Grade 10 Physics
Advance Stream
Academic Year 2024/2025 – Term 2
Chapter-9
Series and Parallel
Circuits
Section No. : 9.1
Section’s Name : Simple Circuits
Period 1
Learning
objectives
• Understand the Series Combination of Resistors.
• Derive the relation for Equivalent Resistance for a
Circuit of Resistors in Series Combination.
Keywords
Series Combination.
Resistors/Resistance.
Equivalent Resistance.
Introduction
Remember that…
Resistance is the opposition presented to electric
current by a material or device.
 Ohm’s law states that …
“Current passing through a wire is directly proportional to the
potential difference between it’s terminals.”
 The SI units for measuring the resistance of a conductor is the Ohm (Ω).
 Ohms () = Volts/Ampere (V/A).
Parallel and Series Connections
Introduction
 Resistors can be connected in series or in Parallel.
 Resistors can be connected together in an unlimited
number of series and parallel combinations to form
complex resistive circuits.
A River Model
 A mountain river can be used to model an electric circuit.
 The river flows downhill to the plains below.
 Some rivers flow downhill in a single stream, other rivers
might split into two or more smaller streams.
 The total amount of water flowing down the mountain
remains unchanged.
Also, note that…
 The distance the river drops is similar to the potential
difference in a circuit.
 The rate of water flow is similar to current in a circuit.
 Large rocks and other obstacles produce resistance to
water flow and are similar to resistors in a circuit.
 The water cycle is similar to a battery or a generator in
an electric circuits
A River Model
Series Circuit
Parallel Circuit
Series Circuits
Now, watch
this video
Resistors in Series
A Series Circuit describes two or more components of a circuit that provide a
single path for current.
Resistors in Series carry the Same current.
Resistors in Series
Activity
Now, click here to start the
activity
https://phet.colorado.edu/en/simulation/circ
uit-construction-kit-dc
Equivalent Resistance of a Series circuit
 Vbattery= V1 + V2
Conservation of energy
 I Requivalent = I R1 + I R2
Ohm’s law
Take I as a common factor to get
 I Requivalent = I (R1 + R2)
Requivalent = R1 + R2
Note:
Same current ( I ) passing
through the circuit.
Resistors in Series, continued
The equivalent resistance in a series circuit
is the sum of the circuit’s resistances.
The equivalent resistance can be used to find
the current in a circuit.
Total Current in series Circuit
 Two or more resistors in the actual circuit have the same effect on the current as one
equivalent resistor.
 The following equation can be used to
calculate the current.
∆V = IReq
 The total current in a series circuit equals to potential difference divided by equivalent resistance.
Evaluation Question
Critical Thinking Question
 A series circuit shows a bulb burned out.
Solution
a) What will happen to the other bulbs?
Answer:
One bulb burning out in a series circuit breaks the circuit.
b) Would this also happen in a parallel circuit?
Answer:
In parallel circuits, each bulb has its own circuit, so all but one
bulb could be burned out, and the last one will still function.
Exercises
Example 1
Resistors in Series
A 9.0 V battery is connected to four light
bulbs, as shown at right.
Find the equivalent resistance for the circuit
and the current in the circuit.
Given:
 ∆V = 9.0 V
 R1 = 2.0 Ω
 R2 = 4.0 Ω
 R3 = 5.0 Ω
 R4 = 7.0 Ω
Unknown:
 Req = ?
 I=?
Diagram
Solution
Because the resistors are connected in series. Thus, the equivalent resistance can
be calculated with the following equation
Req = R1 + R2 + R3 + R4
 Substitute the values into the equation and solve…
R eq = 2.0 Ω + 4.0 Ω + 5.0 Ω + 7.0 Ω
R eq = 18.0 Ω
The following equation can be used to calculate the current.
∆V = I Req
Rearrange the equation
to isolate the unknown:
V
I 
Req
 Substitute the equivalent resistance value into the equation for current…
V
9.0 V
I

 0.50 A
Req 18.0 Ω
Example 2
Three 20 Ω resistors are connected in series across a 120-V generator. What is the
equivalent resistance of the circuit? What is the current in the circuit?
Solution
Given:
 ∆V = 120 V
 R1 = 20 Ω
 R2 = 20 Ω
 R3 = 20 Ω
Unknown:
 Req = ?
 I=?
Example 3
A 10 Ω, 15 Ω, and 5 Ω resistor are connected in a series circuit with a 90 V battery.
What is the equivalent resistance of the circuit? What is the current in the circuit?
Solution
Given:
 ∆V = 90 V
 R1 = 10 Ω
 R2 = 15 Ω
 R3 = 5 Ω
Unknown:
 Req = ?
 I=?
Req=32
I= 2.3 A
Period 2
Chapter-9
Series and Parallel
Circuits
Section No. : 9.1
Section’s Name : Simple Circuits.
Learning
objectives
Calculate the Equivalent Resistance for a
Circuit of Resistors in Series.
Find the Current and Potential Difference
across each Resistor in the Circuit.
Understand the function of the Voltage
dividers.
Keywords
• Voltage dividers.
Example 4
A string of holiday lights has ten bulbs with equal resistances connected in series.
When the string of lights is connected to a 120-V outlet, the current through the
bulbs is 0.06 A.
Solution
R= 195
Example 5
Q45
A 9-V battery is in a circuit with three resistors connected in series.
Solution
Period 3
Chapter-9
Series and Parallel
Circuits
Section No. : 9.1
Section’s Name : Simple Circuits
Learning
objectives
Understand the function of Photoresistors.
Problem solving.
Keywords
• Photoresistor.
Voltage dividers
 A voltage divider is a simple circuit which turns a large voltage into
a smaller one.
 A voltage divider produces a source of potential difference
that is less than the potential difference across the battery.
 Consider the circuit shown in Figure 1. Two resistors (R1 and R2) are
connected in series across a battery with potential difference V The
equivalent resistance of the circuit is
R = R1 + R2
The current is represented by the following equation…
Voltage dividers, Cont.
By substituting in Ohm’s law…
 By choosing the right resistors, you can
produce a potential difference of 5 V across
a portion of an electric circuit.
Photoresistors
Note that…
 Voltage dividers often are used with
sensors, such as Photoresistors.
 A photoresistor’s resistance depends on the
amount of light that strikes it.
 A photoresistor is sensitive to light. Its resistance
decreases when lighting increases.
Photoresistors, Cont.
 This circuit can be used as a light meter,
such as the one shown in Figure.
 In this device, an electronic circuit
detects the potential difference and
converts it to a measurement of
illuminance that can be read on the
digital display.
Light meters used in photography use a
voltage divider. The amount of light
striking the photoresistor sensor
determines the voltage output of the
voltage divider.
Example 1
POTENTIAL DIFFERENCE IN A SERIES CIRCUIT
Two resistors, 47 Ω and 82 Ω, are connected in series across a 45 V battery.
a. What is the current in the circuit?
b. What is the potential difference across each resistor?
c. If you replace the 47 Ω resistor with a 39 Ω resistor, will the current
increase, decrease, or remain the same?
d. What is the new potential difference across the 82 Ω resistor?
Given:
 ∆V = 45 V
 R1 = 47 Ω
 R2 = 82 Ω
Unknown:
 I=?
 ∆V1 =?
 ∆V2=?
Solution
Solution, Cont.
Q47
Example 2
The circuit shown in Example 1 is producing these
symptoms: the ammeter reads 0 A, ΔV1 reads 0 V,
and ΔV2 reads 45 V. What has happened?
Solution
R2 has failed. It has infinite resistance, and the
battery voltage appears across it.
Q48
Example 3
Solution
Given:
 VA = 17.0 V
 RA = 255 Ω
 RB = 292 Ω
Example 3 , Cont.
Given:
 VA = 17.0 V
 RA = 255 Ω
 RB = 292 Ω
Example 3 , Cont.
Given:
 VA = 17.0 V
 RA = 255 Ω
 RB = 292 Ω
Example 3 , Cont.
Answer
Yes. The law of conservation of energy states that energy
cannot be created or destroyed; therefore, the rate at which
energy is converted, or power dissipated, will equal the sum
of all parts.
Given:
 VA = 17.0 V
 RA = 255 Ω
 RB = 292 Ω
The resistor with the lower resistance will dissipate less power, and thus will
be cooler.
Learning
objectives
Problem Solving.
Exercises
Example 1
VOLTAGE DIVIDER
A 9.0 V battery and two resistors, 390 Ω and
470 Ω, are connected as a voltage divider.
What is the potential difference across the
470 Ω resistor?
Given:
 ∆VSource = 9.0 V
 R1 = 390 Ω
 R2 = 470 Ω
Unknown:
 ∆V2 = ?
Diagram
Solution
Q52
Example 2
A 22 Ω resistor and a 33 Ω resistor are connected in series and placed across a 120 V
potential difference.
a. What is the equivalent resistance of the circuit?
Solution
Given:
 ∆V = 120 V
 R1 = 22 Ω
 R2 = 33 Ω
b. What is the current in the circuit?
Example 2 , Cont.
A 22 Ω resistor and a 33 Ω resistor are connected in series and placed across a 120V
potential difference.
c. What is the voltage drop across each resistor?
Solution
Given:
 ∆V = 120 V
 R1 = 22 Ω
 R2 = 33 Ω
d. What is the voltage drop across the two resistors together
Q53
Example 3
Three resistors of 3.3 kΩ, 4.7 kΩ, and 3.9 kΩ are connected in series across a 12 V
battery.
a. What is the equivalent resistance?
Solution
Given:
 ∆V = 12 V
 R1 = 3.3 kΩ
 R2 = 4.7 k Ω
 R3 = 3.9 k Ω
b. What is the current through the resistors?
Example 3 , Cont.
Three resistors of 3.3 kΩ, 4.7 kΩ, and 3.9 kΩ are connected in series across a 12 V
battery.
c. What is the voltage drop across each resistor?
Solution
Given:
 ∆V = 12 V
 R1 = 3.3 kΩ
 R2 = 4.7 k Ω
 R3 = 3.9 k Ω
d. Find the total voltage drop across the three resistors.
Period 4
Learning
objectives
Calculate the Equivalent Resistance for a
Circuit of Resistors in Parallel.
Find the Current and Potential Difference
across each Resistor in the Circuit.
Keywords
Parallel Combination.
Resistors/Resistance.
Equivalent Resistance.
Resistors in Parallel
 A Parallel Circuit describes two or more components of a circuit that
provide separate conducting paths for current because the components
are connected across common points or junctions.
Note that…
 Lights wired in parallel have more than one
path for current.
 Parallel circuits do not require all elements
to conduct.
Resistors in Parallel
Activity
Now, click here to start the
activity
https://phet.colorado.edu/sims/html/circuitconstruction-kit-dc-virtual-lab/latest/circuitconstruction-kit-dc-virtual-lab_en.html
Equivalent Resistance for Resistors in Parallel
ITotal = I1 + I2
– Ohm’s law:
I
“Conservation of charge”
Vbattery
Req
V1
V2


R1
R2
I1
 Vbattery= V1 = V2
Potential energy loss is the same across all parallel resistors.
V
V
V


VReq VR1 VR2
• Because Vbattery= V1 = V2,
the equation above reduces as
follows:
I
1
1 1


Req R1 R2
I2
Resistors in Parallel, continued..
Resistors in parallel have the same potential
differences across them.
The sum of currents in parallel resistors equals
the total current.
The equivalent resistance of resistors in parallel
can be calculated using this relationship…
Resistors Connected in Series (or) Parallel
Example
Resistors in Parallel
A 9.0 V battery is connected to four resistors in parallel,
as shown at right.
Find the equivalent resistance for the circuit and the
total current in the circuit.
Given:
 ∆V = 9.0 V
 R1 = 2.0 Ω
 R2 = 4.0 Ω
 R3 = 5.0 Ω
 R4 = 7.0 Ω
Unknown:
 Req = ?
 I=?
Diagram
Solution
Because each resistor is connected in parallel, the equivalent resistance can be
calculated with the equation…
 Substitute the values into the equation and solve…
Solution, Cont.
The following equation can be used to calculate the current.
∆V = I Req
Rearrange the equation
to isolate the unknown:
V
I 
Req
 Substitute the equivalent resistance value into the equation for current…
Period 3
Week 4
Chapter-4
Series and Parallel
Circuits
Section No. : 9.1
Section’s Name : Simple Circuits
“
Learning
Problem solving.
objectives
Q55
Exercises
Example 1
Three 15.0 Ω resistors are connected in parallel and placed across a 30.0 V battery .
Solution
Given:
 ∆VSource = 30 V
 R1 = 15 Ω
 R2 = 15 Ω
 R3 = 15 Ω
Unknown:
 Req= ?
a. What is the equivalent resistance of the parallel circuit?
Example 1, Cont.
Three 15.0 Ω resistors are connected in parallel and placed across a 30.0V battery.
Solution
Given:
 ∆VSource = 30 V
 R1 = 15 Ω
 R2 = 15 Ω
 R3 = 15 Ω
Unknown:
 Req= ?
b. What is the current through the entire circuit?
c. What is the current through each branch of the circuit?
Q56
Critical Thinking
Example 2
Suppose that one of the 15.0 Ω resistors in the previous problem is replaced by a 10.0 Ω resistor.
Solution
a. Does the equivalent resistance change? If so, how?
Yes, it gets smaller.
b. Does the amount of current through the entire circuit
change? If so, in what way
Yes, it gets larger.
c. Does the amount of current through the other 15.0 Ω
resistors change? If so, how?
No, it remains the same. Currents are independent.
Q57
Example 3
A 120.0 Ω resistor, a 60.0 Ω resistor, and a 40.0 Ω resistor are connected in parallel and
placed across a 12.0 V battery.
Solution
Given:
 ∆VSource = 12 V
 R1 = 120 Ω
 R2 = 60 Ω
 R3 = 40 Ω
a. What is the equivalent resistance of the parallel circuit?
b. What is the current through the entire circuit?
Example 3 , Cont.
A 120.0 Ω resistor, a 60.0 Ω resistor, and a 40.0 Ω resistor are connected in parallel
and placed across a 12.0 V battery.
Solution
Given:
 ∆VSource = 12 V
 R1 = 120 Ω
 R2 = 60 Ω
 R3 = 40 Ω
c. What is the current through each branch of the circuit?
Period 4
Week 4
Chapter-4
Series and Parallel
Circuits
Section No. : 9.1
Section’s Name : Simple Circuits
“Kirchhoff’s Rules”
Learning
objectives
Understand the Kirchhoff’s Rules.
Keywords
Loop.
Junction.
Kirchhoff’s Rules
Introduction
 Kirchhoff was a German physicist who
formulated two rules that govern electric
circuits—the loop rule and the junction rule.
 You can use these two rules to analyze
complex electric circuits.
Both rules are based on fundamental
scientific laws.
Kirchhoff’s Rules
The Loop Rule
 The loop rule is based on the law of
conservation of energy.
Kirchhoff's Loop Rule states that the sum of the
voltage differences around the loop must be
equal to zero.
In other words,
The sum of increases in electric potential around a loop in an electric
circuit equals the sum of decreases in electric potential around that loop.
Kirchhoff’s Rules
The Junction Rule
 The junction rule describes currents and is based on
the law of conservation of charge.
Note that..
 Recall that the law of conservation of charge states that
charge can neither be created nor destroyed.
 Kirchhoff's Junction Rule states that In an electric circuit, the total
current into a section of that circuit must equal the total current out
of that same section.
Kirchhoff’s Rules
The Junction Rule, Cont.
A junction is a location where three or
more wires are connected together.
 The circuit shown in Figure has two such junctions—
one at point A and another at point B.
 The given figure shows an application of Kirchhoff’s junction rule.
 At junction A the total current entering (𝑰𝟏 ) equals the total
current leaving (𝑰𝟐 + 𝑰𝟑 ).
 At junction B, the total current entering is 𝑰𝟐 + 𝑰𝟑 , and the total
current leaving is 𝑰𝟏 .
Kirchhoff’s Rules
The Junction Rule, Cont.
In the given Figure…
𝑰1 = 𝑰𝟐 + 𝑰𝟑 at junction A
And
𝑰𝟐 + 𝑰𝟑 = 𝑰1 at junction B.
For example…
if 𝑰𝟐 = 0.3 𝐴
and 𝑰𝟑 = 0.7 𝐴
then
𝑰1 = 0.3 + 0.7 = 1.0 𝐴
Q 61
Example 2
Total Current A series circuit has four resistors. The current through
one resistor is 810 mA. How much current is supplied by the source?
Solution
Q 60
810 mA. Current is the same everywhere in a series circuit.
Example 3
A parallel circuit has four branch currents: 120 mA, 250 mA, 380
mA, and 2.1 A. How much current is supplied by the source?
𝐼𝑇 = 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4
Solution
= 120 𝑚𝐴 + 250 𝑚𝐴 + 380 𝑚𝐴 + 2.1 𝐴
= 0.12 𝐴 + 0.25 𝐴 + 0.38 𝐴 + 2.1 𝐴 = 𝟐. 𝟗 𝑨
Q 62
Example 1
Exercises
A switch is connected in series with a 75 W bulb to a source of 120 V.
Solution
a. What is the potential difference across the switch when it is
closed (turned on)?
0 V; V = IR with R = 0
b. What is the potential difference across the switch if another
75-W bulb is added in series?
0 V; V = IR with R = 0
. Current is the same everywhere in a series circuit.
a- 0 V; V IR with R = 0
b- 0 V; V IR with R = 0
Q 65
Critical Thinking Question
Evaluation Question
The circuit in Figure has four identical resistors. Suppose that a wire is added to
connect points A and B. Answer the following questions, and explain your reasoning.
Solution
a. What is the current through the wire?
0 A; the potentials of points A and B are the same
b. What happens to the current through each resistor?
nothing
c. What happens to the current drawn from the battery
nothing
d. What happens to the potential difference
across each resistor?
nothing
Period 5
Week 4
Self Learning- Safety Devices- p 231-232
Learning
objectives
Recognize the meaning of Short Circuit.
Understand how do fuses, circuit breakers, and
ground-fault interrupters protect household wiring.
Keywords
Short circuit.
Fuse.
Circuit breakers.
Ground-Fault Interrupters (GFIs).
Safety Devices
Note that…
 You might have circuit breakers, fuses, and ground-fault interrupters (GFIs) in your
home.
 All of these devices are added to circuits to prevent damage.
 In an electric circuit, fuses and
circuit breakers prevent circuit
overloads that can occur when
too many appliances are turned
on at the same time or when a
short circuit occurs in one
appliance.
Safety Devices
A short circuit occurs when a circuit with
very low resistance is formed.
Let’s understand why short circuit occurred !
 When appliances are connected in parallel, each
additional appliance placed in operation reduces the
equivalent resistance in the circuit and increases the
current through the wires.
 This additional current might produce enough thermal
energy to melt the wiring’s insulation, cause a short
circuit, or even begin a fire.
Safety Devices
A fuse is a short piece of metal that acts as a safety device by melting
and stopping the current when too large a current passes through it.
Note that…
 Fuses are designed to melt before other elements in a circuit are
damaged.
Safety Devices
Activity
Fuses
Now, click here to start the
activity
https://phet.colorado.edu/sims/html/circuitconstruction-kit-dc/latest/circuitconstruction-kit-dc_en.html
Safety Devices
 A circuit breaker is an automatic switch that acts as a safety device by stopping the
current if the current gets too large and exceeds a threshold value.
Note That
 When there is too much current through the bimetallic strip in a circuit breaker, the
heat that is generated causes the strip to bend and release the latch. The handle
moves to the off position, causing the switch to open and the current to stop.
Safety Devices
A ground-fault interrupter (GFI) is a
device that contains an electronic circuit
that detects small current differences
between the two wires in the cord
connected to an appliance.
 An extra current path, such as one through
water, could cause this difference.
 The GFI stops the current when it detects
such differences. This often protects a
person from electrocution.
Evaluation Questions
Example 1
Explain how a fuse functions to protect an electric circuit
Solution
The purpose of a fuse is to prevent conductors from being overloaded with
current, causing fires due to overheating. A fuse is simply a short length of wire
that will melt from the heating effect if the current exceeds a certain maximum.
Example 2
What is a short circuit? Why is a short circuit dangerous?
Solution
 A short circuit is a circuit that has extremely low resistance.
 A short circuit is dangerous because any potential difference will produce
a large current. The heating effect of the current can cause a fire.