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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
A cylindrical copper rod has resistance R. It is
re-formed to three times its original length
with no change of volume. Its new resistance
is:
A. R
B. 3R
C. 9R
D. R/9
E. R/3
(27 – 1)
In order to create a current through a resistor, a potential
difference must be created across its terminals. One way
(27 – 3)
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 "seat of an emf " or an "emf device". Here emf
The arrow points from the negative to the positive
terminal of the device. When the emf device is
connected to a circuit its internal mechanism transports
stands for: electromotive force. Examples of emf devices:
a battery, an electric generator, a solar cell, a fuel cell, etc
positive charges from the negative to the positive terminal
These devices act like "charge pumps" in the sense that they move positive charges
and sets up a charge flow (a.k.a. current) around the
circuit. In doing so the emf device does work dW on
from the low potential (negative) terminal to the high potential (positive) terminal.
a charge dq which is given by the equation:
dW = Edq. The required energy comes
A mechanical analog is given in the figure below.
High (+)
reservoir
pump
Low (-)
reservoir
In this mechanical analog a water pump transfers
water from the low to the high reservoir. The water
from chemical reactions in the case of a battery; in the case
of a generator it comes from the mechanical force that
returns from the high to the low reservoir through a
pipe which is the analog of the resistor.
The emf (symbol E ) is defined as the potential
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
Ei−iR = 0
An emf device is said to be ideal if the voltage V across its
terminals a and b does not depend on the current i that flows
through the emf device. V = E
V =E
rotates the generator shaft; in the case of a solar cell it
difference between the terminals of the emf device
when no current flows through it.
(27 – 2)
Ideal and real emf devices
Current in a single loop circuit
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 i flows
through the circuit in the clockwise direction.
Ideal emf device
In a time interval dt a charge dq = idt passes through the circuit. The battery is
V
doing work dW = Edq = Eiidt.i Using energy conservation we can set this amount
E
i
V = E − ir
Real emf device
E
Notation : The polarity of an emf device is
indicated by an arrow with a small circle at its tail.
V
i
An emf device is said to be real if the voltage V across its
terminals a and b decreases with current i according to
the equation: V = E − ir.
The parameter r is known as the device's "internal resistance"
of the emf device.
(27 – 4)
of work equal to the rate at which heat is generated on R. Eiidt = Ri 2 dt →
Ei= Ri → Ei−iR = 0
Kirchhoff put the equation above in the form of a rule known as Kirchhoff's loop rule
(KLR for short)
KLR : The algerbraic sum of the changes in potential encountered in a complete
traversal of any loop in a circuit is equal to zero.
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)
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R
i
Resistance Rule :
∆V = -iR
For a move through a resistance
motion
in the direction of the current, the change in the
R
i
∆V = +iR
motion
potential ∆V = −iR
For a move through a resistance in the direction
opposite to that of the current, the change in the
potential ∆V = +iR
KLR example : Consider the circuit of fig.a. The battery is real with internal
resistance r. We apply KLR for this loop starting at point a and going counterclockwise:
E
E
. We note that for an ideal battery r = 0 and i =
R+r
R
Note : The internal resistance r of the battery is an integral part of the battery
internal mechanism. There is no way to open the battery and remove r.
E − ir − iR = 0 → i =
-
EMF Rule :
+
motion
∆V = +E
For a move through an ideal emf device
in the direction of the emf arrow,
the change in the potential ∆V = +E
For a move through an ideal emf device in a direction
+
-
∆V = -E
opposite to that of the emf arrow,
the change in the potential ∆V = −E
motion
In fig.b we 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 ∆V in the battery
is positive because we go from the negative to the positive terminal. The change
∆V across the two resistors is negative because we chose to traverse the loop
(27 – 6)
in the direction of the current. The current flows from high to low potential
(27 – 7)
Potential difference between two points :
Consider the circuit shown in the figure. We wish to
calculate the potential difference Vb − Va between
i
V
point b and point a.
Vb − Va = sum of all potential changes ∆V along the path from point a to point b
We choose a path in the loop that takes us from the initial point a to the final point b.
V f − Vi = sum of all potential changes ∆V along the path.
There are two possible paths: We will try them both.
Equivalent Resistance :
Consider the combination of resistors
shown in the figure . We can substitute
these combinations of resistors with a
single resistor Req that is
"electrically equivalent"
to the resistor group it substitutes.
This means that if we apply the same voltage V across the
resistors in fig.a and across Req the same current i is provided by the battery.
Alternatively, if we pass the same current i through the circuit in fig.a
and through the equivalent resistance Req , the voltage V across them is identical.
Left path: Vb − Va = E − ir
This can be stated in the following manner: If we place the resistor
Right path: Vb − Va = iR
combination and the equivalent resistor in separate black boxes,
by doing electrical mesurements we cannot distinguish between the two.
Note : The values of Vb − Va we get from the two paths are the same.
(27 – 9)
(27 – 8)
Resistors in series : e e
Multiloop circuits :
Consider the three resistors connected in series
Consider the circuit shown in the figure. There
are three brances in it: Branch bad, bcd, and bd.
(one after the other) as shown in fig.a. These resistors have
the same current i but different voltages V1 , V2 , and V3
The net voltage across the combination is the sum V1 + V2 + V3
We will apply KLR for the loop in fig.a starting at point a,
and going around the loop in the counterclockwise direction:
of a particular current the calculation will yield
a negative value and thus provide us with a warning.
E
E − iR1 − iR2 − iR3 = 0 → i =
(eqs.1)
R1 + R2 + R3
We asign current i1 for branch bad, current i2 for branch bcd, and current i3
We will apply KLR for the loop in fig.b starting at point a,
for branch bd. Consider junction d. Currents i1 and i3 arrive, while i2 leaves.
and going around the loop in the counterclockwise direction:
Charge is conserved thus we have: i1 + i3 = i2 . This equation can be fomulated
E − iReq = 0 → i =
Req = R1 + R2 + R3
We assign currents for each branch and define the
current directions arbitrarily. The method is self
correcting. If we have made a mistake in the direction
E
(eqs.2)
Req
as a more general principle knwon as Kirchhoff's junction rule (KJR)
If we compare eqs.1 with eqs.2 we get: Req = R1 + R2 + R3
KJR : The sum of the currents entering any junction is equal to the sum of the currents
leaving the junction
For n resistors connected in series the quivalent resistance is:
Req = R1 + R2 + ... + Rn
n
Req = ∑ Ri = R1 + R2 + ... + Rn
i =1
(27 – 10)
(27 – 11)
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Resistors in parallel
In order to determine the currents i1 , i2 , and i3
Consider the three resistrors shown in the figure
in the curcuit we need three equations. The first
"In parallel" means that the terminals of the resistors
equation will cone from KJR at point d:
are connected together on both sides. Thus resistrors
in parallel have the same potential applied across them.
KJR/junction d: i1 + i3 = i2 (eqs.1)
In our circuit this potential is equal to the emf E of the
battery. The three resistors have different currents
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: E1 − i1R1 + i3 R3 = 0 (eqs.2) Now we go around the right loop (bcd)
flowing through them. The total current is the sum
of the individual currents. We apply KJRat point a.
i = i1 + i2 + i3
starting at point and going in the counterclockwise direction:
KLR/loop bcd: − i3 R3 − i2 R2 − E2 = 0 (eqs.3)
1
1
1
1
= +
+
Req R1 R2 R3
we get:
not provide any new information.
KLR/loop abcd: E1 − i1 R1 − i2 R2 − E2 = 0 (eqs.4)
(27 – 12)
Ammeters and Voltmeters
An ammeter is an istrument that measures current.
i2 =
E
R2
i3 =
E
→
R3
1
1 1
1
= +
+
Req R1 R2 R3
(27 – 13)
RC circuits : Charging of a capacitor
i
Consider the circuit shown in the figure. We assume
that the capacitor is initially uncharged and that at
In order to measure the current that flows through
+
a conductor at a certain point we must cut the
conductor at this point and connect the two ends
t = 0 we throw the switch S from the middle position
- to position a. The battery will charge the capacitor C
through the resistor R.
of the conductor to the ammeter terminals so that
the current can pass through the ammeter.
An example is shown in the figure, where
E
R1
1
1
1 
i = E +
+  (eqs.1) From fig.b we have:
 R1 R2 R3 
E
i=
(eqs.2) If we compare equations 1 and 2
Req
We have a system of three equations (eqs.1,2 and 3) and three unknowns ( i1 , i2 , and i3 )
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 equation (KLRfor the outer loop abcd) but this equation does
i1 =
Our objective is to examine the charging process as function of time.
the ammeter has been inserted between points a and b.
We will write KLR starting at point b and going in the the counterclockwise direction.
q
dq
dq
q
= 0 The current i =
→E−
R − = 0 If we rearrange the terms
C
dt
dt
C
dq
q
we have:
R + = E This is an inhomogeneous, first order, linear differential
dt
C
equation with initial condition: q(0) = 0. This condition expresses the fact that
E − iR −
It is essential that the ammeter resistance RA be much smaller than the other
resistors in the circuit. In our example: RA
R1 and RA
R2
A voltmeter is an instrument that measures the potential difference between two point
in a circuit. In the example of the figure we use a voltmeter to measure the potential
across R1. The voltmeter terminals are connected to the two points c and d.
at t = 0 the capacitor is uncharged.
It is essential that the voltmeter resistance RV be much lerger than the other
resistors in the circuit. In our example: RV
R1 and RV
R2
dq
q
R+ =E
dt
C
Intitial condition: q(0) = 0
Differential equation:
Solution: q = CE (1 − e −t /τ )
(27 – 14)
(27 – 15)
τ = RC
i
Here: τ = RC
The constant τ is known as the "time constant" of the circuit.
If we plot q versus t we see that q does not reach its terminal
value CE but instead increases from its initial value and it
reaches the terminal value at t = ∞. Do we have to wait for
an internity to charge the capacitor? In practice no.
q
qo
τ = RC
q(t = τ ) = ( 0.632 ) CE
q(t = 3τ ) = ( 0.950 ) CE
q(t = 5τ ) = ( 0.993 ) CE
If we wait only a few time constants the charge, for all
practical purposes has reached its terminal value CE..
dq  E  −t /τ
=  e
If we plot i versus t
dt  R 
we get a decaying exponential (see fig.b)
(27 – 16)
The current i =
RC circuits : Discharging of a capacitor
Consider the circuit shown in the figure. We assume
t
O
+ that the capacitor at t = 0 has charge qo and that at
- t = 0 we throw the switch S from the middle position
to position b. The capacitor is disconnected from the
battery and looses its charge through resistor R.
We will write KLR starting at point b and going in
q
the counterclockwise direction: − − iR = 0
C
dq
dq
q
Taking into account that i =
we get:
R+ =0
dt
dt
C
This is an homogeneous, first order, linear differential equation with initial condition:
q(0) = qo The solution is: q = qo e-t /τ
Where τ = RC. If we plot q versus t we
get a decaying exponetial. The charge becomes zero at t = ∞. In practical terms we
only have to wait a few time constants.
q(τ ) = ( 0.368) qo ,
q(3τ ) = ( 0.049 ) qo ,
q(5τ ) = ( 0.007 ) qo
(27 – 17)
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In the diagram R1 > R2 > R3. Rank
the three resistors according to the
current in them, least to greatest.
A. 1, 2, 3
B. 3, 2, 1
C. 1, 3, 2
D. 3, 1, 3
E. 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. 0.75V
B. 0.8V
C. 1.25V
D. 12V
E. 20V
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