1

Copyright reserved

PHISICAL SCIENCES GRADE 11 PHYSICS

ELECTRIC CIRCUITS THEORY & EXERCISES

COMPILED BY:

G. IZQUIERDO RODRIGUEZ

2020

NORTHERN CAPE DEPARTMENT OF EDUCATION

2

Copyright reserved

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:

1. Electrical current, I, is defined as the rate of flow of charge through a circuit. (𝐼 =𝑄

∆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

which is then lost moving through the circuit. (𝑉 =𝑊

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).

3

Copyright reserved

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 of cells

Voltage, V (V)

Current, I (A)

1

2

3

4

4

Copyright reserved

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 of resistors

Voltage, V (V)

Current, I (A)

1

2

3

4

5

Copyright reserved

Possible answers for table 1 and 2

Part 1 (Varying the voltage):

Number of cells

Voltage, V (V)

Current, I (A)

1 1,5 0,002

2 3 0,0035

3 4,5 0,005

4 6 0,0065

If we draw graph of current vs. voltage we will get:

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.

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0 1 2 3 4 5 6 7

I (A

)

V (V)

Electric current vs Voltage

6

Copyright reserved

Part 2 (Varying the current):

Number of resistors

Voltage, V (V)

Current, I (A)

1 4,5 0,004

2 2,5 0,002

3 1,5 0,001

4 1,0 0,0005

If we draw a table of voltage vs. current we get:

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.

00.5

11.5

22.5

33.5

44.5

5

0 0.001 0.002 0.003 0.004 0.005

V (

V)

I (A)

Voltage vs Electric current

7

Copyright reserved

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 = I2R∆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 =W

∆t

Where:

CONDUCTORS

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

Examples:

- Reristors

- Nichrome wire

Non-Ohmic conductors

Resistance changes as their temperature changes.

Examples:

- Light bulb

- Diodes

- Transistors

8

Copyright reserved

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 chargeQ

WV

. If we rearrange the formula for calculate the work:

𝑊 = 𝑉𝑄

We can substitute in the equation to calculate power:

𝑃 =𝑊

∆𝑡=

𝑉𝑄

∆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

∆𝑡=

𝑉

1×

𝑄

∆𝑡= 𝑉𝐼

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.

9

Copyright reserved

Now doing the substitution of 𝐼 =𝑉

𝑅 in the formula to calculate power:

𝑃 = 𝑉𝐼 =𝑉𝑉

𝑅=

𝑉2

𝑅

So we can calculate the power with the following formula:

P =V2

R

Now doing the substitution of 𝑉 = 𝐼𝑅 in the formula to calculate power:

𝑃 = 𝐼𝑅𝐼 = 𝐼2𝑅 So we can calculate the power with the following formula:

𝑃 = 𝐼2𝑅 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:

P =W

∆t

P = VI P =

V2

R

P = I2R

10

Copyright reserved

PHYSICS

ELECTRIC CIRCUITS

CURRENT EMF & POTENTIAL DIFFERENCE

SERIES PARALLEL Ohm’s Law Energy used by a component (Work) W

Power

The total charge that passes through a conductor per unit of time.

𝐼 =𝑄

Δ𝑡

Measured in amperes (A)

Work done (energy transferred) per unit charge to move the charge from the negative electrode to the positive electrode in the battery.

𝜀 =𝑊

𝑞

Potential defference is the work done per unit charge between two points in a circuit.

𝑉 =𝑊

𝑄

Measured in volts (V)

321 RRRRT

321 III

321 VVVVT

(Resistors act as potential dividers.)

321

1111

RRRRE

321 VVV

321 IIIIT

(Resistors act as current dividers.)

The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature.

I

VR

General:

VItW

Series:

RtIW 2

Parallel:

tR

VW

2

Whole circuit:

ItW Measured in joules (J)

The rate at which electrical work is done or electrical energy is transferred.

𝑷 =𝑾

∆𝒕

𝑷 = 𝑽𝑰

𝑷 =𝑽𝟐

𝑹

𝑷 = 𝑰𝟐𝑹

Measured in watts (W)

1R2R 3R

2R

1R

3R

I

11

Copyright reserved

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. 1.1. Calculate the reading of the ammeter.

1.2. 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 R2 and R3 are in parallel

1

𝑅23=

1

𝑅2+

1

𝑅3

1

𝑅23=

1

10+

1

10

R23= 5 Ω

S

10 Ω

30 V 10 Ω

A

15 Ω

10 Ω

R1

R2 R3

R4

A

12

Copyright reserved

R1and R23are in series

𝑅123 = 𝑅1 + 𝑅23 𝑅123 = 10 + 5 = 15 Ω R123 and R4are in parallel

1

𝑅123=

1

𝑅123+

1

𝑅4

1

𝑅23=

1

15+

1

15

R23= 7,5 Ω 𝐼

𝑡𝑜𝑡𝑎𝑙=𝑉𝑅

𝐼𝑡𝑜𝑡𝑎𝑙=

307,5

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 = 𝐼2𝑅

𝑃 = 22 × 15 = 50 𝑊 1.3. Decreases Resistance increases Potential difference remains the same Current is inversely proportional to resistance

13

Copyright reserved

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?

(2)

V

I

(B)

V

I

(A)

V

I

(C)

V

I

(D)

A)

14

Copyright reserved

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 V1?

A Stays the same (2)

B Decreases

C Increases

D Becomes zero

15

Copyright reserved

1.3 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

B

C

D

I2

3

I

I2

I2

1

(2)

1.4 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

B C D

Light bulb, resistor, cell Resistor, light bulb, cell Cell, light bulb, variable resistor Cell, variable resistor, light bulb

(2)

16

Copyright reserved

1.5 When the potential difference across a bulb is doubled, the power will...

A

increase four times

(2)

B

double

C

stay unchanged

D decrease four times

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

I

2

1.

(2)

B

I .

C

I2 .

D I4 .

17

Copyright reserved

1.7

In the circuit represented below, switch S is closed, and the internal resistance of the cells is negligible. 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

B C D

48 kWh 4,8 kWh 0,48 kWh 480 kWh

(2)

A

V

S

18

Copyright reserved

• •

•

•

R

S

Q

P

T

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 READING ON V2

A decreases increases

B increases increases

C stays the same decreases

D increases decreases

(2)

1.10 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.

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 Increases Increases Increases

D Decreases Increases Decreases

(2)

[20]

19

Copyright reserved

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.

2.1 State Ohm's law in words (2)

2.2 Calculate the

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.

Explain the answer.

(4)

2.3.2 Reading on the voltmeter? Choose from INCREASES, DECREASES

or REMAINS THE SAME.

Explain the answer.

(4)

[19]

A

5 Ω

R1 = 4 Ω

12 V

V

R2 = 4 Ω

S

R

20

Copyright reserved

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.

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 S2 is closed?

Write down only INCREASES, DECREASES or REMAINS THE SAME.

Explain the answer.

(3) [11]

S2

R1

12 V

V

A

S1

R4

R2 R3

21

Copyright reserved

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) [15]

22

Copyright reserved

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.

5.1 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)

5.2 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)

V

A

S

3 Ω

2 Ω

4 Ω

23

Copyright reserved

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.

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.

5.3.1 Write down TWO variables that the learners would have controlled (keep

constant) in each of the experiments.

(2) 5.3.2 Which one (P or Q) is a better conductor? Explain your answer. (3)

[21]

WIRE

V

A

Potential difference (V)

C

urr

ent

(A)

Wire Q

Wire P

24

Copyright reserved

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.

6.1 State Ohm’s law in words (2)

6.2 Calculate:

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. (3)

6.4 How will the reading of the ammeter be affected if switch S is opened?

Write down only INCREASES, DECREASES or REMAINS THE SAME.

Explain the answer.

(3)

[18]

Rx

B A

12 V

A

S

25

Copyright reserved

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) [12]

26

Copyright reserved

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? (2)

8.2 The graph deviates from Ohm's law at some point.

8.2.1 Write down the coordinates of the plotted point on the graph beyond

which Ohm's law is not obeyed.

(2) 8.2.2 Write down a possible reason for the deviation from Ohm's law as

shown in the graph. Assume that all measurements are correct.

(2) 8.3 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) [10]

27

Copyright reserved

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:

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)

[16]

Number of experiment

Strength of electric current

(A)

Potential difference

(V)

1 0,02 6

2 0,04 12

3 0,06 18

4 0,08 24

28

Copyright reserved