Chapter 18 Direct Current Circuits 18.1 Sources of emf  The source that maintains the current in a closed circuit is called a source.

Chapter 18
Direct Current Circuits
18.1 Sources of emf
The source that maintains the current in
a closed circuit is called a source of emf
Emf=electromotive force
Any devices that increase the potential
energy of charges circulating in circuits are
sources of emf
 Examples include batteries and generators
Emf and Internal Resistance
A real battery has
some internal
 Therefore, the
terminal voltage is
not equal to the emf
More About Internal
The schematic shows
the internal resistance,
The terminal voltage is
V = ε – Ir (source
minus the internal
 For the entire circuit,
ε = IR + Ir =I (R+r)
Internal Resistance and emf,
ε is equal to the terminal voltage when
the current is zero (open circuit)
Also called the open-circuit voltage
R is called the load resistance
The current depends on both the
resistance external to the battery and
the internal resistance
18.2 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 resistors because
any charge that flows through one resistor
flows through the other
 The sum of the potential differences across
the resistors is equal to the total potential
difference across the combination
Resistors in Series, cont
Voltages add
 V =IR1 + IR2
 V =I (R1+R2)
 Req= R1+R2
 The equivalent
resistance Req has the
same effect on the
circuit as the original
combination of
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 resistance ( see parallel
connection of capacitors!)
Equivalent Resistance – Series
An Example
Four resistors are replaced with their
equivalent resistance
18.3 Resistors in Parallel
The potential difference 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 = I1 + I2
The currents are generally not the same
Consequence of Conservation of Charge
Equivalent Resistance –
Parallel, cont
 I1=V/R1
and I2=V/R2. The
complete current provided by
the source is given by,
 1/Req=1/R1+1/R2
 Req is the equivalent resistance for a
parallel circuit
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 –
Equivalent Resistance
 
Req R1 R2 R3
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 ( see series connection
of capacitors)
The equivalent is always less than the smallest
resistor in the group
Problem-Solving Strategy, 1
When two or more resistors are
connected in series, they carry the
same current, but the potential
differences across them are not the
The resistors add directly to give the
equivalent resistance of the series
Problem-Solving Strategy, 2
When two or more resistors are connected
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
The equivalent resistance is always less than
the smallest individual resistor in the
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
Replace any resistors in series or in parallel using
steps 1 or 2.
Sketch the new circuit after these changes have
been made
Continue to replace any series or parallel
Continue until one equivalent resistance is found
Problem-Solving Strategy, 4
If the current in or the potential difference
across a resistor in a 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
Compute the equivalent resistance and
the current of the network (a) below.
Rp=Req for parallel connection
and Rs=Req for series connection
1/(3 W)+1/(6 W)=1/Rp  Rp=2 W (b)
4 W+2 W=Rs  Rs=6 W (c)
I=V/Rs=18 V/6 W=3 A (d)
V=(3 A)(4 W+2 W)12 V+6 V=18 V (e)
6 V/6 W=1 A and 6 V/3 W=2 A (f)
n parallel equal resistors, the
equivalent Rp is expressed by
 Rp=R1/n
 The same shortcut is valid for n
 For
equal capacitors in series
 Cs=C1/n
18.4 Kirchhoff’s Rules
There are ways in which
resistors and batteries
can be connected so that
the circuits formed cannot
be reduced to a single
equivalent resistor (two
examples are shown on
the right)
 Two rules, called
Kirchhoff’s Rules can be
used instead
Statement of Kirchhoff’s Rules
Junction Rule ( I = 0)
The sum of the currents entering any junction
must equal the sum of the currents leaving that
A statement of Conservation of Charge
Loop Rule ( U = 0)
The sum of the potential differences across all the
elements around any closed circuit loop must be
A statement of Conservation of Energy
More About the Junction Rule
 I1 = I2 + I3
 From Conservation
 Diagram (b) shows a
mechanical analog
More About the Loop Rule
The voltage across a
battery is taken to be
positive (a voltage rise) if
traversed from – to + and
and negative if traversed in
the opposite direction.
(b) The voltage across a
resistor is taken to be
negative (a drop) if the
loop is traversed in in the
direction of the assigned
current and positive if
traversed in the opposite
Setting Up Kirchhoff’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 transversing the loop
Record voltage drops and rises as they
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
 Apply the loop rule to as many loops as are
needed to solve for the unknowns
 Solve the equations simultaneously for the
unknown quantities.
Example: Find the current I, r and e.
Junction rule at a:
2 A+1 A-I=0
I=3 A
Loop (1):
12 V-Ir -(3 W)(2 A)=0
r=12 V/(3 A)-6 V/(3 A)
r=2 W
Loop (2):
-e+(1 W)(1 A)-( 3 W)(2 A)=0
e=-5 V (polarity is opposite!)
Check with loop (3):
12 V-(2 W)(3 A)-(1 W)(1 A)+e =0
e=-5 V

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