29.1. Electromotive Force
Figure 29.1. Electron in electronic circuit.
To keep a current flowing in an electronics circuit we need a
source of electric potential. Consider the circuit shown in Figure 29.1. The
electrons in the conductor move from the end with the low (negative) potential
towards the end with the high (positive) potential. The velocity of the
electron is limited by the friction it experiences while traveling through the
conductor, and on average will not change. However, the electron will lose
potential energy, and its total mechanical energy is therefore reduced when it
arrives at the end of the circuit (positive terminal of the source). In order
to sustain the continuous flow of electrons around the circuit, the source must
force the electrons from the terminal with the positive potential to the
terminal with the negative potential. During this process, the source does
work on the electrons, and increases their total mechanical energy. The
strength of the source is measured in terms of the
electromotive force
(emf). The emf of a source is defined as the amount of electric energy
delivered by the source per Coulomb of positive charge as charge passes through
the source from the low-potential terminal to the high-potential terminal. A
steady current will flow through the circuit in Figure 29.1 if the increase of
potential energy of the electrons in the source is equal to the change in the
potential energy of the electrons along their path through the external
circuit. The unit of emf is the Volt (V) and usually the emf is simply called
the voltage of the source. The symbol for emf is [epsilon]. The most
important sources of emf are:
1.
Batteries: Batteries convert chemical energy into electric
energy. In a lead-acid battery the following reactions take place when the
battery delivers a current:
(29.1)
and
(29.2)
These reactions deposit electrons on the negative electrode and absorb
electrons from the positive electrodes. The reactions in eq.(29.1) and
eq.(29.2) will continue until the sulfuric acid is depleted. At this point the
battery is discharged. The battery can be recharged by forcing a current
through the battery in the reverse direction. The reverse of the reactions
listed in eq.(29.1) and eq.(29.2) will then occur.
2.
Electric Generators: Electric generators convert mechanical
energy into electric energy. The principle of operation of electric generators
will be discussed in Chapter 31.
3.
Fuel cells: A mixture of chemicals are combined in the fuel
cell. The chemical reactions that occur are
(29.3)
and
(29.4)
The net result of these reactions is the conversion of hydrogen and oxygen
into water which is removed from the fuel cell in the form of water vapor. A
fuel cell therefore burns fuel. The efficiency of the best fuel cells is about
45%.
4.
Solar Cell: A solar cell converts the energy of the sunlight
directly into electric energy. The emf of a silicon solar cell is 0.6 V.
However, the amount of current that can be extracted is rather small.
27.2. Single-loop currents
A source with a time-independent emf is represented by the symbol shown in
Figure 29.2. The long line in Figure 29.2 represents the positive terminal of
the source, while the short line represents the negative terminal.
Figure 29.2. Schematic symbol of source of emf.
An example of a simple circuit in which a resistor is
connected between the terminals of the emf source is shown in Figure 29.3. The
current through this circuit can be determined by applying
Kirchhoff's
rule, which states:
Around a closed loop in a circuit, the sum of all the emfs and all the
potential changes across resistors and other circuit elements must equal
zero.
In this sum, the emf of a source is reckoned as positive if the current flows
through the source in the forward direction (from the negative terminal to the
positive terminal) and negative if it flows in the backward direction. (Note:
the direction of the current indicates the direction of the positive charge
carriers).
Figure 29.3. Simple single-loop circuit.
The current flowing in the circuit shown in Figure 29.3 flows
through the source from the negative terminal to the positive terminal. The
emf of the source is therefore positive. The potential drop [Delta]V across
the resistor can be determined using Ohm's law, and is equal to I R. Applying
Kirchhoff's rule one obtains the following equation:
(29.5)
The current in the circuit is therefore equal to
(29.6)
Figure 29.4. Real source consisting of internal resistance
Ri and ideal emf [epsilon].
In a real circuit we will have to take the internal resistance
R
i of the source into account. The internal resistance
R
i of the source can be regarded as connected in series to an ideal
emf (see Figure 29.4). If a current I is flowing though the source, the emf
across the external terminals of the source is equal to [epsilon] - I
. R
i. The external emf therefore depends on the current
delivered by the source.
29.3. Multi-loop circuits.
The procedure used to calculate the current flowing through a complicated
circuit (with several resistors and emfs) is called the
loop
method. This procedure can be summarized as follows (see Figure 29.5
and Problem 29.17):
Figure 29.5. Problem 29.17
1. Regard the circuit as a collection of several closed current loops. The
loops may overlap, but each loop must have at least one portion that does not
overlap with other loops. In Figure 29.5 there are clearly 3 current loops.
2. Label the current in the loops I
1, I
2, I
3,
....., and arbitrarily assign a direction to each of the currents. A useful
policy can be to define the direction of the current in a loop as going from
the positive terminal of the emf to the negative terminal of the emf (if an emf
is present in the loop). This policy defines the direction of the current in
loop 1, loop 2 and in loop 3 (see Figure 29.6).
3. Apply Kirchhoff's rule to each loop.
Figure 29.6. Current loops in problem 29.17.
For loop 1 Kirchhoff's law states that
(29.7)
For loop 2 we find
(29.8)
Finally, for loop 3 we find
(29.9)
Using eq.(29.8) we can rewrite eq.(29.9) in the following manner
(29.10)
or
(29.11)
The current I
2 can be obtained by substituting eq.(29.11) into
eq.(29.8):
(29.12)
The current I
1 can be obtained from eq.(29.7):
(29.13)
An alternative method to obtain the currents in a multi-loop circuit is the
branch method which is based on
Kirchhoff's first
rule:
The sum of all currents entering a branch point of a circuit (where three or
more wires merge) must be equal to the sum of the currents leaving the branch
point.
The branch method involves the following steps (and is illustrated by
discussing its application to problem 29.17):
Figure 29.7. Branch method applied to problem 29.17.
1. Regard the given circuit as a collection of branches which begin and end at
the points where wires merge. The circuit in Figure 29.5 has five branches.
2. Label the currents in each branch I
1, I
2,
I
3, ....., and arbitrarily assign a direction to each of these
currents. The currents in the five branches of circuit 29.5 are indicated in
Figure 29.7.
3. Apply Kirchhoff's second law to each loop.
4. Apply Kirchhoff's first law to each branch point.
For loop 1 of the circuit shown in Figure 29.7 Kirchhoff's law dictates that
(29.14)
For loop 2 we obtain
(29.15)
For loop 3 we obtain
(29.16)
Apply Kirchhoff's first law to the three branch points of the circuit shown in
Figure 29.7:
(29.17)
(29.18)
(29.19)
This procedure produces six equations with 5 unknown and the system is
over-defined.
The current I
1 can be obtained from eq.(29.14):
(29.20)
Equation (29.15) can be used to determine I
4:
(29.21)
The current I
3 can be obtained from eq.(), using the solution for
I
4 just derived (eq.(29.21)):
(29.22)
The current I
2 can be obtained from eq.(29.17):
(29.23)
The current I
5 can be obtained from eq.(29.19)
(29.24)
Note: since the loop method involves fewer unknown and fewer
equations, it is usually quicker to use than the branch method.
Example: Problem 29.10
Consider the circuit shown in Figure 29.8. Given that
[epsilon]
1 = 6.0 V, [epsilon]
2 = 10.0 V, and
R
1 = 2.0 [Omega], what must be the value of the resistance
R
2 if the current through this resistance is to be 2.0 A ?
Figure 29.8. Problem 29.10.
Consider the two loops shown in Figure 29.8. We will assume that the current
in each loop flows in a counter clockwise direction (as shown in Figure 29.8).
The voltage drop across resistor R
1 is equal to R
1
. (I
1 - I
2). The sum of all emfs and all
potential drops across the resistors in loop 1 is
(29.25)
The sum of all emfs and all potential drops across the resistors in loop 2 is
equal to
(29.26)
Equation (29.26) can be rewritten as
(29.27)
where we have used eq.(29.25). The current I
2 is thus equal to
(29.28)
The problems states that [epsilon]
1 = 6.0 V, [epsilon]
2 =
10.0 V and I
2 = 2.0 A. These values combined with eq.(29.28) can be
used to determine R
2:
(29.29)
29.4. Energy in circuits
To keep a current flowing in a circuit, work must be done on the circulating
charges. If a charge dq passes through a battery with emf [epsilon], the work
done dW will be equal to
(29.30)
The rate at which the emf source does work is given by
(29.31)
The rate of work is called the power P, and the unit of power is the Watt (W, 1
W = 1 VA).
The moving charges dissipate some of their energy when passing through
resistors. Suppose the potential drop across a resistor is [Delta]V. For an
electron moving through the resistor the loss of potential energy is equal to e
. [Delta]V. The energy lost is converted into heat, and the rate at
which energy is dissipated is equal to
(29.32)
or
(29.33)
the conversion of electric energy into thermal energy is called
Joule
heating.
Example: Example 8
A high-voltage transmission line that connects a city to a power plant
consists of a pair of copper cables, each with a resistance of 4 [Omega]. The
current flows to the city along one cable, and back along the other.
a) The transmission line delivers to the city 1.7 x 10
5 kW of power
at 2.3 x 10
5 V. What is the current in the transmission line ? How
much power is lost as Joule heat in the transmission line ?
b) If the transmission line delivers the same 1.7 x 10
5 kW of power
at 110 V, how much power would be lost in Joule heat ? Is it more efficient to
transmit power at high voltage or at low voltage ?
a) The power delivered to the city, P
delivered, is equal to 1.7 x
10
5 kW. The voltage delivered, [Delta]V
delivered, is
equal to 2.3 x 10
5 V. The current through the cables can determined
by applying eq.(29.32):
(29.34)
This is also the current flowing through the transmission cables. The electric
energy dissipated in the cables is equal to
(29.35)
The power generated by the power plant must therefore be equal to
(29.36)
Comparing the power generated with the power delivered, we conclude that 98% of
the generated power is delivered to the city.
b) If the voltage delivered to the city is 110 V, than current through the
transmission cables must be equal to
(29.37)
This current is roughly 2000 times the current flowing through the transmission
cables if the power is delivered at high voltage. The power dissipated in the
transmission cables is
(29.38)
The power generated by the power plant must therefore be equal to
(29.39)
Comparing the power generated with the power delivered, we conclude that only
0.0009% of the generated power is delivered to the city. Clearly, our
conclusion should be that the transmission of electric energy at high voltage
is much more efficient that the transmission at low voltage. The voltage of
the transmission line is therefore reduced to 110 V as close as possible to the
house of the customer.
Example: Problem 29.24.
A 12-V battery of internal resistance R
i = 0.20 [Omega] is
being charged by an external source of emf delivering 6.0 A.
a) What must be the minimum emf of the external source ?
b) What is the rate at which heat is developed in the internal resistance of
the battery ?
a) The circuit describing this problem is shown in Figure 29.9. In Figure 29.9
[epsilon]
1 is the emf of the battery being recharged, and
[epsilon]
2 is the emf of the recharger. Applying Kirchhoff's second
rule to the single loop circuit we obtain the following relation between the
charging current and the emfs:
(29.40)
Equation (29.40) can be rewritten as
(29.41)
The emf of the battery to be recharged is [epsilon]
1 = 12.0 V and it
has an internal resistance R
i = 0.20 [Omega]. If the recharger is
to deliver a current I = 6.0 A then the emf of the recharger must be equal to
(29.42)
Figure 29.9. Problem 29.24.
b) The power dissipation in the internal resistance of the
battery can be calculated by using eq.(29.33)
(29.43)
Example: Problem 29.32
Suppose that a 12-V battery has an internal resistance of 0.40
[Omega].
a) If this battery delivers a steady current of 1.0 A into an external circuit
until it is completely discharged, what fraction of the initial stored energy
is wasted in the internal resistance ?
b) What if the battery delivers a current of 10.0 A ? Is it more efficient to
use the battery at low current or at high current ?
a) Suppose the battery as an emf [epsilon]
b and delivers a current
I. The internal resistance of the battery is equal to R
i. The
voltage drop across the internal resistance is equal to I R
i. The
external voltage of the battery if thus equal to ([epsilon]
b - I
R
i). The power delivered by the battery to the external circuit is
therefore equal to
(29.44)
The total power delivered by the emf is equal to
(29.45)
The fraction of the total delivered power that is dissipated in the internal
resistance of the battery is equal to
(29.46)
The ratio in eq.(29.46) is proportional to the current I. Using the values of
the parameters specified in the problem we can calculate the ratio:
(29.47)
29.5. The RC circuit
The current through the circuits discussed so far have been time-independent,
as long as the emf of the source is time-independent. The currents in these
circuits can be determined by applying Kirchhoff's first and/or second rule.
A simple circuit in which the current is time dependent is the RC circuit
which consists of a resistor R connected in series with a capacitor C (see
Figure 29.10). Applying Kirchhoff's second rule to the current loop I gives
(29.48)
However, the current I through the resistor and the charge Q on the capacitor
are related:
(29.49)
Substituting eq.(29.49) into eq.(29.48) we obtain
(29.50)
Figure 29.10. Simple RC circuit.
Equation (29.50) is a simple differential equation which can be
solved for Q. The first step is to rewrite eq.(29.50) in the following
manner:
(29.51)
The second step is to integrate each side of eq.(29.51):
(29.52)
where Q
0 is the charge on the capacitor at time t = 0. After
evaluating both integrals in eq.(29.52) we obtain
(29.53)
Equation (29.53) can be rewritten as
(29.54)
or
(29.55)
Let us consider the case in which the capacitor is discharged at time t = 0
(that is Q
0 = 0 C). Equation (29.55) can then be rewritten as
(29.56)
The charge on the capacitor will increase as function of time and the final
charge Q
f on the capacitor is equal to
(29.57)
The time constant of the charging process of the capacitor is equal to RC.
When the charged capacitor is connected in series across a resistor, the charge
on the capacitor will decrease. The time constant for this process is also RC.
Once we know the charge on the capacitor as function of time we can immediately
find the current as function of time by applying eq.(29.48):
(29.58)
The current at time t = 0 is equal to [epsilon]/R and it decreases to zero with
increasing time.
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