NORTHERN CAPE DEPARTMENT OF EDUCATION
PHISICAL SCIENCES
GRADE 11
PHYSICS
ELECTRIC CIRCUITS
THEORY & EXERCISES
COMPILED BY:
G. IZQUIERDO RODRIGUEZ
2020
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ELECTRIC CIRCUITS: Current, Ohm’s law, electromotive force, energy and power in
an electric circuit.
LEARNING OBJECTIVES:





Define current in words.
State Ohm's law in words.
Determine the relationship between current, potential difference and resistance at
constant temperature using a simple circuit.
State the difference between ohmic conductors and non-ohmic conductors and give
an example of each.
Define power in words.
In Grade 10 we learned 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:
𝑄
1. Electrical current, I, is defined as the rate of flow of charge through a circuit. (𝐼 = ∆𝑡)
2. The conventional direction of the current in a circuit is from the positive pole of the battery
(cell) around the circuit to the negative pole of the battery (cell).
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
𝑊
which is then lost moving through the circuit. (𝑉 = 𝑄 )
3. Resistance, R, is an internal property of a circuit element that opposes the flow of charge
and it is the ratio of the voltage applied across a piece of material to the current that flows
in that material. Work must be done for a charge to move through a resistor.
Ohm’s Law
The potential difference across a conductor is directly proportional to the current in
the conductor at constant temperature. (𝑉 ∝ 𝐼)
R=
V
I
OR
V = IR
OR
The current in a conductor is directly proportional to the potential difference across
the conductor at constant temperature. (𝐼 ∝ 𝑉)
I=
V
R
Where:
I is the current through the conductor
V is the voltage across the conductor
R is the resistance of the conductor.
In the Ohm’s law relationship the proportionality constant is the resistance of the
conductor and it is measured in ohms (Ω).
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Experiment 1:
Aim: To determine the relationship between the current going through a resistor and the
potential difference (voltage) across the same resistor.
Apparatus:
 4 cells
 4 resistors
 an ammeter
 a voltmeter
 connecting wires
Method:
This experiment has two parts.
In the first part we will vary the applied voltage across the resistor and measure the
resulting current through the circuit.
In the second part we will vary the current in the circuit and measure the resulting
voltage across the resistor.
After obtaining both sets of measurements, we will examine the relationship between the
current and the voltage across the resistor.
Part 1 (Varying the voltage):
Step 1: Set up the circuit according to the following circuit diagram.
Step 2: Measure the voltage across the resistor using the voltmeter, and the current
in the circuit using the ammeter.
Step 3: Add one more 1,5 V cell to the circuit and repeat your measurements.
Step 4: Repeat until you have four cells.
Step 5: Recall your result in the following table.
Number
cells
1
2
3
4
of
Voltage, V
(V)
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Current,
(A)
I
Part 2 (Varying the current):
Step 1: Set up the circuit according to the following circuit diagram.
Step 2: Measure the current and measure the voltage across the single resistor.
Step 3: Now add another resistor in series in the circuit and measure the current and
the voltage across only the original resistor again. Continue adding resistors
until you have four in series, but remember to only measure the voltage
across the original resistor each time.
Step 4: Enter the values you measure into the following table.
Number
resistors
1
2
3
4
of
Voltage, V
(V)
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Current,
(A)
I
Possible answers for table 1 and 2
Part 1 (Varying the voltage):
Number
cells
1
2
3
4
of
Voltage, V
(V)
1,5
3
4,5
6
Current, I
(A)
0,002
0,0035
0,005
0,0065
If we draw graph of current vs. voltage we will get:
Electric current vs Voltage
0.007
0.006
I (A)
0.005
0.004
0.003
0.002
0.001
0
0
1
2
3
4
5
6
7
V (V)
We can see that the graph is a straight line which shows that as the voltage (V) across a
metal conductor (resistor) increases the current (I) also increases.
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Part 2 (Varying the current):
Number
resistors
1
2
3
4
of
Voltage, V
(V)
4,5
2,5
1,5
1,0
Current,
(A)
0,004
0,002
0,001
0,0005
I
If we draw a table of voltage vs. current we get:
V (V)
Voltage vs Electric current
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
0.001
0.002
0.003
0.004
0.005
I (A)
This graph is also a straight line which shows that as the current (I) increases in a metal
conductor (resistor) the voltage increases too.
The results obtained verify Ohm's Law because the current (I) through a metal conductor,
at a constant temperature, in a circuit is proportional to the voltage (V) across the
conductor and the potential difference across a conductor is directly proportional to the
current in the conductor at constant temperature.
Oohmic conductors and non-ohmic conductors
Electrical conductors can either be ohmic conductors or non-ohmic conductors.
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CONDUCTORS
Ohmic conductors
Non-Ohmic conductors
Have a constant resistance when the
voltage is varied across them or the
current through them is increased.
A graph of current vs. voltage will be
straight-line
Resistance changes as their
temperature changes.
Examples:
- Light bulb
- Diodes
- Transistors
Examples:
- Reristors
- Nichrome wire
Electromotive force (emf)
Emf is the work done (energy transferred) per unit charge to move the charge from the
negative electrode to the positive electrode in the battery.
OR
The emf of an emf device is the work per unit charge that the device does in moving charge
from its low-potential terminal to its high-potential terminal.
Energy and power in an electric circuit
The component of a circuit converts electrical energy from a battery or the mains to other
forms of energy such as heat, light, mechanical and chemical energy.
The energy (W) used by a component in a circuit depends on the potential difference across
the component (V), the current (I) and the time the current flows (∆t). Energy is measured in
joules (J)
W = Vq OR W = VI∆t OR W = I2 R∆t
OR
W=
V2
R
∆𝑡
Power of a device or appliance is the rate at which electrical energy is transformed or
converted in an electrical circuit.
OR
Power is the rate at which work is done.
We can write this mathematically:
P=
Where:
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W
∆t
𝑃 is power measure in watts (W). An appliance use one watt of power when it converts
one joule of energy in one second. (1 𝑊 = 𝐽/𝑠).
𝑊 is the work done and is measure in joules (J).
∆𝑡 is the interval of time measure in (s).
This expression of power can also be written as:
W = P∆t
In Grade 10 we learned that the potential difference is the work done per unit charge V 
W
Q
. If we rearrange the formula for calculate the work:
𝑊 = 𝑉𝑄
We can substitute in the equation to calculate power:
𝑃=
𝑊 𝑉𝑄
=
∆𝑡 ∆𝑡
P=
VQ
∆t
In grade 10 we also learned that current is the rate of flow of charge (charge per unit time):
𝐼=
𝑄
∆𝑡
We can substitute in the equation of power:
𝑃=
𝑊 𝑉 𝑄
= ×
= 𝑉𝐼
∆𝑡 1 ∆𝑡
So in electric circuits, power is a function of both voltage and current and we talk about the
power dissipated in a circuit element.
P = VI
This equation gives us the power converted by any device, where I is the current that passes
through it and V is the potential difference across it.
We also substitute Ohm's Law in the equation to calculate power.
We know that I
=
V
R
and so V
= IR.
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𝑉
Now doing the substitution of 𝐼 = 𝑅 in the formula to calculate power:
𝑉𝑉 𝑉 2
𝑃 = 𝑉𝐼 =
=
𝑅
𝑅
So we can calculate the power with the following formula:
V2
P=
R
Now doing the substitution of 𝑉 = 𝐼𝑅 in the formula to calculate power:
𝑃 = 𝐼𝑅𝐼 = 𝐼 2 𝑅
So we can calculate the power with the following formula:
𝑃 = 𝐼2 𝑅
Any of these equation for power can be used to solve a problem, it depends on what is given
and what is been asked.
We have the following equations for power:
W
P=
∆t
V2
P=
R
P = VI
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P = I2R
PHYSICS
ELECTRIC CIRCUITS
CURRENT
The total
charge that
passes through
a conductor per
unit of time.
𝑄
𝐼=
Δ𝑡
Measured in
amperes (A)
EMF &
POTENTIAL
DIFFERENCE
Work done (energy
transferred) per unit
charge to move the
charge from the
negative electrode to
the positive
electrode in the
battery.
𝑊
𝜀=
𝑞
Potential
defference is the
work done per unit
charge between two
points in a circuit.
𝑊
𝑉=
𝑄
SERIES
R1
PARALLEL
R2
R3
Ohm’s Law

R1
Energy used by
a component
(Work) W
General:
W  VIt
Series:
W  I 2 Rt
R2
I
R3
RT  R1  R2  R3
I1  I 2  I 3
VT  V1  V2  V3
(Resistors act as
potential dividers.)
Measured in
volts (V)
1
1
1
1



RE R1 R2 R3
V1  V2  V3
I T  I1  I 2  I 3
(Resistors act as current
dividers.)
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Parallel:
The potential difference
across a conductor is
directly proportional to
the current in the
conductor at constant
temperature.
R
V
I
W
V2
t
R
Whole circuit:
W  It
Measured in
joules (J)
Power
The rate at
which
electrical
work is
done or
electrical
energy is
transferred.
𝑾
𝑷=
∆𝒕
𝑷 = 𝑽𝑰
𝑽𝟐
𝑷=
𝑹
𝑷 = 𝑰𝟐 𝑹
Measured
in watts
(W)
EXAMPLE
QUESTION 1
1. In the circuit diagram below the resistance of the battery, wires and ammeter are
negligible.
The emf of the battery is 30 V.
S
15 Ω
10 Ω
10 Ω
30 V
10 Ω
A
1.1.
1.2.
Calculate the reading of the ammeter.
Calculate the power dissipated by the 15 Ω resistor.
1.3.
How will the reading on the ammeter be affected if the switch S is opened? Write
down only INCREASES, DECREASES or REMAINS THE SAME. Briefly explain
the answer.
SOLUTION
QUESTION 1
R1
R4
R2
R3
A
R2 and R3 are in parallel
1
1
1
=
+
𝑅23 𝑅2 𝑅3
1
1
1
=
+
𝑅23 10 10
R23= 5 Ω
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R1and R23are in series
𝑅123 = 𝑅1 + 𝑅23
𝑅123 = 10 + 5 = 15 Ω
R123 and R4are in parallel
1
1
1
=
+
𝑅123 𝑅123 𝑅4
1
1
1
=
+
𝑅23 15 15
R23= 7,5 Ω
𝐼𝑡𝑜𝑡𝑎𝑙=𝑉
𝑅
𝐼𝑡𝑜𝑡𝑎𝑙= 30
7,5
𝐼𝑡𝑜𝑡𝑎𝑙=4 𝐴
𝐼𝑇 = 𝐼1 + 𝐼4
R123 = R4= 15 Ω
𝐼1 = 𝐼4
𝐼𝑇 = 2𝐼1
𝐼1 = 𝐼4 = 2 𝐴
𝐼1 = 𝐼2 + 𝐼3
𝐼2 = 𝐼3
2 = 2𝐼2
𝐼2 = 1 𝐴
𝐼𝑜 = 𝐼2 + 𝐼4
𝐼𝑜 = 1 + 2 = 3 𝐴
1.2 𝑃 = 𝐼 2 𝑅
𝑃 = 22 × 15 = 50 𝑊
1.3. Decreases
Resistance increases
Potential difference remains the same
Current is inversely proportional to resistance
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QUESTION 1: MULTIPLE-CHOICE QUESTIONS
Four options are provided as possible answers to the following questions. Each question
has only ONE correct answer. Choose the answer and write only the letter (A–D) next
to the question number (1.1–1.10) in the ANSWER BOOK, for example 1.11 D.
1.1
Which ONE of the following graphs correctly represents the relationship between
potential difference and current for an ohmic conductor?
V
V
I
I
(B)
(A)
V
V
I
I
(C)
(D)
A)
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(2)
1.2
The circuit diagram shows two light bulbs of resistance 4 ῼ and 2 ῼ each
connected in parallel in the circuit. The two resistors of resistance 4 ῼ and 6 ῼ
each are connected in series to the circuit.
If the 2 ῼ light bulb burns out, what happens to the reading on V 1?
A
Stays the same
B
Decreases
C
Increases
D
Becomes zero
(2)
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1.3
1.4
The three resistors in the circuit diagram shown below are identical. If the
reading on the ammeter A1 is I what will the reading be on A2?
A
3
I
2
B
I
C
2I
D
1
I
2
(2)
Consider the three circuit components represented below.
Which ONE of the options below best represents the names of the
components in the correct sequence, from left to right?
A
Light bulb, resistor, cell
B
Resistor, light bulb, cell
C
Cell, light bulb, variable resistor
D
Cell, variable resistor, light bulb
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(2)
1.5
When the potential difference across a bulb is doubled, the power will...
A
increase four times
B
double
C
stay unchanged
D
decrease four times
(2)
1.6
Two resistors of equal resistance are connected in SERIES to a battery with
negligible internal resistance. The current through the battery is I.
When the two resistors are connected in PARALLEL to the same battery, the
current through the battery will be …
A
B
1
I.
2
I.
C
2I .
D
4I .
(2)
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1.7
In the circuit represented below, switch S is closed, and the internal resistance
of the cells is negligible.
A
S
V
How will the ammeter and voltmeter readings change if switch S is opened?
AMMETER READING
VOLTMETER READING
A
increase
Increase
B
Increase
No change
C
Decrease
No change
D
Decrease
Decrease
(2)
1.8
How much energy in kWh is needed to keep a 60 W light on for 8 hours
A
48 kWh
B
4,8 kWh
C
0,48 kWh
D
480 kWh
(2)
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1.9
In the following circuit diagram, V1 and V2 are voltmeters.
When switch S in the diagram is closed, the readings on V1 and V2 respectively
change as follows:
READING ON V1
1.10
READING ON V2
A
decreases
increases
B
increases
increases
C
stays the same
decreases
D
increases
decreases
(2)
The four resistors P, Q, R and T in the circuit below are identical. The cell has
an emf  and negligible internal resistance. The switch is initially CLOSED.
P
S •
•
•

Q
R
•
T
Switch S is now OPENED. Which ONE of the following combinations of
changes will occur in P, R and T?
CURRENT IN P
CURRENT IN R
CURRENT IN T
A
Decreases
Remains the same
Decreases
B
Increases
Remains the same
Increases
C
D
Increases
Decreases
Increases
Increases
Increases
Decreases
(2)
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QUESTION 2
The battery in the circuit diagram below has an emf of 12 V and internal resistance is
negligible.Resistor R has an unknown resistance. The reading of the ammeter is 2 A when
the switch S is OPEN.
A
12 V
R
V
5Ω
S
R1 = 4 Ω
R2 = 4 Ω
2.1
State Ohm's law in words
2.2
Calculate the
(2)
2.2.1
Total resistance.
(3)
2.2.2
Reading on the voltmeter.
(6)
Switch S in now CLOSED.
2.3
How does this change affect the:
2.3.1
Reading on the ammeter? Choose from INCREASES, DECREASES
or REMAINS THE SAME.
(4)
Explain the answer.
2.3.2
Reading on the voltmeter? Choose from INCREASES, DECREASES
or REMAINS THE SAME.
(4)
Explain the answer.
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QUESTION 3
In the circuit diagram below the emf of the battery is 12 V and the internal resistance is
negligible. The resistance of resistors R1, R2 and R3 is the same equal to 6 Ω, the
resistance of resistor R4 is 3 Ω. Switch S1 is closed and S2 is open. Ignore the resistance
of the wires.
V
12 V
S1
R4
S2
A
R1
R2
R3
3.1
State Ohm's law in words.
(2)
3.2
Calculate the reading of the ammeter.
(6)
3.3
How will the reading on the ammeter be affected if switches S 2 is closed?
Write down only INCREASES, DECREASES or REMAINS THE SAME.
(3)
[11]
Explain the answer.
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QUESTION 4
In the following circuit diagram the internal resistance of the battery and the resistance
of the conductors can be neglected. The emf of the battery is 12 V.
4.1
Calculate the equivalent resistance of the circuit.
(4)
4.2
What is the reading of ammeter A2?
(4)
4.3
Calculate the power dissipated in resistor R1?
(3)
4.4
Switch S is now opened. Will the power dissipated in RESISTOR R1 be
affected? Write down only YES or NO.
Explain the answer without doing calculations.
(4)
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QUESTION 5
Three resistors, 3 Ω, 2 Ω and 4 Ω are connected to a battery of 12 V as shown in circuit
diagram below. The resistance of the battery and that of the connecting wires are
negligible.
V
A
2Ω
S
3Ω
4Ω
5.1
5.2
Calculate :
5.1.1
the total resistance of the circuit.
(5)
5.1.2
the reading on the ammeter.
(4)
5.1.3
the voltage across the parallel resistors.
(4)
How will the reading on the ammeter be affected if switch S is opened? Write
down only INCREASES, DECREASES or REMAINS THE SAME.
Explain your answer.
(3)
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5.3
Learners investigate the conducting ability of two metal wires P and Q, made
of different materials. They connect one wire at a time in a circuit as shown
below.
A
WIRE
V
The potential difference across each wire is increased in equal increments and
resulting current through this wires is measured. Using the measurements, the learners
obtained the following sketch graphs for each of the wires.
Wire Q
Current (A)
Wire P
Potential difference (V)
5.3.1
5.3.2
Write down TWO variables that the learners would have controlled (keep
constant) in each of the experiments.
Which one (P or Q) is a better conductor? Explain your answer.
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(2)
(3)
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QUESTION 6
Two resistors A of 3 Ω and B of 2 Ω, a 12 V battery with a negligible internal resistance, an
unknown resistor Rx and an ammeter are connected as indicated in the circuit diagram below.
The ammeter reading is 2 A.
12 V
S
A
A
B
Rx
6.1
State Ohm’s law in words
6.2
Calculate:
(2)
6.2.1
The total resistance in the circuit.
(3)
6.2.2
The resistance of the unknown resistor Rx
(4)
6.2.3
The strength of the current flowing through resistors A and B.
(3)
6.3
Calculate the power dissipated in the unknown resistor Rx.
6.4
How will the reading of the ammeter be affected if switch S is opened?
(3)
Write down only INCREASES, DECREASES or REMAINS THE SAME.
Explain the answer.
(3)
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QUESTION 7
The circuit diagram shows a 12 V battery connected to an ammeter, three resistors and
a voltmeter as shown in the diagram. The battery has no internal resistance.
7.1
Calculate the equivalent (effective) resistance of the parallel resistors.
(3)
7.2
The reading on the ammeter is 2 A. What is the reading on the voltmeter?
(3)
7.3
Determine the magnitude of the current through the 4 resistor.
(3)
7.4
How much energy is dissipated in the 12 Ω resistor in 1 minute?
(3)
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QUESTION 8
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.
8.1
Which one of the measured quantities is the dependent variable?
8.2
The graph deviates from Ohm's law at some point.
8.2.1
8.2.2
8.3
(2)
Write down the coordinates of the plotted point on the graph beyond
which Ohm's law is not obeyed.
(2)
Write down a possible reason for the deviation from Ohm's law as
shown in the graph. Assume that all measurements are correct.
(2)
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)
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QUESTION 9
A grade 12 learner, is investigating the relationship between potential difference and current
using a resistor with unknown resistance. He set up the circuit shown in the diagram. Thabo
adjusts the current in the circuit using the rheostat. He takes the ammeter reading every time
while the voltmeter is disconnected. He then measures the potential difference across the
unknown resistor for each current value I. He drafts the following table:
Number of
experiment
1
2
3
4
Strength of
electric current
(A)
0,02
0,04
0,06
0,08
Potential
difference
(V)
6
12
18
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9.1. Write an investigative question for this investigation.
(2)
9.2. What could Thabo’s hypothesis have been for this experiment?
(2)
9.3. Write down the:
9.3.1 dependent variable
9.3.2 independent variable?
(2)
9.4. Plot his experimental data on a graph.
(4)
9.5. Calculate the gradient of the graph.
(3)
9.6. What quantity does the gradient of the graph represent?
(1)
9.7. Write a conclusion for this experiment.
(2)
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