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OCR AA-Level Physics

10. Electrical Circuits2019-2020

Name:________________

Physics Teacher:______________

Year 12

Spec Ref.

Checklist Textbook Page

Reviewed Practiced Confident

4.3.1 a Can you explain Kirchhoff’s second law and the conservation of energy?

170

4.3.1 b Can you describe Kirchhoff’s first and second laws applied to electrical circuits?

170-172

4.3.1 c Can you determine the total resistance of two or more resistors in series using R = R1 + R2 + …?

174

4.3.1 dCan you determine the total resistance of two or

more resistors in parallel using

1R

= 1R1

+ 1R2

+. ..?

174-175

4.3.1 e Can you analyse circuits with components both in series and in parallel?

177-180

4.3.1 f Can you analyse circuits with more than one source of e.m.f.?

177-180

4.3.2 a Can you define source of e.m.f. and internal resistance?

181

4.3.2 b Can you define terminal p.d. and ‘lost volts’? 181

4.3.2 c i

Can you use the equations Ɛ = I (R + r) and Ɛ = V + I r?

182

4.3.2 c ii

Can you list and describe techniques and procedures used to determine the internal resistance of a chemical cell or other source of e.m.f.?

183-184

4.3.3 a Can you analyse a potential divider circuit with components?

186-187

4.3.3 b Can you use potential divider circuits with variable components, e.g., LDRs and thermistors?

189-190

4.3.3 c i

Can you use potential divider equations

V out=R2

R1+R2×V in

and

V 1

V 2=R1

R2 ?

186-187

4.3.3 c ii

Can you list and describe techniques and procedures used to investigate potential divider circuits which may include a sensor such as a thermistor or a LDR?

189-190

210. Electrical Circuits

Module 4: Electrons Waves and Photons 10. Electric Circuits In this topic you will gain a deeper knowledge of circuits and how they work. You will learn another of Kirchhoff’s Laws and analyse how resistance, p.d and current behave in both series and parallel circuits. You would also learn in detail about internal resistance and the behaviour of potential divider circuits both with and without sensing components such as the Thermistor and LDR. PAGs 3 and 4 will also be completed in this topic

Lesson Content overview Specification reference:Learners should be able to demonstrate and apply their knowledge and understanding of:

1 Kirchhoff’s Laws 4.3.1 Series and Parallel Circuits

(a) Kirchhoff ’s second law; the conservation of energy

(b) Kirchhoff ’s first and second laws applied to electrical circuits

2 Combining Resistors

4.3.1 Series and Parallel Circuits

(c) total resistance of two or more resistors in series; R = R1 + R2 + …

(d) total resistance of two or more resistors in parallel;

1R

= 1R1

+ 1R2

+…

3/4 Analysing Circuits 4.3.1 Series and Parallel Circuits

(e) analysis of circuits with components, including both series and parallel

(f) analysis of circuits with more than one source of e.m.f.

5 Internal Resistance

4.3.2 Internal Resistance

(a) source of e.m.f.; internal resistance

(b) terminal p.d.; 'lost volts'

(c) (i) the equations = I (R + r) and = V + Ir

(c) (ii) techniques and procedures used to determine the internal resistance of a chemical cell or other source of e.m.f.

6 Potential Divider Circuits

4.3.3 Potential Dividers

(a) potential divider circuit with components

(c) (i) potential divider equations e.g.

V out=R2

R1+R2×V ¿ and

V 1

V 2=R1

R2

7 Sensing Circuits 4.3.3 Potential Dividers

(b) potential divider circuits with variable components e.g. LDR and thermistor

(c) (ii) techniques and procedures used to investigate potential divider circuits which may include a sensor such as a thermistor or an LDR.

Reminder: Notes in blue dashed boxes like this contain suggested follow-up/consolidation work. If these are not done in lessons, you should be working through these to ensure you have grasped the content from lessons.


Kirchhoff’s Laws

Kirchhoff’s First Law (Recap)

Kirchhoff’s First Law states that:

‘For any point on an electrical circuit, the sum of the currents into that point is equal to the sum of the currents coming out of that point.’

This can be expressed mathematically as:

∑ I ¿=∑ I out

In the following example this would mean that:

I 1+ I 2=I 3+ I 4 + I 5

This law is a consequence of the conservation of charge. Like energy, charge cannot be created or destroyed. Kirchhoff’s Second Law

Kirchhoff’s Second Law states that:

‘in any circuit the sum of the electromotive force is equal to the sum of the potential difference in a closed loop.’

This can be expressed mathematically as:

∑ ε=∑V∈a closed loop.

In the following example this would mean that:

This law is a consequence of the conservation of energy. This is in part due to e.m.f. and p.d. being energy transferred per unit charge.

Important reminder: Electromotive force is not a force


Worked Example (in class)

Suggested Follow-up Work

Summary Questions on pages 172 and 173 (Blue Textbook) Worksheets 1 and 2 on pages 23-25 of this booklet.


Combining Resistors

In this course, you need to be able to work out the overall resistance of both series and parallel circuits.

Resistance in Series

Due to Kirchhoff’s Second Law we know that the sum of the p.d across each resistor in series will be equal to the total e.m.f:

ε=V 1+V 2+…

By substituting in V=I ×R we get the expression:

IR=I1 R1+ I 2R2+…

Since the resistors are in series, the same value of current is running though each of them ( I=I 1=I 2=…). This yields the relationship for resistors connected in series:

R=R1+R2+…

Resistance in Parallel

Due to Kirchhoff’s First Law we know that the sum of the currents through each resistor in parallel is equal the the current through the power supply:

I=I 1+ I 2+…

By substituting in V=I ×R we get the expression:

VR

=V 1

R1+V 2

R2+…

Since the resistors are in parallel, Kirchhoff’s second law implies that the p.d across each resistor will be equal to the e.m.f. This yields the relationship for resistors connected in parallel:

1R

= 1R1

+ 1R2

+…

In some circuits you may need to use a combination of both relationships.


Worked Examples

Equations Used:

V=IR R=R1+R2+… 1R

= 1R1

+ 1R2

+…

Use the diagram of an electric circuit to answer the questions.

1. What is the total resistance of the circuit?1R p

=13+ 1

16 Rp=2.526…RT=12+2.526…RT=14.5Ω

2. What is the value of the current through the 3 resistor?

I T=VRT

I T=16

14.5=1.1 AV 12=1.1×12V 12=13.2V V 16=V 3=16−13.2V 3=2.8 I 3=

2.83

I 3=0.9A


Summary Questions on page 176 (Blue Textbook) Worksheets 3 and 4 on pages 26-29 of this booklet.


Now find the total resistance

Calculate the resistance of the parallel resistors first.

16

3

12

16v

Calculate the current through the power supply

Calculate what share of the e.m.f the 12 resistor gets

Calculate what share of the e.m.f the 3 resistor gets

Finally use I=VR to find the

current in the 3 resistor

Worked Example (In Class)

Equations Used:

V=IR R=R1+R2+… 1R

= 1R1

+ 1R2

+…

Calculate the current and potential difference for each resistor.


Analysing Circuits

When analysing circuits, you will need to utilise a variety of equations and laws from your tool box. Theses are:

a. Kirchhoff’s Laws:

∑ I ¿=∑ I out at any point∈acircuit∑ ε=∑V∈a closed loop.

b. Resistor Combinations:series :R=R1+R2+…

¿ : 1R

= 1R1

+ 1R2

+…

c. Key Electricity Equations:

V=I ×R∆Q=I ×∆ t P=I ×V W=Q×V

Top Tips for Analysing Circuits

Sketch a quick table of what you have and highlight what you need. o You will not always be able to get to the answer in one step. Filling in gaps in a table can

help you make steps towards your answer. Use subscripts to denote values for each component (e.g. R1, VLDR)

o Don’t lose silly marks by getting your numbers all mixed up! Know your units

o Watch out for prefixes as well as labelled values.


Summary Questions on page 180 (Blue Textbook) Worksheets 5 and 6 on pages 30-38 of this booklet

Worked Example (in class)

Equations Used:

V=IR P=IV Q=¿W=QV R=R1+R2+… 1R

= 1R1

+ 1R2

+…

Complete the table of values for the circuit

Component Current [A] P.d. [V] Resistance [] Power [W]

Battery

Resistor A


A

B

C

D

Resistor B

Resistor C

Resistor D

Total Circuit

Working Space:


Internal Resistance

All sources of e.m.f has some internal resistance. This means that not all of the energy transferred to the charge carriers is available to the circuit, as some is transferred to the internal resistance of the cell (r). We represent this in circuit using the following symbol:

The internal resistance of the power supply results in a difference between the measured p.d. across the terminals of the power supply (terminal p.d.) and the actual e.m.f. of the cell, which is referred to as the ‘lost volts’.

lost volts=electromotive force−terminal p .d .

The lost volts are equal to the p.d. across the internal resistor: lost volts=I × r

EMF: Energy changed from some form to electrical energy per unit charge (to drive charge round a complete circuit). For a battery: Energy transfer per unit charge from chemical to electrical form

Internal resistance: The voltage across the battery’s terminals V drops below the battery’s e.m.f. when a current 𝐼 is drawn. This can be modelled by treating the battery as having a resistance of its own, 𝑟, which

is referred to as the internal resistance. Some energy is transferred into thermal energy in the battery (when it drives charge).

Things to note:

For very small values of current: ε ≈V As the current in the circuit increases, the ‘lost volts’ will also increase. The terminal p.d. is always smaller that the e.m.f.

EquationsFrom the relationship between e.m.f and terminal p.d. (V) we can get the equation:

ε=V +lost volts ε=V + Ir

where is the electromotive force in Volts [V], V is the terminal p.d. in Volts [V], I is the current in Amps [A] and r is the internal resistance of the power supply in Ohms [].

Since V=I ×R we can also write:

ε=IR+ Ir ε=I (R+r )

Where R is the resistance of the external circuit in Ohms [].Note that this is essentially V=I ×R for the whole circuit, taking into account r.

Key Experiment – e.m.f and Internal Resistance


A typical way to determine the e.m.f of a supply and its internal resistance is to measure the change in current and terminal p.d. for a circuit with varying resistance. This can be done easily using the circuit to the right.

Taking the equation for internal resistance and rearranging it yields the following:

ε=V + Ir

becomes:V=−rI+ε

Note that if terminal p.d. (V) is plotted against current (I) we have an equations of a straight line ( y=mx+c ¿.

For this graph the y-intercept is the electromotive force (ε) and the gradient is the negative internal resistance (-r).


Summary Questions on page 185 (Blue Textbook) Practice Questions 1, 2, 6 and 7 (pages 191-193 of your blue textbook) Worksheets 7 and 8 on pages 39-45 of this booklet

Worked Example


Equations Used:

V=IR ε=V + Ir

A battery has e.m.f. = 12.5V and is connected to a resistor of 20.0 Ω. The terminal p.d., V, is 12.0 V.

a. Draw a circuit diagram, showing the internal resistance, r, in series with the battery.

b. Calculate:I. the lost volts

lost volts=ε−terminal p .d lost volts=12.5−12.0 lost volts=0.5V

II. the current in the resistor

V=I ×R IR=V R

RRIR=

12.020.0

IR=0.6 A

III. the internal resistance of the battery.

IR=Ir I r=0.6A ε=V +Ir r= ε−VI

r=12.5−12.00.6 r=0.83Ω

Worked Examples (In Class)1. A 12V battery with an internal resistance of 1.5 is connected in series with a resistor. The resistor

draws a current of 3A. What is the terminal p.d. of the battery and what is the resistance of the resistor?


Make use of your analysing circuit toolkit

Using the theory of ‘lost volts’

Rearrange internal resistance equation

2. Sketch a graph of V against I for a power supply with internal resistance. Label on the e.m.f. and the internal resistance. On the same graph sketch a second line to show a power supply with the same e.m.f. but a larger internal resistance.


Stretch Content: Connecting Cells (with Internal Resistance)

Cells (or power sources) can be connected together to produce either higher e.m.f or higher current. Depending on which one you want you would connect them in series or in parallel.

= 3V; r = 1

If cells are connected in series the available e.m.f increases. However, this also causes an increase in internal resistance which limits the current the circuit can produce.

= 1.5V; r = 0.25

If two identical cells are connected in parallel, they produce the same e.m.f as on cell but have a much smaller internal resistance. This means that a larger current can be provided to the circuit.

Some Questions to think about


Potential Divider CircuitsIn a potential divider circuit, the potential difference (p.d.) across the output can be varied when connected to a fixed input (fixed e.m.f.).

This is done by sharing the p.d. across two resistors (one of which is connected to the output). Due to Kirchhoff’s second law we know that the p.d.is shared between components in series. The p.d. in fact splits in a ratio proportional to the resistance of each component.

V A

V B=R A

RB

where VA is the p.d. across resistor of value RA and VB is the p.d. across resistor of value RB.

Let VA be equal to Vout so that RA is equal to R2: and let VB be equal to Vin so that RB is equal to R!+R2:

V out

V ¿=

R2

R1+R2

By rearranging this equation, you yield the potential divider equation:

V out=( R2

R1+R2)×V ¿

Top Tip: Remember that the resistor connected to the output is always R2 in this equation so must go on the top of the ratio.


Summary Questions on page 188 (Blue Textbook) Practice Question 4 (pages 191-192 of your blue textbook) Worksheet on Potentials Divider Circuits (Worksheet 9 – Pages 4647)


Worked Examples

Equations Used:

V out=( R2

R1+R2)×V ¿

1. A 150 resistor and a 250 resistor are connected as part of a potential divider circuit to a 12V supply. The output is connected across the 250 resistor. Calculate the output p.d. of the circuit.

V out=( R2

R1+R2)×V ¿

V out=( 250150+250 )×12V out=( 250

400 )×12

V out=( 250400 )×12

V out=7.5V2. A potential divider is connected to a 300V power source with negligible internal resistance. The

resistance of R1 is 50. Calculate the resistance of R2 is the output p.d. is 100.

V out

V ¿=

R2

R1+R2

V out×(R1+R2)=V ¿×R2

V out R1+V outR2=V ¿R2

V out R1=V ¿R2−V outR2

R2=V out R1

V ¿−V out

R2=100×50

300−100

R2=25Ω

Worked Examples (in class)

Equations Used:

V out=( R2

R1+R2)×V ¿

1. A 600 resistor and a 250 resistor are connected as part of a potential divider circuit to a 80V supply. The output is connected across the 600 resistor. Calculate the output p.d. of the circuit.


Use the potential divider equation

Make sure you correctly identified R2 from the question

Rearrange the potential divider equation for R2

Identify correct quantities from the question,

2. Sketch a potential divider circuit that would produce an output p.d. of 12V for an input p.d of 36V.


Stretch Content: Loading a Potential Divider

We say a potential divider is loaded when a component or circuit is connected across Vout. This would mean you are connecting component(s) in parallel with R2.

On the left is the original potential divider circuit and on the right is the loaded potential divider circuit.

Due to the rules of combining resistors, this causes the the resistance of this part of the circuit to decrease. As a result of this, Vout will now receive a smaller fraction of the overall p.d. of the circuit.

If the load added to the circuit has a high resistance, it has little to no effect on Vout. However, if the load has a small resistance, Vout is significantly reduced.

Some questions to think about


Sensing CircuitsIf either resistor is replaced with a variable resistor, thermistor or an LDR then the output voltage of the potential divider circuit will be affected by the environment around it (by turning a dial, heating the circuit or changing the light levels).

Depending on the position of these resistors (e.g. R1 or R2), the effect on the output p.d. will differ. See the examples below:

Variable Resistor

Increasing the value of the variable resistor in this circuit will increase the output p.d.

Increasing the value of the variable resistor in this circuit will decrease the output p.d.

Thermistor

As the temperature around the thermistor increases, the output p.d. will decrease.

As the temperature around the thermistor increases, the output p.d. will increase.

Light Dependent Resistor

As the light intensity incident on the LDR increases, the output p.d. will decrease.

As the light intensity incident on the LDR increases, the output p.d. will increase.


The PotentiometerLow-voltage circuits tend to vary the output p.d. using a potentiometer rather than a potential divider circuit. Below are some examples of potentiometers (many of which you may have already used).

A potentiometer is a variable resistor with three terminals and a sliding contact. By adjusting the position of the contact you can vary the output p.d. between two of the terminals.

When the contact is moved towards A, R1 decreases and R2 increases. This causes the Vout to increase. When the slider is all the way at A (R1is equal to zero), Vout will be equal to Vin.

When the contact is moved towards B, R1 increases and R2 decreases. This causes the Vout to decrease. When the slider is all the way at B (R2is equal to zero), Vout will be equal to zero

Potentiometers can be made very compact so are often used in portable electronic devices and they can be designed to produced linear changes in output as well as logarithmic changes.


Summary Questions on page 190 (blue textbook) Practice Questions 3, 4, 5 and 8 (pages 191-193 of your blue textbook) Worksheet 10 on pages 48-50 of this booklet


A simple variable resistor is used to control current. As the sliding contact is moved along the rheostat (variable resistor) the resistance increases and the current in the circuit will decrease.

A potentiometer uses all three connections and is used to control voltage. In this case the sliding contact divides the resistor into two and the output voltage can be varied from zero all the way up to the full supply voltage. This is especially useful in a circuit when you want to vary voltage and can be used to make a simple volume control.

Note: the potentiometer is often used in circuits to collect data to plot I-V curves for different components because it allows us to vary the current and voltage over a greater range.

Worked Examples (In Class)1. Sketch a circuit that will increase the output p.d. as the room gets darker. Describe how it works.

2. Below is a graph of resistance against temperature for a thermistor. This thermistor is connect in a potential divider circuit in series with a 10k resistor. Vout is connected across the thermistor. Use the graph and the potential divider equation to calculate Vout at 00C and 250C.


Worksheet 1 – Kirchhoff’s Laws [Score /10]


Worksheet 2 – More Kirchhoff’s Laws


Worksheet 3 – Combining Resistors [Score /20]


Worksheet 4 – Kirchhoff’s Laws and Combining Resistors [Score /12]

1 Calculate the total resistance for the following combinations of resistors.

(2 marks)

2

For the circuit shown, calculatea the total resistance in the circuit (2 marks)

b the current through the 7.0 Ω resistor (1 mark)


c the p.d. across the 7.0 Ω resistor (1 mark)

d the p.d. across the 4.0 Ω resistor (2 marks)

e the p.d. across the 12 Ω resistor (1 mark)

f the current through the 4.0 Ω resistor (2 marks)

g the current through the 12 Ω resistor. (1 mark)


Worksheet 5 – Analysing Circuits 1 [Score /35]

1 Figure 1 shows a circuit in which two resistors in series, X and Y, are connected in parallel with a third resistor, Z, and a battery. Electrons transfer energy from the battery to the resistors.

Figure 1

a i Describe the relationship between the current in resistors X and Y. Give reasons for your answer. (2 marks)

………………………………………………………………………………………………………..………………………………………………………………………………………………………..………………………………………………………………………………………………………..………………………………………………………………………………………………………..

ii Describe the relationship between the current in the battery and the currents in X, Y and Z. Give reasons for your answer. (2 marks)

………………………………………………………………………………………………………..………………………………………………………………………………………………………..………………………………………………………………………………………………………..………………………………………………………………………………………………………..

b i Describe the relationship between the p.d. across the battery and the p.d. across Z. Give reasons for your answer. (3 marks)

………………………………………………………………………………………………………..………………………………………………………………………………………………………..………………………………………………………………………………………………………..………………………………………………………………………………………………………..

ii Describe the relationship between the p.d. across the battery, the p.d. across X and the p.d. across Y. Give reasons for your answer. (3 marks)

………………………………………………………………………………………………………..………………………………………………………………………………………………………..………………………………………………………………………………………………………..………………………………………………………………………………………………………..


Y

Z

X

2. In the circuit in Figure 1, the resistances of X, Y, and Z are 3.0 Ω, 6.0 Ω and 18.0 Ω respectively, and the battery has an e.m.f. of 12.0 V.

a Draw the circuit diagram showing the resistances of X, Y and Z. (1 mark)

b i Calculate the total resistance of X and Y in series. (1 mark)

ii Show that the total resistance of the circuit is 6.0 Ω. (1 mark)

c Calculate the current in:i the battery (1 mark)

ii Z (1 mark)

iii X (2 marks)

d Calculate the p.d. across X. (1 mark)


3. In the circuit in Figure 1, the resistances of Y and Z are changed to 5.0 Ω and 12.0 Ω respectively.a Draw a circuit diagram showing the resistances of X, Y and Z. (1 mark)

b i Explain why the p.d. across Z is 12 V. (1 mark)

………………………………………………………………………………………………………..………………………………………………………………………………………………………..

ii Calculate the current in Z. (1 mark)

c i Calculate the resistance of X and Y in series. (1 mark)

ii Calculate the current in X and Y. (1 mark)

iii Calculate the p.d. across X. (1 mark)

d Calculate the battery current. (1 mark)


4. Figure 2 shows a 3.0 Ω resistor, a 6.0 Ω resistor, and an 18.0 Ω resistor in a circuit with a battery that has an e.m.f. of 12.0 V. The battery has negligible internal resistance (i.e. its resistance can be assumed to be zero).

Calculate the current and p.d. for each resistor. (4 marks)

Figure 2


12.0 V

3.0

6.0

18.0

5. When a diode D conducts, the p.d. across it is 0.60 V. The diode is connected in its forward direction, as shown in Figure 3, in series with a 300 Ω resistor and a 1.5 V cell of negligible internal resistance (i.e., its resistance can be assumed to be zero).

Figure 3

a Calculate:i the p.d. across the resistor (1 mark)

ii the current in the resistor. (1 mark)

b In the circuit in Figure 3, a resistor R of resistance 400 Ω is connected across the diode. The p.d. across the diode remains the same.i Explain why the current in the 300 Ω resistor is unchanged. (1 mark)

………………………………………………………………………………………………………..………………………………………………………………………………………………………..………………………………………………………………………………………………………..

ii Calculate the current in the 400 Ω resistor. (2 marks)

iii Calculate the current in the diode. (1 mark)


300 ΩD

Worksheet 6 – Analysing Circuits 2 [Score /26]


Worksheet 7 – EMF and Internal Resistance 1 [Score /25]

1 A cell has an e.m.f. of 2.0 V and an internal resistance of 0.1 Ω.Calculatea the p.d. across the terminals when it is supplying a current of 5.0 A (2 marks)

b the external resistance in the circuit. (2 marks)

2 The e.m.f. of a cell is 2.0 V. When a resistor is connected across the terminals of the cell, the p.d. across the terminals drops to 1.5 V and a current of 2.0 A flows.Calculatea the internal resistance of the cell (2 marks)

b the resistance of the resistor (2 marks)

c the energy wasted in the cell when 1.0 C of charge passes round the circuit. (2 marks)


3 For the circuit shown in Figure 4, when the switch is open the voltmeter reads 10.0 V and when it is closed the voltmeter reads 8.0 V. The resistance R 4.0 Ω.Calculate the internal resistance, r. (3 marks)

Figure 4

4 A bulb in a torch is powered by two identical cells connected in series, each of e.m.f. 1.5 V. The bulb dissipates power at the rate of 625 mW and the p.d. across the bulb is 2.5 V. Calculatea the internal resistance of each cell (2 marks)

b the energy dissipated in each cell in one minute. (2 marks)


Exam-style question5 A 1.5 V cell of internal resistance 2.0 Ω is connected in two different ways to two 4.0 Ω resistors, as

shown in Figure 5.

Figure 5a For the series combination calculate

i the current through both resistors (2 marks)

ii the terminal p.d. (2 marks)

b For the parallel combination calculatei the current from the cell (2 marks)

ii the current through each resistor. (2 marks)


Worksheet 8 – EMF and Internal Resistance 2 [Score /31]

1 Calculate the missing values of p.d. in Figure 2 (assume the internal resistance is negligible). (4 marks)

Figure 2

2 A battery of e.m.f. 9.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor. Calculate:a the current through the resistor (2 marks)

b the p.d. across it (1 mark)

c the lost volts. (1 mark)


3 A battery of e.m.f. 12.0 V and internal resistance 0.50 Ω is connected in series to a 10.0 Ω resistor and an unknown resistor R. The p.d. across R is 6.0 V. Calculate:a the current through the circuit (2 marks)

b the p.d. across the 10.0 Ω resistor (1 mark)

c the lost volts. (1 mark)

4 a Use these values for a battery to plot a graph: (I 2.0 A, V 11.0 V); (I 4.0 A, V 10.0 V); (I 8.0 A, V 8.0 V); (I 12.0 A, V 6.0 V); and (I 18.0 A, V 3.0 V). (4 marks)

b Use your graph from part a to calculate the e.m.f. and the internal resistance. (4 marks)


Worksheet 9 – Potential Divider Circuits [Score /25]


Worksheet 10 – Potential Dividers and Sensing Circuits [Score /24]

1 The input p.d. to a potential divider is 9.0 V.a What is the output p.d. across a 1.0 kΩ resistor if the second resistor is 5 kΩ? (1

mark)

b What is the output p.d. across a 330 Ω if the second resistor is 990 Ω? (1 mark)

c If the output p.d. across a 680 Ω resistor is 7.0 V, what is the resistance of the other resistor? (1 mark)

2 A potential divider circuit, like that shown in Figure 2, is used to switch on a light when the light intensity drops below a certain value.

The supply p.d. is 12 V. The light must switch on at twilight, when the resistance of the LDR has a value of 5.0 kΩ. A p.d. of 4.0 V is required to switch on the light.

Calculate the fixed resistance required for the circuit. (2 marks)


3 Figure 3 shows a length of uniform resistance wire AB connected in a circuit with a 12.0 V power supply to make a potentiometer.

Figure 3What will be the value of the output p.d. when:a the length AX half the length of AB (2 marks)

b the length AX

13 the length of AB (2 marks)

c the length of AX

34 the length of AB. (2 marks)

4 A heater switches on when the p.d. is 5.0 V. You have a supply p.d. of 9.0 V and a thermistor with a resistance that decreases with temperature, which has a resistance of 450 Ω at a temperature of 20°C.

Calculate the value of the series resistor that should be used in the circuit, shown in Figure 4, to switch on the heater when the temperature falls to 20°C. (3 marks)

Figure 4


Past Paper Questions

1.


2.


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