Series and parallel resistor networks
(Revision)
In Grade 10 and Grade 11 you learnt about electric circuits and we introduced three
quantities which are fundamental to dealing with electric circuits. These quantities are
closely related and are current, voltage (potential difference) and resistance. To recap:
1. Electrical current, I, is defined as the rate of flow of charge through a circuit.
2. Potential difference or voltage, V, is related to the energy gained or lost per unit charge
moving between two points in a circuit. Charge moving through a battery gains energy
which is then lost moving through the circuit.
3. Resistance, R, is an internal property of a circuit element that opposes the flow of
charge. Work must be done for a charge to move through a resistor.
These quantities can be related, in circuit elements whose resistance remains constant, by
Ohm's law.
Note:
Ohm's Law
For a resistor at constant temperature the ratio VI is constant. This ratio we call resistance.
This is equivalent to saying that the amount of electric current through a metal conductor, at
a constant temperature, in a circuit is proportional to the voltage across the conductor and
can be described by
I=VR
In other words, at constant temperature, the resistance of the conductor is constant,
independent of the voltage applied across it or current passed through it.
We have focused on the properties of a single component. Now we need to look at a
collection of components in a circuit.
important:
You will often hear people switch between using the terms voltage and potential
difference to describe the same quantity. This is correct but very important to note.
Circuits don't consist of a single element and we've learnt about how voltage, current and
resistance are affected in circuits with multiple resistors. There are two basic layouts we
consider for a network of resistors, series and parallel. Resistors in series and resistors in
parallel have different features when talking about current, voltage and equivalent
resistance.
Example 1: Ohm's Law [NSC 2011 Paper 1]
Question
Learners conduct an investigation to verify Ohm's law. They measure the current through a
conducting wire for different potential differences across its ends. The results obtained are
shown in the graph below.
1. 9.1.
Which ONE of the measured quantities is the dependent variable?
(1 mark)
2. 9.2.
The graph deviates from Ohm's law at some point.
1. 9.2.1.
Write down the coordinates of the plotted point on the graph beyond which Ohm's
law is not obeyed.
(2 marks)
2. 9.2.2.
Give a possible reason for the deviation from Ohm's law as shown in the graph.
Assume that all measurements are correct.
(2 marks)
3. 9.3.
Calculate the gradient of the graph for the section where Ohm's law is obeyed.
Use this to calculate the resistance of the conducting wire.
(4 marks)
Answer
Question 1
Current OR I
(1 mark)
Question 2
The graph deviates from Ohm's law at some point.
1. 9.2.1.
(4,0 ; 0,64)
(2 marks)
2. 9.2.2.
Temperature was not kept constant.
(2 marks)
Question 3
(4 marks)
[TOTAL: 9 marks]
important:
A circuit may consist of a combination of parallel and series networks that can in turn be in
parallel or series. We can treat parts of the total circuit independently by applying Ohm's
Law to each of the components.
Note:
Resistors connected in series
Resistors are in series if they are consecutive elements in the sequence of the circuit and
there are no branches between them. For n resistors in series the equivalent resistance is:
Rs=R1+R2+R3+…+Rn
For n resistors in series the potential difference is split across the resistors:
VTotal=V1+V2+V3+…+Vn
The current is constant through the resistors.
ITotal=I1=I2=I3=…=In
It makes sense to remind ourselves of why these are consistent with other topics we have
covered previously:

Conservation of charge: we have learnt that charges are not created or destroyed.
This is consistent with the current being the same throughout a resistor network that is
in series. Charges aren't being added or lost or bunching up and, therefore, the rate at
which charge moves past each point should be the same.

Conservation of energy: we have learnt that energy isn't created or destroyed but
transferred through work. The voltage across a resistor is the energy per unit charge
(work) required to move through the resistor. The total work done to move through a
network of resistors in series should be the sum of the work done to move through
each individual resistor.
Example 2: Ohm's Law, all components in series
Question
Two ohmic resistors (R1 and R2) are connected in series with a cell with negligible internal
resistance. Find the resistance of R2, given that the current flowing through R1 and R2 is
0,25 A and that the potential difference across the cell is 1,50 V. R1 =1,00 Ω.
Answer
Draw the circuit and fill in all known values.
Determine how to approach the problem.
We can use Ohm's Law to find the total resistance R in the circuit, and then calculate the
unknown resistance using:
R=R1+R2
because R1 and R2 are connected in series.
Find the total resistance
R=VI=1,50,25=6 Ω
Find the unknown resistance
We know that:
R=6,00 Ω
and that
R1=1,00 Ω
Since
R=R1+R2
R2=R−R1
Therefore,
R1=5,00 Ω
Example 3: Ohm's Law, series circuit
Question
In the case of the circuit shown, calculate:
1. the potential difference of V1, V2 and V3 across the resistors R1, R2, and R3
2. the resistance of R3.
Answer
Determine how to approach the problem
We are given the potential difference across the cell and the current in the circuit, as well as
the resistances of two of the three resistors. We can use Ohm's Law to calculate the
potential difference across the known resistors. Since the resistors are in a series circuit the
potential difference is V=V1+V2+V3 and we can calculate V3. Now we can use this
information to find the potential difference across the unknown resistor R3.
Calculate potential difference across R1
Using Ohm's Law:
R1I⋅R1V1V1=V1I=I⋅V1I=I⋅R1=2⋅2=4,00 V
Calculate potential difference across R2
Use Ohm's Law:
R2I⋅R2V2V2=V2I=I⋅V2I=I⋅R2=2⋅6=12,00 V
Calculate potential difference across R3
Since the potential difference across all the resistors combined must be the same as the
potential difference across the cell in a series circuit, we can find V3 using:
VV3V3=V1+V2+V3=V−V1−V2=36−4−12=20,00 V
Find the resistance of R3
We know the potential difference across R3 and the current through it, so we can use
Ohm's Law to calculate the value for the resistance:
R3R3=V3I=202=10,00 Ω
Write the final answer




V1=4,00 V
V2=12,00 V
V3=20,00 V
R3=10,00 Ω
Note:
Equivalent resistance in a parallel network
A parallel configuration is when the current splits into a number of branches which contain
components (resistors in our case). A branch may contain multiple resistors in series and
still be part of the parallel configuration. For nbranches of resistors in parallel, the equivalent
resistance can be calculated from the total resistance of each branch and is:
1Rp=1RB1+1RB2+1RB3+…+1RBn
For n branches of resistors in parallel the potential difference is the same across each of
the branches:
VTotal=VB1=VB2=VB3=…=VBn
The current is split through the branches.
ITotal=I1+I2+I3+…+In
Let's take a moment to see if our conservation laws still make sense:

Conservation of charge: we have learnt that charges are not created or destroyed.
This is consistent with the current splitting between the branches. Charges aren't being
added or lost or bunching up and, therefore, the total number of charges going through
the branches must be the same as the number entering the point where the circuit
branches.

Conservation of energy: we have learnt that energy isn't created or destroyed but
transferred through work. Energy per unit charge doesn't change unless work is done
therefore it makes sense that the energy per unit charge in each branch should be the
same.
Example 4: Ohm's Law, resistors connected in
parallel
Question
Calculate the current (I) in this circuit if the resistors are both ohmic in nature.
Answer
Determine what is required
We are required to calculate the total current flowing in the circuit.
Determine how to approach the problem
Since the resistors are ohmic in nature, we can use Ohm's Law. However, there are two
resistors in the circuit and we need to find the total resistance.
Find the equivalent resistance in the circuit
Since the resistors are connected in parallel, the total (equivalent) resistance R is:
1R=1R1+1R2.
1RTherefore, R=1R1+1R2=12+14=2+14=34=43=1,33 Ω
Apply Ohm's Law
RR⋅IRII=VI=VI⋅IR=VR=V⋅1R=(12)(34)=9,00 A
Write the final answer
The total current flowing in the circuit is 9,00 A.
Example 5: Ohm's Law, parallel network of resistors
Question
An 18,00 V cell is connected to two parallel resistors of 2,00 Ω and 6,00 Ω respectively.
Calculate the current through each of the ammeters when the switch is closed and when it
is open.
Answer
Determine how to approach the problem
We need to determine the current through the cell and each of the parallel resistors. We
have been given the potential difference across the cell and the resistances of the resistors,
so we can use Ohm's Law to calculate the current.
There are two alternative approaches we could adopt:

we could use the fact that the potential difference across each of the resistors is the
same as the potential difference across the battery because they are in a parallel
configuration and then use Ohm's Law; or

we could determine the equivalent resistance of the circuit and the total current and
then use that to determine the current through each of the resistors.
important:
Both methods will result in the correct answer if you don't make any calculation errors but
one is shorter.
Now determine the current through one of the parallel resistors
We know that for a configuration with just two resistors in parallel and a cell as in this case,
the potential difference across the cell is the same as the potential difference across each of
the resistors in parallel. For this circuit:
V=V1=V2=18,00 V
Let's start with calculating the current through R1 using Ohm's Law:
R1I1I1=V1I1=V1R1=18,002,00=9,00 A
Calculate the current through the other parallel resistor
We can use Ohm's Law again to find the current in R2:
R2I2I2=V2I2=V2R2=18,006,00=3,00 A
Calculate the total current
The current through each of the parallel resistors must add up to the total current through
the cell:
II2=I1+I2=9,00+3,00=12,00 A
When the switch is open
The branch through R1 is not complete so no current can flow through it. This means we
can ignore it completely and consider a simple circuit with a battery and a single
resistor, R2, in it.
We can use Ohm's Law again to find the current in R2:
R2I2I2=V2I2=V2R2=18,006,00=3,00 A
Write the final answer
When the switch is closed:

The current through the cell is 12,00 A.

The current through the 2,00 Ω resistor is 9,00 A.

The current through the 6,00 Ω resistor is 3,00 A.
When the switch is open:

The current through the 6,00 Ω resistor is 3,00 A.
Informal experiment 1: Series and parallel networks
Aim
To investigate the changes in current and voltage when branches of circuits are open or
closed..
Apparatus
You will need the following items for this investigation:

batterys / cells

electric leads

a set of resistors and/or light bulbs

ammeters

voltmeters
Method
For this investigation, configure a circuit with resistors in both series and in parallel. For
example, try:

including parallel branches with different numbers of light bulbs in each branch

changing the numbers of light bulbs or resistors in each branch

try adding a resistor in series with the parallel network
In each branch include an ammeter and a switch. Make notes about what happens when
you remove a branch by opening the switch in the branch. What happens to the current in
the other branches. Try to predict what will happen before opening or closing a switch and
before adding or removing any light bulbs or resistors.
Discussion
Be sure to note and discuss:

whether adding a new branch increases or decreases the total current in the circuit,

whether adding a new branch increases or decreases the current in the original
branches,

whether adding a resistor in series with a parallel network increases or decreases the
current, and

compare what happens when you add a resistor in series with adding another branch
to the parallel network.
Exercise 1: Series and parallel networks
Problem 1:
The diagram shows an electric circuit consisting of a battery and four resistors.
Figure 1
The potential difference (voltage) over the battery is VA=2,8 V
The resistors are rated as follows:

R1=7,2 Ω



R2=4,3 Ω
R3=7,5 Ω
R4=4,1 Ω
Assume that positive charge is flowing in the circuit (conventional current).
Using the concepts of Ohm's law, and electric circuits, determine the following:
1. What type of circuit is shown in the diagram?
2. What is the total equivalent resistance Req of the circuit?

round your answer to 1 digit after the decimal comma

use the values for any physical constants you might need, as listed here
3. What is the potential difference (voltage) across R1, or V1 ?

round your answer to 3 digits after the decimal comma

use the values for any physical constants you might need, as listed here
4. What is the potential difference (voltage) across R2, R3, and R4, or V2, V3, and V4?

round your answers to 3 digits after the decimal comma

use the values for any physical constants you might need, as listed here
Practise more questions like this
Answer 1:
1. We need to determine whether the circuit is a series, parallel, or combination type of
circuit. We do this by looking at how current flows through a circuit.
Recall that current is the movement of electric charge from a higher potential to a lower
potential.
We assume that the flowing charge is positive (conventional current). This means that
the charges start at the positive terminal of our power source (battery). Here the
charges have lots of electrical potential energy.
The charges then move towards the negative terminal through the path that is created
by the components and wires of the circuit. During this process, the electrical potential
energy is converted into thermal energy by the resistors. Therefore, the charges have
less electrical potential energy when they reach the negative terminal of the power
source (battery).
The flow of current is indicated by the arrows in the diagram below:
Figure 2
From the diagram it is clear that there is only one path for the current to flow, since the
circuit does not split into two or more paths. This means that the current flows through
all the components, one after the other (in series). We will label this current IA.
The circuit shown in the diagram is a series circuit.
0
2. We know that the circuit in the diagram is a series type circuit.
Recall that in a series circuit the equivalent resistance can be calculated by summing
the resistance values of the individual resistors:
Req,S=R1+R2+R3+…
(1)
We can rewrite (1) for the circuit in question as:
Req=R1+R2+R3+R4
(2)
We can represent this by drawing an equivalent circuit:
Figure 3
We are given the following information:

the resistance of R1=7,2 Ω

the resistance of R2=4,3 Ω

the resistance of R3=7,5 Ω

the resistance of R4=4,1 Ω
Substituting the above values in (2), we get:
Req=R1+R2+R3+R4=7,2+4,3+7,5+4,1=23,1 Ω
Therefore, the total equivalent resistance in the circuit, or Req=23,1 Ω
23.1
3. Potential difference, or voltage is a way to describe the difference in electrical potential
energy across a component in a circuit. Remember that a resistor converts electrical
potential energy into thermal energy, so the electrical potential is higher on the "in" side
compared to the "out" side of a resistor.
Ohm's law describes the relationship between the total current I through an ohmic
conductor, its resistance R, and the potential difference V across it:
I=VR
(3)
Figure 4
For the equivalent circuit, we can rewrite (3) as:
IA=VAReq
(4)
Figure 5
We are asked to calculate the potential difference V1 over R1, as shown in the circuit
diagram above, so we use Ohm's law and rewrite (3) to get:
V1=I1R1
(5)
Since the circuit in question is a series circuit, we know that the current I1 flowing
through R1, is the same as the current IA. This is represented as:
I1=IA
(6)
Substituting (6) into (5) we get:
V1=IAR1
(7)
Substituting (4) into (7) we get:
V1=IAR1=(VAReq)R1=VAR1Req
(8)
Recall from Question 2:
Req=R1+R2+R3+R4
(2)
Substituting (2) into (8) we get:
V1=VAR1R1+R2+R3+R4
(9)
We are given the following information:

the resistance of R1=7,2 Ω

the resistance of R2=4,3 Ω

the resistance of R3=7,5 Ω

the resistance of R4=4,1 Ω

the potential difference (voltage) over the battery VA=2,8 V
Substituting the above values in (9), we get:
V1=VAR1R1+R2+R3+R4=(2,8)(7,2)7,2+4,3+7,5+4,1=0,87273…
,873 V (rounded)
(calculated)≈0
Therefore, the potential difference (voltage) across R1, or V1≈0,873 V.
0.873
4. We are asked to calculate the three remaining potential differences, V2, V3, and V4.
This is shown in the diagram below:
Figure 6
5. We will use the same method to obtain three expression for V2, V3, and V4 as was
used in Question 3for V1. We get:
6. V2=VAR2R1+R2+R3+R4
(10)
7. V3=VAR3R1+R2+R3+R4
(11)
8. V4=VAR4R1+R2+R3+R4
(12)
9. We are given the following information:

the resistance of R1=7,2 Ω

the resistance of R2=4,3 Ω

the resistance of R3=7,5 Ω

the resistance of R4=4,1 Ω

the potential difference (voltage) over the battery VA=2,8 V
Substituting the above values in (10), (11), and (12) we get:
V2=VAR2R1+R2+R3+R4=(2,8)(4,3)7,2+4,3+7,5+4,1=0,52121…
,521 V (rounded)
(calculated)≈0
V3=VAR3R1+R2+R3+R4=(2,8)(7,5)7,2+4,3+7,5+4,1=0,90909…
,909 V (rounded)
(calculated)≈0
V4=VAR4R1+R2+R3+R4=(2,8)(4,1)7,2+4,3+7,5+4,1=0,49697…
,497 V (rounded)
(calculated)≈0
Therefore, the potential difference (voltage) across R2, or V2≈0,521 V.
The potential difference (voltage) across R3, or V3≈0,909 V.
The potential difference (voltage) across R4, or V4≈0,497 V.
0.521 0.909 0.497
Problem 2:
For the following circuit, calculate:
1. the current through the cell
2. the potential difference across R4
3. the current through R2
Practise more questions like this
Answer 2:
1. To find the current we first need to find the equivalent resistance. We start by
calculating the equivalent resistance of the parallel combination:
1RpRp=1R2+1R3=11+11=2=0,5 Ω
Now we have a circuit with two resistors in series so we can calculate the equivalent
resistance:
Rs=R1+R4+Rp=2+1,5+0,5=4 Ω
So the current through the cell is:
I=VR=104=2,5 A
2. The current through all the resistors is 2,5 A. (The current is the same through series
combinations of resistors and we can consider the entire parallel set of resistors as one
series resistor.)
Using this we can find the potential difference through R4:
V=I⋅R=(2,5)(1,5)=3,75 V
3. The current through all the resistors is 2,5 A. (The current is the same through series
combinations of resistors and we can consider the entire parallel set of resistors as one
series resistor.)
Using this we can find the current through R2.
We first need to find the potential difference across the parallel combination:
V=I⋅R=(2,5)(0,5)=1,25 V
Now we can find the current through R2 using the fact that the potential difference is
the same across each resistor in the parallel combination:
I=VR=1,251=1,25 A
Problem 3:
Calculate the equivalent resistance of:
1. three 2 Ω resistors in series;
2. two 4 Ω resistors in parallel;
3. a 4 Ω resistor in series with a 8 Ω resistor;
4. a 6 Ω resistor in series with two resistors (4 Ω and 2 Ω) in parallel.
Practise more questions like this
Answer 3:
1. three 2 Ω resistors in series:
RT=R1+R2+R3=(2)+(2)+(2)=6 Ω
2. two 4 Ω resistors in parallel:
1RT1RT1RTRTRT=1R1+1R2=14+14=24=42=2 Ω
3. a 4 Ω resistor in series with a 8 Ω resistor:
RT=R1+R2=(4)+(8)=12 Ω
4. a 6 Ω resistor in series with two resistors (4 Ω and 2 Ω) in parallel:
First determine the equivalent resistance of the two resistors in parallel:
1RP1RP1RPRPRP=1R1+1R2=14+12=34=43=1,33 Ω
The parallel pair are in series with the third resistor, therefore the total resistance is:
RT=R1+RP=(6)+(1,33)=7,33 Ω
6 Ω; 2 Ω ; 12 Ω; 7,33 Ω
Problem 4:
Calculate the total current in this circuit if both resistors are ohmic.
Practise more questions like this
Answer 4:
There are two resistors in parallel so we calculate the equivalent resistance of the
combination:
1Rp=1R1+1R2
1Rp=13+16=26+16=36
Rp=2 Ω
Now we can use Ohm's law to find the current:
I=VR
I=92=4,5 A
Problem 5:
Two ohmic resistors are connected in series. The resistance of the one resistor is 4 Ω. What
is the resistance of the other resistor if a current of 0,5 A flows through the resistors when
they are connected to a voltage supply of 6 V
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Answer 5:
We first use Ohm's law to find the total resistance:
R=VI
R=60,5=12 Ω
Now we can find the resistance of the other resistor:
Rs=R1+R2
12=4+R2
R2=8 Ω
Problem 6:
Determine the equivalent resistance of the following circuits:
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Answer 6:
We first determine the resistance of the parallel component:
1Rp=1R1+1R2=14+12=34
Rp=43
Now we can calculate the total resistance:
Rs=R3+Rp=2+43=103 Ω
Problem 7:
Practise more questions like this
Answer 7:
We first calculate the equivalent resistance of the parallel part:
1Rp=1R1+1R2=12+11=32
Now we calculate the total resistance:
Rs=Rp+R3+R4=32+4+6=11,5 Ω
Problem 8:
The diagram shows an electric circuit consisting of a battery and four resistors.
Figure 7
The potential difference (voltage) over the battery is VA=1,2 V
The the resistors are rated as follows:




R1=4,2 Ω
R2=2,9 Ω
R3=3,8 Ω
R4=3,5 Ω
Assume that positive charge is flowing in the circuit (conventional current).
Using the concepts of Ohm's law, and electric circuits, determine the following:
1. What type of circuit is shown in the diagram?
2. What is the total equivalent resistance Req of the circuit?

round your answer to 1 digit after the decimal comma

use the values for any physical constants you might need, as listed here
3. What is the potential difference (voltage) across R1, or V1 ?

round your answer to 3 digits after the decimal comma

use the values for any physical constants you might need, as listed here
4. What is the potential difference (voltage) across R2, R3, and R4, or V2, V3, and V4?

round your answers to 3 digits after the decimal comma

use the values for any physical constants you might need, as listed here
Practise more questions like this
Answer 8:
1. We need to determine whether the circuit is a series, parallel, or combination type of
circuit. We do this by looking at how current flows through a circuit.
Recall that current is the movement of electric charge from a higher potential to a lower
potential.
We assume that the flowing charge is positive (conventional current). This means that
the charges start at the positive terminal of our power source (battery). Here the
charges have lots of electrical potential energy.
The charges then move towards the negative terminal through the path that is created
by the components and wires of the circuit. During this process, the electrical potential
energy is converted into thermal energy by the resistors. Therefore, the charges have
less electrical potential energy when they reach the negative terminal of the power
source (battery).
The flow of current is indicated by the arrows in the diagram below:
Figure 8
From the diagram it is clear that there is only one path for the current to flow, since the
circuit does not split into two or more paths. This means that the current flows through
all the components, one after the other (in series). We will label this current IA.
The circuit shown in the diagram is a series circuit.
0
2. We know that the circuit in the diagram is a series type circuit.
Recall that in a series circuit the equivalent resistance can be calculated by summing
the resistance values of the individual resistors:
Req,S=R1+R2+R3+…
(1)
We can rewrite (1) for the circuit in question as:
Req=R1+R2+R3+R4
(2)
We can represent this by drawing an equivalent circuit:
Figure 9
We are given the following information:

the resistance of R1=4,2 Ω

the resistance of R2=2,9 Ω

the resistance of R3=3,8 Ω

the resistance of R4=3,5 Ω
Substituting the above values in (2), we get:
Req=R1+R2+R3+R4=4,2+2,9+3,8+3,5=14,4 Ω
Therefore, the total equivalent resistance in the circuit, or Req=14,4 Ω
14.4
3. Potential difference, or voltage is a way to describe the difference in electrical potential
energy across a component in a circuit. Remember that a resistor converts electrical
potential energy into thermal energy, so the electrical potential is higher on the "in" side
compared to the "out" side of a resistor.
Ohm's law describes the relationship between the total current I through an ohmic
conductor, its resistance R, and the potential difference V across it:
I=VR
(3)
Figure 10
For the equivalent circuit, we can rewrite (3) as:
IA=VAReq
(4)
Figure 11
We are asked to calculate the potential difference V1 over R1, as shown in the circuit
diagram above, so we use Ohm's law and rewrite (3) to get:
V1=I1R1
(5)
Since the circuit in question is a series circuit, we know that the current I1 flowing
through R1, is the same as the current IA. This is represented as:
I1=IA
(6)
Substituting (6) into (5) we get:
V1=IAR1
(7)
Substituting (4) into (7) we get:
V1=IAR1=(VAReq)R1=VAR1Req
(8)
Recall from Question 2:
Req=R1+R2+R3+R4
(2)
Substituting (2) into (8) we get:
V1=VAR1R1+R2+R3+R4
(9)
We are given the following information:

the resistance of R1=4,2 Ω

the resistance of R2=2,9 Ω

the resistance of R3=3,8 Ω

the resistance of R4=3,5 Ω

the potential difference (voltage) over the battery VA=1,2 V
Substituting the above values in (9), we get:
V1=VAR1R1+R2+R3+R4=(1,2)(4,2)4,2+2,9+3,8+3,5=0,35000…
,350 V (rounded)
(calculated)≈0
Therefore, the potential difference (voltage) across R1, or V1≈0,350 V.
0.350
4. We are asked to calculate the three remaining potential differences, V2, V3, and V4.
This is shown in the diagram below:
Figure 12
5. We will use the same method to obtain three expression for V2, V3, and V4 as was
used in Question 3for V1. We get:
6. V2=VAR2R1+R2+R3+R4
(10)
7. V3=VAR3R1+R2+R3+R4
(11)
8. V4=VAR4R1+R2+R3+R4
(12)
9. We are given the following information:

the resistance of R1=4,2 Ω

the resistance of R2=2,9 Ω

the resistance of R3=3,8 Ω

the resistance of R4=3,5 Ω

the potential difference (voltage) over the battery VA=1,2 V
Substituting the above values in (10), (11), and (12) we get:
V2=VAR2R1+R2+R3+R4=(1,2)(2,9)4,2+2,9+3,8+3,5=0,24167…
,242 V (rounded)
(calculated)≈0
V3=VAR3R1+R2+R3+R4=(1,2)(3,8)4,2+2,9+3,8+3,5=0,31667…
,317 V (rounded)
(calculated)≈0
V4=VAR4R1+R2+R3+R4=(1,2)(3,5)4,2+2,9+3,8+3,5=0,29167…
,292 V (rounded)
(calculated)≈0
Therefore, the potential difference (voltage) across R2, or V2≈0,242 V.
The potential difference (voltage) across R3, or V3≈0,317 V.
The potential difference (voltage) across R4, or V4≈0,292 V.
0.242 0.317 0.292