative charge is removed from the left plate of C 2, leaving it
with an excess positive charge All of the right plates gain charges
of Q and all the left plates have charges of +Q
Slide 40
More About Capacitors in Series An equivalent capacitor can be
found that performs the same function as the series combination The
potential differences add up to the battery voltage Capacitors in
series all have the same charge, Q, as does their equivalent
capacitance
Slide 41
Capacitors in Series The equivalent capacitance of several
capacitors in series The equivalent capacitance of a series
combination is always less than any individual capacitor in the
combination
Slide 42
Example 6 The circuit shown in the figure consists of a 12-V
battery and three capacitors. It is redrawn from Fig. 12.27a in the
book. Determine both the voltage across and charge on each
capacitor after the switch S is closed and electrostatic
equilibrium is established. Find the equivalent capacitance of the
network. Given: C 1 = 2.0 F, C 2 = 2.0 F, C 3 = 5.0 F, and V = 12 V
Find: C, V 1, V 2, V 3, Q 1, Q 2, and Q 3 + 2.0 F 12 V 2.0 F 5.0 F
C3C3 C1C1 C2C2
Slide 43
Example 6 Combining C 1 and C 2 which are in series + 12 V 1.0
F 5.0 F C3C3 C 1 + C 2
Slide 44
Example 6 Combining C 3 and (C 1 + C 2 ) which are in parallel
+ 12 V 6.0 F C The equivalent capacitance of the network
Slide 45
Example 6 There is 12 V across C 3 so Q 3 = C 3 V 3 = (5.0
F)(12 V) = 60 C + 2.0 F 12 V 2.0 F 5.0 F C3C3 C1C1 C2C2 There is 12
V across the combination of the two 2.0 mF capacitors so there is a
potential difference of 6.0 V across each Q 1 = Q 2 = (2.0 F)(6.0
V) = 12 C
Slide 46
Problem-Solving Strategy Be careful with the choice of units
Combine capacitors following the formulas When two or more unequal
capacitors are connected in series, they carry the same charge, but
the potential differences across them are not the same The
capacitances add as reciprocals and the equivalent capacitance is
always less than the smallest individual capacitor
Slide 47
Problem-Solving Strategy, cont Combining capacitors When two or
more capacitors are connected in parallel, the potential
differences across them are the same The charge on each capacitor
is proportional to its capacitance The capacitors add directly to
give the equivalent capacitance
Slide 48
Problem-Solving Strategy, final Repeat the process until there
is only one single equivalent capacitor A complicated circuit can
often be reduced to one equivalent capacitor Replace capacitors in
series or parallel with their equivalent Redraw the circuit and
continue To find the charge on, or the potential difference across,
one of the capacitors, start with your final equivalent capacitor
and work back through the circuit reductions
Slide 49
20.3
Slide 50
Household Circuits The utility company distributes electric
power to individual houses with a pair of wires Electrical devices
in the house are connected in parallel with those wires The
potential difference between the wires is about 120V
Slide 51
Household Circuits A meter and a circuit breaker are connected
in series with the wire entering the house Wires and circuit
breakers are selected to meet the demands of the circuit If the
current exceeds the rating of the circuit breaker, the breaker acts
as a switch and opens the circuit Household circuits actually use
alternating current and voltage
Slide 52
Household Usage Electricity usage is measured in kilowatt-hours
(kWh) Watts are units of Power P = VI (for DC) Usually measured in
Kilowatts or Horsepower 1hp = 746W 1kWh = 3.6 x 10 6 J
Slide 53
Types of Current Direct Current (DC) charge flows uniformly in
ONE direction Alternating Current (AC) charge flows in opposite
directions alternating in a regular, periodic way with a given
frequency. Peak vs. Average Voltage in AC the difference is the
Average Voltage coming out of the wall is a percentage of the Peak
Voltage supplied. **Alternating Current is both easier to generate
AND to transmit long distances**
Slide 54
Power in AC Power in AC circuits is calculated in the same way
as in DC circuits but using average voltage values instead of peak.
Peak values for current (I) can be found using Peak Voltage when
resistance (R) is known. I freakin HATE Circuit analysis!!
Slide 55
Reactance in AC In AC circuits, where the current is constantly
(and ver rapidly) reversing, if you have anything other than simple
resistance (Capacitors or Inductors), the response time or
Reactance of those components will create a lag time in voltage
response to the current shift. Inductive Reactance Capacitive
Reactance
Slide 56
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
Slide 57
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 just a few seconds, can be fatal
Slide 58
Ground Wire Electrical equipment manufacturers use electrical
cords that have a third wire, called a case ground Prevents
shocks