Download Combinations of resistors

The 4th of FEBRUARY
Teacher:
Altynbekov Ayan
• A simple circuit.
• Combination of resistors.
• The source of electromotive force.
• Kirchhoff’s rules.
• Circuits with back electromotive
force.
• Efficiency of a circuit element.
• Measurement of current and voltage
A simple circuit contains a source of electrical
energy, an energy consuming unit (a resistor)
and wires that connects the circuit elements to
each other.
Electric circuits usually contain combinations of
resistors in many configurations. This is required in
order to obtain specific currents or to maintain a
required potential difference across a circuit element.
In this section we will examine combinations of
resistors. A single resistor can always be found which
would replace a combination of resistors. The value
of this single resistor is called the equivalent
resistance of the combination.
The electrons leave the negative terminal and
pass through resistors R1 and R 2 to travel to
the positive terminal of the battery. This kind of
combination is called as series combination of
resistors.
In series combinations, we have the following
characteristics in terms of potential difference,
current and resistance relationship:
1) Since the electrons must follow a single
path, the current passing through each resistor
is the same.
I1 = I2 = IT
2) The total current flowing through the circuit
is
UT
IT =
R eq
3) The total energy supplied by the battery to
each charge is dissipated (used) by the resistors
R1 and R 2 .
Total energy=q ∙ UT = q1 ∙ U1 + q2 ∙ U2
So, the total potential difference across all the
resistors is equal to the sum of the potential
differences across each resistor.
UT = U1 + U2
4) The equivalent resistance of the resistors, R1
and R 2 can be found, since UT = R eq IT and
UT = U1 + U2
R eq IT = R1 I1 + R 2 I2
R eq = R1 + R 2
Three resistors, 2Ω, 3Ω and 5Ω are connected in
series to the terminals of a 20 V battery, as
shown in the figure.
Determine:
a) The equivalent resistance of the circuit.
b) The total current.
c) The current flowing through each resistor.
d) The potential difference across each resistor.
Solution
a) Since the resistors are connected in series,
R eq = R1 + R 2 + R 3
R eq = 2Ω + 3Ω + 5Ω
R eq = 10Ω
b) If we apply Ohm’s law to the whole circuit,
UT
20 V
IT =
=
R eq 10Ω
IT = 2A
c) Since current passing through each resistor is
equal to
I1 = I2 = I3 = IT = 2A
d) If we apply Ohm’s law for the resistor R1
U1 = I1 R1
U1 = 2A 2Ω = 4V
For the resistor R 2
U 2 = I2 R 2
U2 = 2A 3Ω = 6V
For the resistor R 3
U 3 = I3 R 3
U3 = 2A 5Ω = 10V
Remember that total potential difference is equal
to the sum of the potential differences across
each resistor. To check the result,
UT = U1 + U2 + U3 = 4V + 6V + 10V
UT = 20V
When resistors are connected to the same two
points or to any two circuit points of two
different wires, with negligible resistance, they
are said to be connected in parallel.
Resistors R1 and R 2 are connected to the same
points, A and B. Electrons from the negative
side will arrive at point B, then separate into
different branches.
If R1 is smaller than R 2 , a greater number of
electrons will flow through R1 as it has less
opposition to flow of charge than R 2 . Amount
of charge passing through a resistor is
proportional to the amount of current.
The characteristics of parallel combinations
can be listed as follows in terms of U, I and R.
- Since resistors are connected to the same
points or potential levels, they will have the
same potential difference.
𝐔𝐓 = 𝐔𝟏 = 𝐔𝟐
- When current flows through a resistor, the
number of electrons does not change along
the resistor, but the energy they carry from
the electric field of the battery is dissipated.
Thus, the current in the main circuit should
be equal to the sum of the currents through
the branches.
𝐈𝐓 = 𝐈𝟏 + 𝐈𝟐
- The equivalent resistance of a parallel
circuit can be found, since
UT = U1 = U2 and IT = I1 + I2
UT
U1 U2
=
+
R eq R1 R 2
𝟏
𝟏
𝟏
=
+
𝐑 𝐞𝐪 𝐑 𝟏 𝐑 𝟐
Consequently in parallel circuits with n
resistors
IT = I1 + I2 +. . +I3
UT = U1 = U2 =. . = U3
1
1
1
1
=
+
+. . +
R eq R1 R 2
R3
Three resistors, 2Ω, 3Ω and 6Ω, are connected
in parallel to a 15 V source, as shown in figure.
Calculate:
a) The equivalent resistance between points K
and L.
b) The total current.
c) The potential difference across each
resistor.
d) The currents I1 , I2 and I3 .
Solution
a) The equivalent resistance of resistors R1 and
R 2 is,
1
1
1
1
=
+
+
R eq R1 R 2 R 3
1
1
1
1
=
+
+
R eq 2Ω 3Ω 6Ω
R eq = 1Ω
b) If we apply Ohm’s law to the whole circuit,
UT
IT =
R eq
12V
IT =
1Ω
IT = 12 A
c) The potential difference across each resistor is
equal to total potential difference between points
K and L.
U1 = U2 = U3 = UT = 12V
d) If we apply Ohm’s law to each arm of the
circuit we obtain,
U1
I1 =
R1
12V
I1 =
2Ω
I1 = 6 A
d) If we apply Ohm’s law to each arm of the
circuit we obtain,
U2
I2 =
R2
12V
I2 =
3Ω
I2 = 4A
d) If we apply Ohm’s law to each arm of the
circuit we obtain,
U3
I3 =
R3
12V
I3 =
6Ω
I3 = 2A
If we check the total current passing through the
circuit, it equals the sum of the currents flowing
through each resistor
IT = I1 + I2 + I3
IT = 6A + 4A + 2A
IT = 12A
A device such as a battery or generator that
converts chemical or mechanical energy into
electrical energy is called a source of
electromotive force, symbolized by ε.
Electromotive force (emf) is not a force, it is
the work done on each charge by a battery to
move them around the circuit. Emf is energy
produced per unit charge by the source.
work
emf =
charge
W
ε=
q
The SI unit of emf is
Joule
Coulomb
or the Volt
From the given equation we can write W = qε
since q = It
𝐖 = 𝛆𝐈𝐭
Emf not only drives the current through the external
resistor but it also drives the current through the
battery itself. The atoms inside a power supply are
resistant to the motion of electrons inside the power
supply. This resistance is called internal resistance of a
battery and is denoted by r.
The internal resistance is indicated in a circuit
diagram as shown in figure.
We can represent the relationship between
potential difference across the terminals of
battery U, and emf ε, as follows: since energy is
conserved, energy produced by the battery is
completely converted into heat.
𝐖𝐞𝐥𝐞𝐜𝐭𝐫𝐢𝐜𝐚𝐥 = 𝐖𝐡𝐞𝐚𝐭
Welectrical = Wheat
2
εIt = I R + r t
ε=I R+r
For the external circuit U = IR, thus
ε − Ir = U
Note; if Ir is negligibly small, then
𝛆≅𝐔
The potential difference across the terminals is 15
V when the battery is not connected to a circuit.
The internal resistance of the battery is 1Ω.
a) Find the terminal voltage of the battery, when
an external resistance of 2Ω is connected.
b) Calculate the power dissipated by the external
resistor, and power dissipated by the internal
resistor.
Solution
a) To find current
ε
I=
R+r
15V
I=
2Ω + 1Ω
I = 5A
To find terminal voltage
U = ε − Ir
U = 15V − 5A 1Ω
U = 10V
b) Power dissipated by external resistor
PR = I 2 R = 5A 2 2Ω
PR = 50W
Power dissipated by internal resistor
Pr = I 2 r = 5A
2
Pr = 25W
1Ω
There are circuits that cannot be reduced to
simple series and parallel circuits, thus Ohm’s
Law cannot be used, one such circuit is shown
in figure.
Gustav Robert Kirchhoff (1824-1887) produced
a set of rules which offer a general method to
solve circuit problems.
The sum of the currents flowing into a junction
(any point that joins three or more conductors
in a circuit) is equal to the sum of the currents
leaving that junction.
𝐈 = 𝐈𝟏 + 𝐈𝟐
𝐈 = 𝐈𝟏 + 𝐈𝟐
In figure the points A and B are junctions. The
currents entering either junction, must be equal to the
currents leaving the junction. This rule comes from
the law of conservation of electric charge. The latter
states that electric charge entering the junction cannot
be created or destroyed.
The second rule is applied to any closed path
made by the conductors of the circuits. The
closed path is called a loop. Kirchhoff’s second
rule states: The algebraic sum of the emf’s and
the potential differences around a closed loop is
zero.
𝛆+
𝐈𝐑 = 𝟎
In applying Kirchhoff’s rules, the following
rules should be noted:
1. Select a direction for current and show it on
each branch of the circuit. The direction can be
chosen arbitrarily. If the current is in the
opposite direction, the result will have a minus
sign in the solution.
In applying Kirchhoff’s rules, the following
rules should be noted:
2. Apply the first rule paying attention to the
directions of currents.
3. For each loop draw a circular arrow
indicating the direction which describes the
path you will follow while applying the second
rule.
4. Apply the second rule.
a) for a resistor the sign of potential drop (IR)
- is negative if the direction of the loop and
direction of the current are the same.
4. Apply the second rule.
b) for a resistor the sign of potential drop (IR)
- is positive if the direction of the loop and
direction of the current are opposite.
4. Apply the second rule.
a) for a battery the sign of potential drop;
- is positive if the direction of the loop is
moving from the negative to the positive
terminal of the battery.
4. Apply the second rule.
b) for a battery the sign of potential drop;
- is negative if the direction of the loop is
moving from the positive to the negative
terminal of the battery.
A resistor of 4Ω is connected to a series combination
of two batteries, 8 V and 4 V.
Calculate:
a) The current I.
b) The potential difference Uba .
c) The potential difference Uba , when switch S is
open.
Solution:
a) Using Kirchhoff’s second rule
ε+
IR = 0
Starting from point ‘a’
ε2 + ε1 + −IR − Ir − Ir = 0
ε2 + ε1
12V
I=
=
= 2A
R+r+r
6Ω
b) The potential difference Uba .
Uba = Va − Vb =?
Vb + ε2 − Ir + ε1 − Ir = Va
ε2 + ε1 − 2Ir = Va − Vb
12V − 2A ∙ 2Ω = Va − Vb
Uba = 8V
c) The potential difference Uba , when switch S is
open.
Uba = Va − Vb =?
Vb + ε2 + ε1 = Va
ε2 + ε1 = Va − Vb
Uba = 12V
Back emf of a motor is the energy per unit charge
converted into mechanical energy by the motor.
Since the motor uses electrical energy, electric
potential decreases across the motor in the
direction of the current.
If an electric motor is connected to a circuit, it
produces mechanical energy, as well as thermal
energy.
𝐄𝐦 =
/
𝛆 𝐈𝐭
𝐄𝐭 =
𝟐
/
𝐈 𝐫 𝐭
/
𝐄𝐦 = 𝛆 𝐈𝐭
/
𝟐 /
𝐄𝐭 = 𝐈 𝐫 𝐭
ε − the back emf of the motor
/
r − the internal resistance of the motor
Electrical devices use electrical energy, this
energy is converted not only to the desired type
of energy but also to some other types of
energy.
- an electric motor converts electrical energy
into mechanical energy but also some energy is
converted into heat.
- a battery produces an emf, this emf is not
given directly to the external resistor, because
some potential is dropped by the internal
resistor.
As a result efficiency is defined as the ratio of
useful energy converted by the device Ec , to
the whole energy spent by the device Es .
Efficiency is denoted by the letter η , that is
𝐮𝐬𝐞𝐟𝐮𝐥 𝐞𝐧𝐞𝐫𝐠𝐲 𝐜𝐨𝐧𝐯𝐞𝐫𝐭𝐞𝐝
𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐜𝐲 =
𝐰𝐡𝐨𝐥𝐞 𝐞𝐧𝐞𝐫𝐠𝐲 𝐬𝐩𝐞𝐧𝐭
𝐄𝐜
𝛈=
𝐄𝐬
An ammeter is a device used to measure
electric current and is shown in figure. An
ammeter is connected in series with a circuit.
A voltmeter measures potential drop across any
circuit element, or the potential difference
between any two points of a circuit.
A voltmeter is connected in parallel to a circuit
element.
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Electricity and Magnetism, Zambak publishing,
Ahmet Aki, Salim Gur