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CHAPTER 13: ELECTRICAL QUANTITIES IN
CIRCUITS

A CIRCUIT
DIAGRAM
is a way of drawing an electric circuit using standard symbols.
Symbols are used to represent the different types of components that are linked together in a
circuit.

In a SERIES
CIRCUIT
the flow of electrons follows only one path. This is achieved by
connecting the loads in a chain, one after another in one continuous loop. For example, when you
plug in a toaster and an electric kettle into the same outlet, the toaster and kettle are two loads
on the same circuit. The toaster and kettle are two loads in series. If the fuse for that outlet
blows, neither device will work.

In a PARALLEL
CIRCUIT
electrons can flow more than one way. The loads are on at least
two different branches of wires that connect to an energy source. When the electrons reach a
branch, they separate. Some electrons will follow one branch, while the others will follow the
other branch. These separate paths then merge before returning to the energy source. For
example, newer Christmas lights – if one bulb blows, all the other lightbulbs still are on.
Practice Drawings…
a) Draw a circuit diagram showing an two-cell battery, a open switch, and three lamps connected in
series.
b) Draw a circuit diagram showing a three-cell battery, and three lamps connected in parallel. Show a
closed switch to each lamp.

ELECTRIC CURRENT (I)
is a measure of the rate of electron flow past a given point in a
circuit. It is measured in amperes (A).

An AMMETER is a device used to measure electric current. It must be connected in series
with a load to measure the current flow through the load. This ensures that all the electrons
that flow through the lamp will also flow through the ammeter.

One amperes (A) indicates 6.2 x 1018 electrons are flowing through a circuit per second!
The Human Body and Electrical Shock

The human body uses electrical signals sent through the nervous system. However, even a small
electrical shock through the body can be dangerous.

An electric current of about 0.001A passing through the body can give you a tingling sensation.

A current of 0.050 to 0.150 A can cause muscles to contract or convulse out of control. This is
sometimes called the “let go” threshold because beyond this value, you can no longer let go of the
object that is shocking you.

A current of 1.0 to 4.3 A passing through your chest will stop your heart.

A wall outlet that powers a computer can deliver 15 A.
Potential Difference

POTENTIAL DIFFERENCE
(also known as Voltage, V) is the difference in electric
potential energy per unit of charge measures at two points. The unit for voltage if the volt (V).

There is potential difference between the two terminals of an electric cell. Electrons leave the
negative terminal with electric potential energy that can be used to operate a motor. As a result,
the electrons return to the positive terminal of the cell with less electric potential energy than
they started with, because some of their energy was used to run the motor. Once inside the cell,
chemical reactions “re-energize” the electrons and send them out of the negative terminal again.

When an electrician troubleshoots a circuit, the voltage as well as the current at different parts
of the circuit must be measured. A VOLTMETER is the device designed to measure potential
difference. Unlike an ammeter, a voltmeter must be connected in parallel with a load or an energy
source. The reason for this is that voltage is relative to two points. There is always a drop in
voltage across a load or energy source. For example, to measure the voltage across a lamp,
connect the voltmeter in parallel to the lamp. Since the lamp is the only load in the circuit, the
voltage displayed on the voltmeter will be the same as the voltage across the two terminals of the
battery.
Resistance in Circuits

You might have noticed that when you are recharging your cellphone, that the adaptor gets warm.
This warmth is caused by the resistance experienced by the electric current flowing through the
adaptor.

ELECTRICAL RESISTANCE (R)
is the ability of a material to oppose the flow of electric
current, and is measured in ohms (Ω).

When electrons flow through a material that has many “bumps” along its path, there will be more
resistance than if the material is “smooth”. For example, insulators tend to minimize the amount
of electron flow, so the internal resistance of an insulator is quite high. However, a conductor
has a very low internal resistance. That is why electrons flow so easily through copper wiring (a
conductor).

All materials have some internal resistance. The greater the resistance, the lower the current,
and the warmer the material becomes when current flows through it. This happens because, as
electrons move through the material, they bump into atoms that make up the material. In the
process, electrical energy is converted into thermal energy (heat).

There are 4 factors that affect internal resistance:
1) Type of material: The ability of a material to conduct electricity is determined by how freely
electrons can move within the material. For instance, conductors allow electrons to flow freely,
but insulators do not.
2) Cross-sectional area: The thicker the wire (the length of its diameter) the less internal
resistance there is – this is because the electrons flowing through the thicker wire have more
space to move freely.
3) Length: As you increase the length of a wire, its internal resistance increases. This happens
because the electrons have to travel through more material. We see this when we use extension
wires – the longer the extension cord, the warmer it will get while being used. Therefore, to
reduce the chance of fire, manufacturers of extension cords use larger-diameter wires which will
lower the electrical resistance.
4) Temperature: When a wire gets warmer, the atoms that make up the wire gain energy and
vibrate faster. The increased vibration results in more collisions between the atoms and the
free-flowing electrons in the current. Therefore, resistance increases with temperature.

An OHMMETER is a device used to measure resistance. Just like a voltmeter, an ohmmeter
must be connected in parallel with a load. However, you do not power up the circuit to measure
resistance because the ohmmeter is powered and provides an electric current through the load.
Ohm’s Law

German physicist Georg Ohm discovered a mathematical relationship between potential
difference and current.

He noticed that as the length of the wire increased, the current decreased.

If you were to plot this information, the steeper the slope of the straight line, the greater the
resistance.

The relationship among voltage, current, and resistance can be written as:
R=V
I
Where R is the resistance measured in ohms (Ω), V is the voltage (or potential difference) in volts (V), and
I is the current in amperes (A). This relationship is called Ohm’s Law.

Ohm’s Law states that as the potential difference across a load increases, so does the current.
Example A: Calculating the Resistance of a Load

Question #1: A load has 1.2 A of current flowing through it. The voltage across the load is 6.0
V. Calculate the resistance of the load.

Given:
I = 1.2 A and V = 6.0 V

Required:
R (resistance)

Analysis:
R = V/I

Solution:
R = 6.0 V/ 1.2 A

Statement:
The resistance of the load is 5.0 Ω

Question #2: A hair dryer is plugged into a wall outlet that has a voltage of 115 V. When the
R = 5.0 Ω
hair dryer is on, a current of 4.0 A flows through it. Calculate the resistance of the hair dryer.

Given:

Required:

Analysis:

Solution:

Statement:
Example B: Calculating the Current through a Resistor

Question #1: A 110 Ω resistor is connected to a power supply set at 12 V. Calculate the current
going through the resistor.

Given:
R = 110 Ω
and
V = 12 V

Required:
I (electric current)

Analysis:
I = V/R

Solution:
I = 12 V/ 110 Ω

Statement:
The current going through the resistor is approximately 0.11 A

Question #2: A device creates a potential difference of 1500 V across two points at the surface
I = 0.11 A
of human skin. Dry skin has a resistance of 125 000 Ω. Calculate the current moving through the
skin.

Given:

Required:

Analysis:

Solution:

Statement:
Example C: Calculating the Potential Difference across a Resistor

Question #1: A toaster has a 24.0 Ω resistor that has a 5.00 A of current going through it
when the toaster is on. Calculate the potential difference across the resistor.

Given:
R = 24.0 Ω
and I = 5.00 A

Required:
V (potential difference)

Analysis:
V = IR

Solution:
V = (5.00 A)(24.0 Ω)

Statement:
The potential difference across the resistor is 120 V

Question #2: A compact fluorescent lightbulb has a resistance of 1100 Ω. The current going
V = 120 V
through the bulb is 0.11 A. Calculate the potential difference across the lightbulb.

Given:

Required:

Analysis:

Solution:

Statement:
How Series and Parallel Circuits Differ
A) Loads in Series:

When you connect an energy source to a series circuit, a voltage or potential difference is
created and electrons begin to flow. No matter how many loads are connected in series, there is
only on path that the current can flow.

If you have a circuit with one load, the total resistance of the circuit will be different than if you
have two or more loads connected in series. The electrons have only one path to follow and with
two or more loads, they have more “bumps” to deal with along the way. Because of this, the
current flowing through the circuit with multiple loads will be less than the current flowing
through a circuit with only one load.

Therefore, there is less resistance and more current when there are fewer loads.
Sample Problem #1: Comparing the Current in Two Series Circuits:

Question #1: A circuit has two identical lamps. The potential difference across the battery is
10 V. The circuit has a total resistance of 10 Ω. Use the total resistance given to calculate the
current through the circuit.

Given:
RT = 10 Ω
and V = 10 V

Required:
I (electric current)

Analysis:
I = V/RT

Solution:
I = 10 V/ 10 Ω

Statement:
The current through each lamp is 1 A.

Question #2: A series circuit has four identical lamps. The potential difference of the energy
I=1A
source is 60 V. The total resistance of the lamps is 20 Ω. Use the total resistance given to
calculate the current through each lamp.

Given:

Required:

Analysis:

Solution:

Statement:

When lamps are connected in series, you will notice that the brightness of each lamp will
decrease as you connect more lamps.

When the circuit only has one load (a lamp) the electrons pass through the load, and all the
electric potential energy that the electrons received from the battery gets converted into light
and heat.

Voltage is related to electric potential – the voltage drops across the one load will be the same as
the voltage drop across the battery.

When two identical loads (such as two lamps) are connected in series, the electrons will lose
potential energy. The voltage drop across each load decreases as you add more loads.

Vload = Vsource / # of loads
Sample Problem #2: Calculating the Voltage in a Series Circuit

Question #1: A series circuit contains three identical lamps. The potential difference of the
battery is 30 V. Calculate the potential difference across each lamp.

Given:
Vsource = 30 V

Required:
Vload (voltage drop across each lamp)

Analysis:
Vload = Vsource / # of loads

Solution:
Vload = 30 V/ 3

Statement:
The potential difference across each lamp is 10 V.

Question #2: A series circuit contains four identical lamps. The voltage across the energy
Vload = 10 V
source is 96 V. Calculate the voltage across each lamp.

Given:

Required:

Analysis:

Solution:

Statement:
2) Loads in Parallel:

When an energy source is connected to a parallel circuit, the potential difference is created
across the two terminals of the energy source.

The potential difference causes electrons to flow.

Unlike a series circuit, the electrons have different paths to follow. The number of paths
depends on the number of loads connected in parallel.

The quantity (the amount) of current flowing through each path decreases as the number of
paths increases.

Iload = Isource / # of loads
Sample Problem #3: Calculating the Current in a Parallel Circuit

Question #1: The total resistance in the circuit is 2 Ω. The potential difference of the
battery is 18 V. Calculate the current through each lamp, if there are 3 identical lamps.

Given:
RT = 2 Ω
and
V = 18 V

Required:
Iload (current through each lamp)

Analysis:
Isource = V/ RT (to calculate the current coming out of the energy source) and..
Iload = Isource / # of loads (to calculate the current through each identical load)

Solution:
Isource = 18 V/ 2 Ω
Isource = 9 A
Iload = 9 A/ 3
Iload = 3 A

Statement:
The current through each lamp is 3 A.

Question #2: A parallel circuit contains four identical lamps. The potential difference across
the energy source is 48 V. The total resistance of the lamps is 12 Ω. Calculate the current
through each lamp.

Given:

Required:

Analysis:

Solution:

Statement:

If you connect loads in parallel, the total resistance will be less than if the loads were connected
in series.

Since voltage is related to electric potential energy, the voltage drop across each parallel load
will be the same as the voltage drop across the battery. In any parallel circuit, the voltage drop
across each parallel branch will be the same as the voltage across the energy source.
Sample Problem #4: Calculating Voltage in a Parallel Circuit

Question #1: A parallel circuit contains three identical lamps. The current coming out of the
energy source is 2.5 A. The total resistance of the circuit is 6.0 Ω. Calculate the voltage across
the energy source and across each lamp.

Given:
Isource = 2.5 A
and
RT = 6.0 Ω

Required:
Vsource (voltage drop across the energy source) and Vload (voltage across each lamp).

Analysis:
Vsource = Vload (loads are in parallel to the energy source)
Vsource = (Isource)(RT)

Solution:
Vsource = (2.5 A) (6.0 Ω)
Vsource = 15 V
Vload = 15 V

Statement:
The voltage across the energy source is 15 V, and the voltage across each lamp is
Also 15 V.

Question #2: A parallel circuit contains 10 identical lamps. The current through the energy
source is 3.0 A. The total resistance of the circuit is 15 Ω. Calculate the voltage across the
energy source and across each lamp.

Given:

Required:

Analysis:

Solution:

Statement:
Resistance, Current, and Voltage in Circuits
Quantity
Series Circuit
Total resistance
of circuit (RT)
Current through
loads (Iloads)
Voltage across
loads (Vloads)
Increases
Isource decreases as more loads are
added
Vsource splits based on the number
of loads
Parallel Circuit
Decreases
Isource splits among loads based on the number
of branches in parallel
Voltage of each parallel branch is the same as
the Vsource