00971504825082
Current and Resistance
Electric Current
Electric Current
:It is the flow of electric charges in an electrical conductor. The electric charge can be either electrons or ions
According to the International System of Units, electric current is measured in amperes
While electric current is measured with a device called an ammeter
In this simulation, close the circuit and observe the movement of electric charges and how they lose energy in the resistor
In this simulation, the movement of electrons in a simple circuit from the lowest to the highest voltage is represented. Note that the term "electric current" means the movement of positive charges from the higher to the lower voltage
Current Density
It is the current flowing per unit area in a conducting wire at a specific point. Current density is denoted by the symbol
j
It is a vector quantity whose direction is the same as the movement of positive charges (opposite to the movement of electrons)
\[j=\frac{i}{A}\] and is measured in units of \[\frac{A}{m^2}\]
Drift velocity is the directed velocity of electrons in an electrical circuit.
Any conducting wire contains electrons moving randomly at high speed. When a potential difference is applied across the wire, the electrons' movement becomes a directed random velocity called drift velocity, which is very slow, on the order of
\[𝜗_d =10^{-4}\frac {m}{s}\]
We have a conductor with cross-sectional area (A)
(𝜗𝑑) and an electric field is applied to it. The electrons move opposite to the field with a drift velocity
(d t) and during a time interval of
it travels a distance of
\[𝜗𝑑 . dt\] and thus the volume of electrons passing through the cross-section equals
\[A. 𝜗𝑑 . dt\] so the number of electrons in this volume equals
\[n . A. 𝜗𝑑 . dt\]
(-e) and each electron is charged with a charge of
Therefore, the charge flowing through this area
\[dq = - e. n . A. 𝜗𝑑 . dt\]
Then we get the current intensity
\[i =\frac{ dq }{ dt }= - e. n . A. 𝜗𝑑\]
And current density
\[j = \frac{i}{A} = - e. n . 𝜗𝑑\]
\[1 \star \]
A wire with cross-sectional area \[A=5.2×10^{-6}m^2\] carries a current of
\[i=2\;A\] and is made of copper with molar mass and density \[M=63.5\;\; g\;\;\;\;\;\;.𝜌=8960 \frac{Kg}{m^3}\]
There is one free electron per atom. The drift velocity of the electrons equals
Given that \[ q=1.6 ×10^{-19}c\;\;\;\;\;\;Avogadro's\;\; number= 6.022 ×10^{23}\]
\[𝑣_𝑑= 5.6×𝟏𝟎^{−4}\;\; m/s\;\;\;\;\;\;-C\]
\[𝑣_𝑑= 6×𝟏𝟎^{−5}\;\; m/s\;\;\;\;\;\;-A\]
\[𝑣_𝑑= 4.3×𝟏𝟎^{−5}\;\; m/s\;\;\;\;\;\;-D\]
\[𝑣_𝑑= 2.8×𝟏𝟎^{−4}\;\; m/s\;\;\;\;\;\;-B\]
Click here to show solution
Choose the correct answer
Ohm's Law
In this simulation, Ohm studied the relationship between current intensity and voltage across an ohmic resistor. Choose a value for an ohmic resistance and change the voltage each time. Observe what happens to the current intensity. Repeat the experiment with another resistance
Ohm's Law Experiment
Ohm's Law Experiment - Complete the following table
Experiment No.
Voltage (Volt)
Current (Ampere)
Resistance (Ohm)
1
2
3
4
5
Ohm's Law
Ohm's Law is a fundamental principle in electricity, named after the German physicist "Georg Simon Ohm".
It states that the potential difference across a metallic conductor is directly proportional to the current flowing through it.
The constant ratio between voltage and current is defined as electrical resistance. It is noted that the resistance of a conductor is a constant value and does not change with the change in potential difference across its ends.
The equation can be expressed as follows:
\[ V = I . R \]
Test yourself
Resistivity and Resistance
Electrical resistance: It is the extent to which a material opposes the flow of electric current
When a potential difference is applied across a wire
\[∆V \] and a current of intensity \[i\] passes through it,
the wire's opposition to the current is given by the relation according to Ohm's law \[R=\frac {∆V}{i}\]
The unit of electrical resistance is ohm, which equals \[𝝮=\frac{V}{A}\]
Some devices are described in terms of their ability to conduct rather than their ability to oppose current
\[G\] which is measured in Siemens \[G=\frac {i}{∆V}=\frac {1}{R}\;\;\;\;\;\;\;\;\;\;\;\;\;\;S=\frac {A}{V}=\frac {1}{𝝮}\]
The resistance of any wire to current flow depends on the material it is made of, its geometric shape, and temperature.
Each wire has a specific resistance called resistivity \[𝜌\], which is the ratio between the electric field intensity and current density
\[𝜌=\frac {E}{J}\;\;\;\;\;\;\;\;\;\;\;𝜌=\frac {\frac {V}{m}}{\frac {A}{m^2}}=\frac {V.m}{A}=𝝮.m\]
Resistivity and Temperature Coefficient of Resistivity for Some Conductors
\[∝\] Temperature coefficient at temperature
\[20^0c\]\[×10^{-3}\;\;K^{-1}\]
\[ 𝜌\] Resistivity at temperature
\[20^0c\]\[×10^{-8}\;\;𝝮.m\]
Material Name
3.8
1.6
Silver
3.9
1.72
Copper
3.4
2.44
Gold
3.9
2.82
Aluminum
2
3.9
Brass
4.5
5.51
Tungsten
5
9.7
Iron
We know that
\[E=\frac{∆V}{L}\;\;\;\;\;\;\;\;\;J=\frac{i}{A}\]
\[𝜌=\frac {E}{J}=\frac {\frac{∆V}{L}}{\frac{i}{A}}=\frac{∆V.A}{i.L}=\frac{iR.A}{i.L}=\frac{R.A}{L}\]
\[R= 𝜌.\frac{A}{L}\]
Factors Affecting Ohmic Resistance
In this simulation, we observe that the resistance value changes with changes in resistor length, cross-sectional area, and the material the resistor is made of
\[2 \star \]
Four wires of the same type and at the same temperature
One of the following wires has the least resistance
Wire 3-C
Wire 1-A
Wire 4-D
Wire 2-B
Click here to show solution
Choose the correct answer
\[3 \star \]
Two wires of the same material and at the same temperature if
\[\frac{R_1}{R_2}= \frac{2}{3}\] then one of the following answers satisfies this
\[L_1=\frac{1}{2}L_2 , A_1=\frac{1}{3} A_2\;\;\;\;\;\;-C\]
\[L_1=\frac{3}{4}L_2 , A_1=\frac{1}{2} A_2\;\;\;\;\;\;-A\]
\[L_1=3L_2 , A_1=2 A_2\;\;\;\;\;\;-D\]
\[L_1=\frac{4}{3}L_2 , A_1=2 A_2\;\;\;\;\;\;-B\]
Click here to show solution
Choose the correct answer
There is another factor which is temperature. We notice the effect of temperature increase on resistivity.
\[ρ_T= ρ_0[1 +𝛼(T–T_0)]\]\[R_T= R_0[1 + 𝛼(T–T_0)]\]
\[4 \star \]
A copper wire with resistivity
\[1.7 \times 10^{-8} \;Ω.m\]
at 20°C. The temperature was increased by 15°C and its resistivity became
\[1.82 \times 10^{-8} \;Ω.m\]
The temperature coefficient equals
\[𝛼=5.6 × 10^{-3}\;K^{-1}\;\;\;\;\;\;-C\]
\[𝛼=21.7 × 10^{-3}\;K^{-1}\;\;\;\;\;\;-A\]
\[𝛼=4.7 × 10^{-3}\;K^{-1}\;\;\;\;\;\;-D\]
\[𝛼=2.6 × 10^{-3}\;K^{-1}\;\;\;\;\;\;-B\]
Click here to show solution
Choose the correct answer
Resistor Color Codes
:(Four bands) Reading resistor values through color codes
A carbon resistor may have 4 to 6 bands where the first band represents tens, the second band represents units of the resistance value, the third band represents the power raised to 10, and the fourth band represents the tolerance in reading
Electromotive Force
Battery: It does work on electric charges in the circuit and gives them electrical energy and pushes them in the circuit, so it's called electromotive force
When connecting a battery to a circuit containing conducting wires, an electric field is formed that gives energy to the outgoing charges that collide with the charges in the conducting wires and directs their movement
and pushes them in the circuit parts
The battery exerts a force on the charges
\[( f)\] and gives them electrical energy \[( e )\] and pushes and moves them in the circuit
\[( m )\]
Therefore, the potential difference produced by the battery is denoted by
v(emf)
Example 1) The battery provides a constant potential difference while it doesn't provide a constant current. What is the reason?
Click here to show solution
2500 mAh
What is the amount of charge it provides?
Click here to show solution
A student measured the potential difference across a battery terminals
Once with the circuit open and another time with the circuit closed
The measurements were as shown in the figure
To determine the reason, the following experiment was conducted. There are two circuits, one is closed and the other is open
Compare the voltmeter readings for both circuits. In the closed circuit, there was a loss in the source voltage. The reason for this is that there is an internal resistance inside the battery, denoted by the symbol
r
and measured in units of
Ohm
Calculate the value of the battery's internal resistance from the lost voltage and the current in the circuit
\[ r = \frac{V_r}{I}\]
When measuring the potential difference across the battery terminals with the circuit open, it's called electromotive force
The internal resistance value doesn't appear unless current passes through it
While when measuring with the circuit closed, it's called potential difference and is always less than the electromotive force
If the internal resistance is neglected, then the electromotive force equals
the potential difference
Example 3) In the figure, the external resistance is 10 ohms and the battery's internal resistance is 2 ohms. When the circuit is closed, determine the following values:
The voltage across the external resistance of 10 ohms equals
\[V_R=....................\]
The voltage across the internal resistance of 2 ohms equals
\[V_r=....................\]
The battery's electromotive force equals
\[V_{emf}=....................\]
Click here to show solution
\[5 \star \star \]
A circuit connected to a resistor
\[4 Ω\] and a switch. A voltmeter is connected across the circuit and reads when the switch is open
\[12V\] and when the switch is closed
\[10V\]
The value of the battery's internal resistance equals
\[ 𝑟= 0.8\;\; Ω \;\;\;\;\;\;-C\]
\[𝑟= 0.5\;\; Ω \;\;\;\;\;\;-A\]
\[𝑟= 1.2\;\; Ω \;\;\;\;\;\;-D\]
\[𝑟= 1\;\; Ω \;\;\;\;\;\;-B\]
Click here to show solution
Choose the correct answer
Series Connection of Resistors
Series connection: The electric current flows in one path and passes through all circuit components
Characteristics of series connection:
Important Results
Example 1)Three resistors in series \[ R_1 = 10 Ω, R_2 = 5 Ω, R_3 = ? \]
Connected to a battery with voltage \[V= 20 V\] and current \[I=0.5 A\]
Then the value of resistor \[ R_3 =....\]
Click here to show solution
Click here to show solution
Parallel Connection of Resistors
Parallel connection: When electric current flows through more than one path and passes through all circuit components
In parallel connection, the current is distributed through circuit components according to their resistance
The source voltage equals the voltage across each parallel branch
In this simulation, three resistors are connected in parallel. Measure the total current and the current through each resistor, and also measure the total voltage and the voltage across each resistor. Record the results in the table below the experiment
Complete the table data below
\[R_{eq}=?Ω\]
\[R_{3}=-- Ω\]
\[R_{2}=-- Ω\]
\[R_{1}=-- Ω\]
\[I_{tot}=....A\]
\[I_{3}=....A\]
\[I_{2}=....A\]
\[I_{1}=....A\]
\[V_{tot}=....V\]
\[V_{3}=....V\]
\[V_{2}=....V\]
\[V_{1}=....V\]
\[R_{eq} = \frac{V_{tot}}{I_{tot}}\]\[R_{eq} = ......\]
\[R_{eq} = (\frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3})^{-1}\]\[R_{eq} = ......\]
Compare the results you obtained. What do you conclude?\[............\]
Observations\[........................\]
Play and evaluate yourself
Parallel Connection: Electric current flows through more than one path to complete its circuit
Features of Parallel Connection:
Important Results
In this simulation, a circuit measures current and voltage across each resistor using ammeter and voltmeter
You can change resistor values using the icons below each resistor
Example 3) Three resistors in parallel \[ R_1 = 3 Ω, R_2 = 9 Ω, R_3 = ? \]
Connected to a battery with voltage \[V= 10 V\] with total current \[I= 5 A\]
Then the value of resistor \[ R3 =.....\]
Click here to show solution
Example 4
In the adjacent circuit, the ammeter reading for total current equals
Click here to show solution
Example 5
Two resistors, the first
\[R_1=50 Ω\] and the second with unknown resistance were connected and the equivalent resistance was \[R_{eq}=80 Ω\] The connection was (series - parallel)
Click here to show solution
Example 6
Two equal resistors were connected and it was observed that the voltage across the first equals the voltage across the second and equals the sum of voltages across both resistors matching the source voltage, then the connection was
(series - parallel)
Click here to show solution
Example 7
Two unequal resistors were connected and it was observed that the voltage across the first equals the voltage across the second, then the connection was
(series - parallel)
Click here to show solution
Compound Electrical Circuits
Steps to calculate equivalent resistance of a circuit
Identify which resistors are in series and which are in parallel
Example 1) A set of resistors connected as shown with a battery having internal resistance, find the equivalent resistance
Click here to show solution
In this simulation, a set of resistors connected in series and parallel where you can control their values and display current and voltage across each resistor
Example 2) A set of resistors connected as shown, then the current passing through resistor R1 = ?
Click here to show solution
Example 3) A set of resistors connected as shown, then the voltage across resistor R4 = ?
Click here to show solution
Energy and Power
When applying voltage in a circuit, the electromotive force of the battery does work on the charges and gives them electrical energy
\[ dU = dq . ∆𝑉 \]
And we know that
\[dq = I . dt\]
\[dU = I . dt .∆𝑉 \]
\[\frac{𝑑𝑈}{dt}\] which is the rate of energy consumption called power and denoted by
\[P\]
\[P = I . ∆𝑉\]
Measured in watts in the SI system
According to Ohm's law \[R = \frac{∆𝑉}{I}\] power can be written as \[𝑃= I^2. 𝑅\]\[𝑃= \frac{∆𝑉^2}{𝑅}\]
In this simulation, a circuit contains an ohmic resistor that consumes the battery's energy
Power dissipated in a series resistor circuit
Power dissipated in a parallel resistor circuit
Power dissipated in a series-parallel resistor circuit
\[1 \star \star \]
The simplified circuit consists of a 9-volt battery with no internal resistance and three resistors as shown in the figure below.
The total power of the circuit equals
\[P=25.45\;\;W \;\;\;\;\;\;-C\]
\[P=20.25\;\;W \;\;\;\;\;\;-A\]
\[ P=17.85W\;\;W \;\;\;\;\;\;-D\]
\[P=13.75\;\;W \;\;\;\;\;\;-B\]
Click here to show solution
Choose the correct answer
Bulb Brightness
When we talk about bulb brightness, we need to consider
the power - the higher the bulb's power,
the brighter it will be
\[P = V.I = I^2.R =\frac{V^2}{R}\]
\[P=\frac{W}{t}\]
This simulation shows several bulbs connected together - compare their brightness
This simulation shows a compound circuit with a power source and four identical bulbs and three switches. Open and close the switches and make predictions about the voltages across the bulbs, currents through the bulbs, and bulb brightness (which is related to the power each dissipates as heat and light). Use checkboxes to show or hide voltages and currents.
\[2 \star \]
In the adjacent circuit, 4 bulbs are connected as shown in the figure
One of the following options would result in a decrease in the ammeter reading in the circuit
( C ) Connect a wire across bulb -C
( A ) Connect a wire across bulb -A
( D ) Connect a wire across bulb -D
( B ) Bulb -B filament burns out
Click here to show solution
Choose the correct answer
\[3 \star \]
In the adjacent figure, 4 identical bulbs are connected to a battery
One of the following bulbs will be brightest
Click here to show solution
\[4 \star\]
In the figure below, the power dissipated in the circuit equals
\[P=69\;\; W\;\;\;\;\;\;-C\]
\[P=54\;\; W\;\;\;\;\;\;-A\]
\[P=88\;\; W\;\;\;\;\;\;-D\]
\[P=97\;\; W\;\;\;\;\;\;-B\]
Click here to show the solution
Current and Resistance |
Electric Current
Electric Current
:It is the flow of electric charges in an electrical conductor. The electric charge can be either electrons or ions
According to the International System of Units, electric current is measured in amperes
While electric current is measured with a device called an ammeter
In this simulation, close the circuit and observe the movement of electric charges and how they lose energy in the resistor
In this simulation, the movement of electrons in a simple circuit from the lowest to the highest voltage is represented. Note that the term "electric current" means the movement of positive charges from the higher to the lower voltage
Current Density
It is the current flowing per unit area in a conducting wire at a specific point. Current density is denoted by the symbol
j
It is a vector quantity whose direction is the same as the movement of positive charges (opposite to the movement of electrons)
\[j=\frac{i}{A}\] and is measured in units of \[\frac{A}{m^2}\]
Drift velocity is the directed velocity of electrons in an electrical circuit. Any conducting wire contains electrons moving randomly at high speed. When a potential difference is applied across the wire, the electrons' movement becomes a directed random velocity called drift velocity, which is very slow, on the order of \[𝜗_d =10^{-4}\frac {m}{s}\]
We have a conductor with cross-sectional area (A)
(𝜗𝑑) and an electric field is applied to it. The electrons move opposite to the field with a drift velocity
(d t) and during a time interval of
it travels a distance of \[𝜗𝑑 . dt\] and thus the volume of electrons passing through the cross-section equals \[A. 𝜗𝑑 . dt\] so the number of electrons in this volume equals \[n . A. 𝜗𝑑 . dt\]
(-e) and each electron is charged with a charge of
Therefore, the charge flowing through this area
\[dq = - e. n . A. 𝜗𝑑 . dt\] Then we get the current intensity \[i =\frac{ dq }{ dt }= - e. n . A. 𝜗𝑑\]
And current density \[j = \frac{i}{A} = - e. n . 𝜗𝑑\]
A wire with cross-sectional area \[A=5.2×10^{-6}m^2\] carries a current of \[i=2\;A\] and is made of copper with molar mass and density \[M=63.5\;\; g\;\;\;\;\;\;.𝜌=8960 \frac{Kg}{m^3}\] There is one free electron per atom. The drift velocity of the electrons equals Given that \[ q=1.6 ×10^{-19}c\;\;\;\;\;\;Avogadro's\;\; number= 6.022 ×10^{23}\]
|
\[𝑣_𝑑= 5.6×𝟏𝟎^{−4}\;\; m/s\;\;\;\;\;\;-C\] |
\[𝑣_𝑑= 6×𝟏𝟎^{−5}\;\; m/s\;\;\;\;\;\;-A\] |
|
\[𝑣_𝑑= 4.3×𝟏𝟎^{−5}\;\; m/s\;\;\;\;\;\;-D\] |
\[𝑣_𝑑= 2.8×𝟏𝟎^{−4}\;\; m/s\;\;\;\;\;\;-B\] |
Click here to show solution
Choose the correct answer
Ohm's Law
In this simulation, Ohm studied the relationship between current intensity and voltage across an ohmic resistor. Choose a value for an ohmic resistance and change the voltage each time. Observe what happens to the current intensity. Repeat the experiment with another resistance
Ohm's Law Experiment - Complete the following table
| Experiment No. | Voltage (Volt) | Current (Ampere) | Resistance (Ohm) |
|---|---|---|---|
| 1 | |||
| 2 | |||
| 3 | |||
| 4 | |||
| 5 |
Ohm's Law
Ohm's Law is a fundamental principle in electricity, named after the German physicist "Georg Simon Ohm". It states that the potential difference across a metallic conductor is directly proportional to the current flowing through it. The constant ratio between voltage and current is defined as electrical resistance. It is noted that the resistance of a conductor is a constant value and does not change with the change in potential difference across its ends. The equation can be expressed as follows:
\[ V = I . R \]
Test yourself
Resistivity and Resistance
Electrical resistance: It is the extent to which a material opposes the flow of electric current
When a potential difference is applied across a wire \[∆V \] and a current of intensity \[i\] passes through it, the wire's opposition to the current is given by the relation according to Ohm's law \[R=\frac {∆V}{i}\] The unit of electrical resistance is ohm, which equals \[𝝮=\frac{V}{A}\] Some devices are described in terms of their ability to conduct rather than their ability to oppose current \[G\] which is measured in Siemens \[G=\frac {i}{∆V}=\frac {1}{R}\;\;\;\;\;\;\;\;\;\;\;\;\;\;S=\frac {A}{V}=\frac {1}{𝝮}\] The resistance of any wire to current flow depends on the material it is made of, its geometric shape, and temperature. Each wire has a specific resistance called resistivity \[𝜌\], which is the ratio between the electric field intensity and current density \[𝜌=\frac {E}{J}\;\;\;\;\;\;\;\;\;\;\;𝜌=\frac {\frac {V}{m}}{\frac {A}{m^2}}=\frac {V.m}{A}=𝝮.m\]
Resistivity and Temperature Coefficient of Resistivity for Some Conductors
|
\[∝\] Temperature coefficient at temperature \[20^0c\]\[×10^{-3}\;\;K^{-1}\] |
\[ 𝜌\] Resistivity at temperature \[20^0c\]\[×10^{-8}\;\;𝝮.m\] |
Material Name |
|
3.8 |
1.6 |
Silver |
|
3.9 |
1.72 |
Copper |
|
3.4 |
2.44 |
Gold |
|
3.9 |
2.82 |
Aluminum |
|
2 |
3.9 |
Brass |
|
4.5 |
5.51 |
Tungsten |
|
5 |
9.7 |
Iron |
We know that \[E=\frac{∆V}{L}\;\;\;\;\;\;\;\;\;J=\frac{i}{A}\] \[𝜌=\frac {E}{J}=\frac {\frac{∆V}{L}}{\frac{i}{A}}=\frac{∆V.A}{i.L}=\frac{iR.A}{i.L}=\frac{R.A}{L}\] \[R= 𝜌.\frac{A}{L}\]
Factors Affecting Ohmic Resistance
In this simulation, we observe that the resistance value changes with changes in resistor length, cross-sectional area, and the material the resistor is made of
Four wires of the same type and at the same temperature
One of the following wires has the least resistance
Wire 3-C |
Wire 1-A |
Wire 4-D |
Wire 2-B |
Click here to show solution
Choose the correct answer
Two wires of the same material and at the same temperature if \[\frac{R_1}{R_2}= \frac{2}{3}\] then one of the following answers satisfies this
\[L_1=\frac{1}{2}L_2 , A_1=\frac{1}{3} A_2\;\;\;\;\;\;-C\] |
\[L_1=\frac{3}{4}L_2 , A_1=\frac{1}{2} A_2\;\;\;\;\;\;-A\] |
\[L_1=3L_2 , A_1=2 A_2\;\;\;\;\;\;-D\] |
\[L_1=\frac{4}{3}L_2 , A_1=2 A_2\;\;\;\;\;\;-B\] |
Click here to show solution
Choose the correct answer
A copper wire with resistivity \[1.7 \times 10^{-8} \;Ω.m\] at 20°C. The temperature was increased by 15°C and its resistivity became \[1.82 \times 10^{-8} \;Ω.m\] The temperature coefficient equals
\[𝛼=5.6 × 10^{-3}\;K^{-1}\;\;\;\;\;\;-C\] |
\[𝛼=21.7 × 10^{-3}\;K^{-1}\;\;\;\;\;\;-A\] |
\[𝛼=4.7 × 10^{-3}\;K^{-1}\;\;\;\;\;\;-D\] |
\[𝛼=2.6 × 10^{-3}\;K^{-1}\;\;\;\;\;\;-B\] |
Click here to show solution
Choose the correct answer
Electromotive Force
When connecting a battery to a circuit containing conducting wires, an electric field is formed that gives energy to the outgoing charges that collide with the charges in the conducting wires and directs their movement and pushes them in the circuit parts
The battery exerts a force on the charges \[( f)\] and gives them electrical energy \[( e )\] and pushes and moves them in the circuit \[( m )\] Therefore, the potential difference produced by the battery is denoted by
v(emf)
Example 1) The battery provides a constant potential difference while it doesn't provide a constant current. What is the reason?
Click here to show solution
2500 mAh
What is the amount of charge it provides?
Click here to show solution
A student measured the potential difference across a battery terminals
Once with the circuit open and another time with the circuit closed
The measurements were as shown in the figure
To determine the reason, the following experiment was conducted. There are two circuits, one is closed and the other is open
Compare the voltmeter readings for both circuits. In the closed circuit, there was a loss in the source voltage. The reason for this is that there is an internal resistance inside the battery, denoted by the symbol
r
and measured in units of
Ohm
Calculate the value of the battery's internal resistance from the lost voltage and the current in the circuit
\[ r = \frac{V_r}{I}\]
When measuring the potential difference across the battery terminals with the circuit open, it's called electromotive force
The internal resistance value doesn't appear unless current passes through it
While when measuring with the circuit closed, it's called potential difference and is always less than the electromotive force
If the internal resistance is neglected, then the electromotive force equalsthe potential difference
Example 3) In the figure, the external resistance is 10 ohms and the battery's internal resistance is 2 ohms. When the circuit is closed, determine the following values:
The voltage across the external resistance of 10 ohms equals
\[V_R=....................\] The voltage across the internal resistance of 2 ohms equals \[V_r=....................\] The battery's electromotive force equals \[V_{emf}=....................\]
Click here to show solution
A circuit connected to a resistor
\[4 Ω\] and a switch. A voltmeter is connected across the circuit and reads when the switch is open
\[12V\] and when the switch is closed
\[10V\]
The value of the battery's internal resistance equals
\[ 𝑟= 0.8\;\; Ω \;\;\;\;\;\;-C\] |
\[𝑟= 0.5\;\; Ω \;\;\;\;\;\;-A\] |
\[𝑟= 1.2\;\; Ω \;\;\;\;\;\;-D\] |
\[𝑟= 1\;\; Ω \;\;\;\;\;\;-B\] |
Click here to show solution
Choose the correct answer
Series Connection of Resistors
Series connection: The electric current flows in one path and passes through all circuit components
Characteristics of series connection:
Important Results
Example 1)Three resistors in series \[ R_1 = 10 Ω, R_2 = 5 Ω, R_3 = ? \] Connected to a battery with voltage \[V= 20 V\] and current \[I=0.5 A\] Then the value of resistor \[ R_3 =....\]
Click here to show solution
Click here to show solution
Parallel Connection of Resistors
Parallel connection: When electric current flows through more than one path and passes through all circuit components
In parallel connection, the current is distributed through circuit components according to their resistance
The source voltage equals the voltage across each parallel branch
In this simulation, three resistors are connected in parallel. Measure the total current and the current through each resistor, and also measure the total voltage and the voltage across each resistor. Record the results in the table below the experiment
Complete the table data below
\[R_{eq}=?Ω\] |
\[R_{3}=-- Ω\] |
\[R_{2}=-- Ω\] |
\[R_{1}=-- Ω\] |
\[I_{tot}=....A\] |
\[I_{3}=....A\] |
\[I_{2}=....A\] |
\[I_{1}=....A\] |
\[V_{tot}=....V\] |
\[V_{3}=....V\] |
\[V_{2}=....V\] |
\[V_{1}=....V\] |
\[R_{eq} = \frac{V_{tot}}{I_{tot}}\]\[R_{eq} = ......\] |
\[R_{eq} = (\frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3})^{-1}\]\[R_{eq} = ......\] |
Compare the results you obtained. What do you conclude?\[............\] |
Observations\[........................\] |
Play and evaluate yourself
Parallel Connection: Electric current flows through more than one path to complete its circuit
Features of Parallel Connection:
Important Results
In this simulation, a circuit measures current and voltage across each resistor using ammeter and voltmeter
You can change resistor values using the icons below each resistor
Click here to show solution
Example 4
In the adjacent circuit, the ammeter reading for total current equals
Click here to show solution
Example 5
Two resistors, the first \[R_1=50 Ω\] and the second with unknown resistance were connected and the equivalent resistance was \[R_{eq}=80 Ω\] The connection was (series - parallel)
Click here to show solution
Example 6
Two equal resistors were connected and it was observed that the voltage across the first equals the voltage across the second and equals the sum of voltages across both resistors matching the source voltage, then the connection was
(series - parallel)
Click here to show solution
Example 7
Two unequal resistors were connected and it was observed that the voltage across the first equals the voltage across the second, then the connection was
(series - parallel)
Click here to show solution
Compound Electrical Circuits
Identify which resistors are in series and which are in parallel
Example 1) A set of resistors connected as shown with a battery having internal resistance, find the equivalent resistance
Click here to show solution
In this simulation, a set of resistors connected in series and parallel where you can control their values and display current and voltage across each resistor
Example 2) A set of resistors connected as shown, then the current passing through resistor R1 = ?
Click here to show solution
Example 3) A set of resistors connected as shown, then the voltage across resistor R4 = ?
Click here to show solution
Energy and Power
When applying voltage in a circuit, the electromotive force of the battery does work on the charges and gives them electrical energy
\[ dU = dq . ∆𝑉 \]
And we know that
\[dq = I . dt\]
\[dU = I . dt .∆𝑉 \]
\[\frac{𝑑𝑈}{dt}\] which is the rate of energy consumption called power and denoted by \[P\] \[P = I . ∆𝑉\]
Measured in watts in the SI system
According to Ohm's law \[R = \frac{∆𝑉}{I}\] power can be written as \[𝑃= I^2. 𝑅\]\[𝑃= \frac{∆𝑉^2}{𝑅}\]
In this simulation, a circuit contains an ohmic resistor that consumes the battery's energy
Power dissipated in a series resistor circuit
Power dissipated in a parallel resistor circuit
Power dissipated in a series-parallel resistor circuit
The simplified circuit consists of a 9-volt battery with no internal resistance and three resistors as shown in the figure below.
The total power of the circuit equals
\[P=25.45\;\;W \;\;\;\;\;\;-C\] |
\[P=20.25\;\;W \;\;\;\;\;\;-A\] |
\[ P=17.85W\;\;W \;\;\;\;\;\;-D\] |
\[P=13.75\;\;W \;\;\;\;\;\;-B\] |
Click here to show solution
Choose the correct answer
Bulb Brightness
This simulation shows several bulbs connected together - compare their brightness
This simulation shows a compound circuit with a power source and four identical bulbs and three switches. Open and close the switches and make predictions about the voltages across the bulbs, currents through the bulbs, and bulb brightness (which is related to the power each dissipates as heat and light). Use checkboxes to show or hide voltages and currents.
In the adjacent circuit, 4 bulbs are connected as shown in the figure
One of the following options would result in a decrease in the ammeter reading in the circuit
( C ) Connect a wire across bulb -C |
( A ) Connect a wire across bulb -A |
( D ) Connect a wire across bulb -D |
( B ) Bulb -B filament burns out |
Click here to show solution
Choose the correct answer
In the adjacent figure, 4 identical bulbs are connected to a battery
One of the following bulbs will be brightest
Click here to show solution
In the figure below, the power dissipated in the circuit equals
\[P=69\;\; W\;\;\;\;\;\;-C\] |
\[P=54\;\; W\;\;\;\;\;\;-A\] |
\[P=88\;\; W\;\;\;\;\;\;-D\] |
\[P=97\;\; W\;\;\;\;\;\;-B\] |
Click here to show the solution
0 Comments