Chapter 27 Circuits
In this chapter we will cover the following topics:
-Electromotive force (emf) -Ideal and real emf devices -Kirchhoff’s loop rule -Kirchhoff’s junction rule -Multiloop circuits -Resistors in series -Resistors in parallel -RC circuits, charging and discharging of a capacitor
(27 – 1)
A cylindrical copper rod has resistance R. It is reformed to three times its original length with no change of volume. Its new resistance is:
A. RB. 3RC. 9RD. R/9E. R/3
In order to create a current through a resistor, a potential
difference must be created across its terminals. One way
of doing this is to connect the resistor to a battery. A device
that can maintain a potential difference between two terminals
is called a or an . Here emf
stands for: electromotive force. Examples of emf devices:
a battery, an electric generator, a so
"seat of an emf " "emf device"
lar cell, a fuel cell, etc
pump
High (+) reservoir
Low (-) reservoir
These devices act like "charge pumps" in the sense that they move positive charges
from the low potential (negative) terminal to the high potential (positive) terminal.
A mechanical analog is given in the figure below.
In this mechanical analog a water pump transfers
water from the low to the high reservoir. The water
returns from the high to the low reservoir through a
pipe which is the analog of the resistor.
The emf (symbol ) is defined as the potential
difference between the terminals of the emf device
when no current flows through it.
E
(27 – 2)
The polarity of an emf device is
indicated by an arrow with a small circle at its tail.
The arrow points from the negative to the positive
terminal of the device. When the emf device is
co
Notation :
nnected to a circuit its internal mechanism transports
positive charges from the negative to the positive terminal
and sets up a charge flow (a.k.a. current) around the
circuit. In doing so the emf device does work on
a charge which is given by the equation:
. The required energy comes
from chemical reactions in the case of a battery; in the case
of a generator it comes from the mecha
dW q
dq
d
dW
E
nical force that
rotates the generator shaft; in the case of a solar cell it
comes from the sun. In the circuit of the figure
the energy stored in emf device B changes form:
It does mechanical work on the motor. It produces
thermal energy on the resistor. It gets converted into
chemical energy in emf device A
(27 – 3)
An emf device is said to be if the voltage across its
terminals and does depend on the current that flows
through the emf device.
An emf device is s
id
a
V
a b i
V
Ideal and real emf devices
ideal
not
E
to be if the voltage across its
terminals and with current according to
the equation: .
The parameter is known as the device's " "
of the emf device
V ir
V
a b i
r
real
decreases
internal resistance
E
.
V
i
E
Ideal emf device
V
i
E
Real emf device
V E
V ir E
(27 – 4)
Consider the circuit shown in the figure. We assume
that the emf device is ideal and that the connecting
wires have negligible resistance. A current flows
through the c
i
Current in a single loop circuit
ircuit in the clockwise direction.
In a time interval a charge passes through the circuit. The battery is
doing work . Using energy conservation we can set this amount
of work equal to the rate at which heat is
dt dq idt
dW dq idt
E Ei i
2 generated on R.
Kirchhoff put the equation above in the form of a rule known as Kirchhoff's l
0
oop rule
(KLR for short)
idt Ri dt
Ri iR
Ei
E Ei i
0iR Ei
The algerbraic sum of the changes in potential encountered in a complete
traversal of any loop in a circuit is equal to zero.
KLR :
The rules that give us algebraic sign of the charges in potential through a resistor
and a battery are given on the next page.(27 – 5)
Ri
motion
-V iR
Ri
motion
V iR
motion-V E+ -
motionV E+-
For a move through a resistance
in the direction of the current, the change in the
potential
For a move through a resistance in the direction
opposite to that of the current, th
V iR
Resistance Rule :
e change in the
potential V iR
For a move through an ideal emf device
in the direction of the emf arrow,
the change in the potential
For a move through an ideal emf device in a direction
opposi
te to that of the
V
EMF Rule :
E
emf arrow,
the change in the potenti al V E
(27 – 6)
Consider the circuit of fig.a. The battery is real with internal
resistance . We apply KLR for this loop starting at point a and going counterclockwise:
0 . We note tha
r
ir iR iR r
KLR example :
EE t for an ideal battery 0 and
The internal resistance of the battery is an integral part of the battery
internal mechanism. There is no way to open the battery and remove .
In fig.b we
r iR
r
r
Note :
E
plot the potential V of every point in the loop as we start at point a
and go around in the counterclockwise direction. The change in the battery
is positive because we go from the negative to th
Ve positive terminal. The change
across the two resistors is negative because we chose to traverse the loop
in the direction of the current. The current flows from high to low potential
V
(27 – 7)
Consider the circuit shown in the figure. We wish to
calculate the potential difference between
point b and point a.b aV V
Potential difference between two points :
We choose a path in the loop that takes us from the initial point a to the final point b.
sum of all potential changes along the path.
There are two possible paths: We will try them both.
Le
f iV V V
ft path:
Right path:
The values of we get from the two paths are the same.
b a
b a
b a
V V ir
V V iR
V V
Note :
E
sum of all potential changes along the path from point a to point b b aV V V
(27 – 8)
i
V
Consider the combination of resistors
shown in the figure . We can substitute
these combinations of resistors with a
single resistor that is
"electrically equivalent"
to the r
eqR
Equivalent Resistance :
esistor group it substitutes.
This means that if we apply the same voltage across the
resistors in fig.a and across the same current is provided by the battery.
Alternatively, if we pass the same current through the
eq
V
R i
i circuit in fig.a
and through the equivalent resistance , the voltage across them is identical.
This can be stated in the following manner: If we place the resistor
combination and the equivalent
eqR V
resistor in separate black boxes,
by doing electrical mesurements we cannot distinguish between the two.
(27 – 9)
1 2 3
Consider the three resistors connected in series
(one after the other) as shown in fig.a. These resistors have
the but different voltages , , and
The net
i V V Vs
Resistors in ser
ame cu
ies :
rr
ent
e e
1 2 3
1 2 31 2 3
voltage across the combination is the sum
We will apply KLR for the loop in fig.a starting at point a,
and going around the loop in the counterclockwise direction:
0
V V V
iR iR iR iR R R
EE
1
( )
We will apply KLR for the loop in fig.b starting at point a,
and going around the loop in the counterclockwise direction:
0 ( )
If we compare eqs.1 with eqs.2 we get: q
eeq
e
qiR iR
R R
eqs.1
eqs.2E
E
2 3
1 21
For resistors connected in series the quivalent resistance i
..
s:
.
n
eq i ni
R R
R R R
n
R R
1 2 3eqR R R R
(27 – 10)1 2 ...eq nR R R R
Consider the circuit shown in the figure. There
are three brances in it: Branch bad, bcd, and bd.
We assign currents for each branch and define the
current directions arbitrarily.
Multiloop circuits :
The method is self
correcting. If we have made a mistake in the direction
of a particular current the calculation will yield
a negative value and thus provide us with a warning.
1 2 3
1 3 2
1
We asign current for branch bad, current for branch bcd, and current
for branch bd. Consider junction d. Currents and arrive, while leaves.
Charge is conserved thus we have :
i i i
i i i
i 3 2. This equation can be fomulated
as a more general principle knwon as Kirchhoff's junction rule (KJR)
i i
The sum of the currents entering any junction is equal to the sum of the currents
leaving the junction
KJR :
(27 – 11)
1 2 3
1 3 2
In order to determine the currents , , and
in the curcuit we need three equations. The first
equation will cone from KJR at point d:
KJR/junction d: ( )
i i i
i i i eqs.1
1 1 1 3 3
The other two will come from KLR: If we traverse the left loop (bad) starting at b
and going in the counterclockwise direction we get:
KLR/loop bad: 0 ( ) Now we go around the rig hi R i R eqs.2E
3 3 2 2 2
1 2 3
t loop (bcd)
starting at point and going in the counterclockwise direction:
KLR/loop bcd: 0 ( )
We have a system of three equations (eqs.1,2 and 3) and three unknowns , , and
i R i R
i i i
eqs.3E
If a numerical value for a particular current is negative this means that the chosen
direction for this current is wrong and that the current flows in the opposite direction.
We can write a fourth equa
1 1 1 2 2 2
tion (KLRfor the outer loop abcd) but this equation does
not provide any new information.
KLR/loop abcd: 0 ( )i R i R eqs.4E E(27 – 12)
Consider the three resistrors shown in the figure
"In parallel" means that the terminals of the resistors
are connected together on both sides. Thus resistrors
in parallel have the
Resistors in parallel
applied across them.
In our circuit this potential is equal to the emf of the
battery. The three resistors have different currents
flowing through them. The total current is the sum
o
same potential
E
1 2 3 1 2 31 2 3
1 2 3
f the individual currents. We apply KJRat point a.
1 1 1 (eqs.1) From fig.b we have:
(eqs.2) If we compare equations eq
i i i i i i iR R R
iR R R
iR
E E E
E
E
1 2 3
1 1 1 1
1 and 2
we get: eqR R R R
1 2 3
1 1 1 1
eqR R R R
(27 – 13)
An ammeter is an istrument that measures current.
In order to measure the current that flows through
a conductor at a certain point we must cut the
conductor at this point and co
Ammeters and Voltmeters
nnect the two ends
of the conductor to the ammeter terminals so that
the current can pass through the ammeter.
An example is shown in the figure, where
the ammeter has been inserted between points a and b.
1 2
It is essential that the ammeter resistance be much smaller than the other
resistors in the circuit. In our example: and
A voltmeter is an instrument that measures the potential diff
A
A A
R
R R R R
1
erence between two point
in a circuit. In the example of the figure we use a voltmeter to measure the potential
across . The voltmeter terminals are connected to the two points c and d.
It is essenti
R
1 2
al that the voltmeter resistance be much lerger than the other
resistors in the circuit. In our example: and V
V V
R
R R R R (27 – 14)
+
-
iConsider the circuit shown in the figure. We assume
that the capacitor is initially uncharged and that at
0 we throw the switch S from the middle position
to posi
t
RC circuits : Charging of a capacitor
tion a. The battery will charge the capacitor
through the resistor .
C
R
Our objective is to examine the charging process as function of time.
We will write KLR starting at point b and going in the the counterclockwise direction.
0 The current q dq dq q
iR i RC dt dt C
E E 0 If we rearrange the terms
we have: This is an inhomogeneous, first order, linear differential
equation with initial condition: (0) 0. This condition expresses the fact that
a
t
dq qR
dt Cq
t
E
0 the capacitor is uncharged.
(27 – 15)
/
Differential equation:
Intitial condition: (0) 0
Solution: 1 Here:
The constant is known as the "time constant" of the circuit.
If we plot versus we see t
t
dq qR
dt Cq
q C e RC
q t
E
E
hat does not reach its terminal
value but instead increases from its initial value and it
reaches the terminal value at . Do we have to wait for
an internity to charge the capacitor? In pract
q
C
t E
/
ice no.
( ) 0.632
( 3 ) 0.950
( 5 ) 0.993
If we wait only a few time constants the charge, for all
practical purposes has reached its terminal value .
The current If t
q t C
q t C
q t C
C
dqi edt R
E
E
E
E.
Ewe plot i versus t
we get a decaying exponential (see fig.b) (27 – 16)
RC
+
-
i
q
t
qo
O
Consider the circuit shown in the figure. We assume
that the capacitor at 0 has charge and that at
0 we throw the switch S from the middle position
to pos
ot q
t
RC circuits : Discharging of a capacitor
ition b. The capacitor is disconnected from the
battery and looses its charge through resistor .
We will write KLR starting at point b and going in
the counterclockwise direction: 0
Taking int
R
qiR
C
o account that we get: 0 dq dq q
i Rdt dt C
- /
This is an homogeneous, first order, linear differential equation with initial condition:
(0) The solution is: Where . If we plot q versus t we
get a decaying exponetial.
to oq q q q e RC
The charge becomes zero at . In practical terms we
only have to wait a few time constants.
( ) 0.368 , (3 ) 0.049 , (5 ) 0.007
o o o
t
q q q q q q
(27 – 17)
RC
In the diagram R1 > R2 > R3. Rank the three resistors according to the current in them, least to greatest.
A. 1, 2, 3B. 3, 2, 1C. 1, 3, 2D. 3, 1, 3E. All are the same
A. 5 Ω B. 7 Ω C. 2 Ω D. 11 Ω E. NOT
I. In the diagram, the current in the 3-resistor is 4A. The potential difference between points 1 and 2 is:
A. 0.75VB. 0.8VC. 1.25VD. 12VE. 20V