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Class………….
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Electrons, Photons and Waves
Module 2.3:
DC Circuits
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Module 2.3 DC Circuits
1.
Kirchoff’s Laws
2.3.1 Series and parallel circuits
Candidates should be able to:
(a) state Kirchhoff’s second law and appreciate that this is a consequence of
conservation of energy;
(b) apply Kirchhoff’s first and second laws to circuits;
conservation
e.m.f.
p.d.
2.
Resistors in parallel
and series
(c) select and use the equation for the total resistance of two or more resistors
in series;
(d) recall and use the equation for the total resistance of two or more resistors in
parallel;
Resistance, Ohm, PD,
Current, series,
parallel
3.
Internal resistance
(e) solve circuit problems involving series and parallel circuits with one or more
sources of e.m.f.;
(f) explain that all sources of e.m.f. have an internal resistance;
(g) explain the meaning of the term terminal p.d.;
(h) select and use the equations e.m.f. = I (R + r), and e.m.f. = V + Ir .
Series, parallel, emf,
od, internal resistance,
power supply
4.
Potential Dividers
2.3.2 Practical circuits
Candidates should be able to:
(a) draw a simple potential divider circuit;
(b) explain how a potential divider circuit can be used to produce a variable p.d.;
(c)recall and use the potential divider equation Vout = Vin (R1 / R1 +
Circuit, Potential
Difference, Potential
Divider, Resistance
5.
Potential Dividers
(d) describe how the resistance of a light dependent resistor (LDR) depends on
the intensity of light;
(e) describe and explain the use of thermistors and light-dependent resistors in
potential divider circuits;
(f) describe the advantages of using dataloggers to monitor physical changes
(HSW 3).
Circuit, Potential
Difference, Potential
Divider, Resistance
6.
G482 Module 1:
2.3 DC Circuits
Test
Review and assess their knowledge and understanding
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Lesson 15 notes - Kirchhoff's laws
Objectives
(a) state Kirchhoff’s second law and appreciate that this is a consequence of
conservation of energy;
(b) apply Kirchhoff’s first and second laws to circuits;
Kirchhoff's two laws are equations based on conservation of charge and
conservation of energy that can be applied to any electric circuit. The two laws
can be used to work out the current in any branch of a circuit, given the emfs
and resistances in the circuit.
Kirchhoff's first law states that the total current entering a junction is equal to
the total current leaving the junction.
Kirchhoff's second law states that the sum of the emfs round any complete
loop in a circuit is equal to the sum of the potential drops round the loop.
Kirchhoff's first law is a statement of conservation of charge since it means
that the total charge flowing into a junction in a given time is equal to the total
charge leaving the junction in the same time. Kirchhoff’s second law is a
statement of conservation of energy since an emf is where energy is supplied
to charges (i.e. a source) and a potential drop is where charges release
energy (i.e. a sink). The sum of the e.m.f s is therefore the total energy
produced in the loop per unit charge and the sum of the potential drops is the
total energy dissipated in the loop per unit charge.
Kirchhoff's first law
The total current into a junction = the total current out of the junction.
Using the convention that currents leaving a junction are the opposite sign to
currents entering the junction, the first law may be expressed as the following
equation:
I1 + I2 + I3 + … = 0 where I1, I2, I3 etc represent the currents in the branches
connected to the junction.
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Kirchhoff’s 1st law
I2
I1
I3
I4
I1 + I2 + I3 + I4 = 0
currents in are assigned + values
currents out are assigned – values
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Kirchhoff's second law
The sum of the emfs round a loop in a circuit = the sum of the potential drops
round the loop.
For a loop of a circuit containing emfs 1, 2, 3, etc and resistances R1, R2,
R3, etc, the second law may be expressed as the following equation:
1 + 2 + 3 + … = I1R1 + I2R2 + I3R3 + … where I1, I2, I3 … represent the
currents through the resistances R1, R2, R3 ….
Using Kirchhoff’s second law

A
I1 – I2
r1
I1
D
I1 – I2
I1
I2
R
B
C
I2
 = I2R + I1r1
for loop ABCD
Notes:
1.
emfs and currents in the opposite direction to the direction round the
loop are given negative values.
2.
The second law is particularly useful for analysing circuits with more
than one loop. In general, for a circuit with n unknown currents, n loops need
to be considered one by one, each loop giving a new equation. In this way,
the n linear equations formed can be solved for the n unknown currents.
Example:
A circuit consists of a cell of emf 1.6 V in series with a resistance 2.0 
connected to a resistor of resistance 3.0 in parallel with a resistor of
resistance 6.0  Determine the total current drawn from the cell and the
potential difference across the 3.0 resistor.
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1.6 V
I1 + I2
2.0 
3.0 
I1 + I2
I1
6.0 
I2
Solution
Consider the circuit loop consisting of the cell and the 3.0  resistor:
1.6 V = 3 I1 + 2  (I1 + I2).
Thus: 1.6 V = 5  I1 + 2  I2.
Consider the circuit loop consisting of the cell and the 6.0 resistor:
1.6 V = 6  I2 + 2  (I1 + I2).
Thus 1.6 V = 2  I1 + 8  I2.
Subtracting the second equation from the first gives:
0 V = 3  I1 + 6  I2
hence I1 = 2 I2.
Substituting I1 = 2 I2 into the second equation gives:
1.6 V = 12  I2.
Thus I2 = 0.13 A and I1 = 0.27 A.
Current through cell = I1 + I2 = 0.40 A.
pd across 3.0 resistor= I1 × 3.0  (= I2  6.0  ) = 0.8 V.
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Lesson 15 Questions – Kirchoff’s Laws
1)a) State Kirchoff’s Second Law
…………………………………………………………………………………………
……………....…………………………………………………………………………
…………
…………………………………………………………………………………………
(2)
Fig.1.1 shows a circuit:
fig 1.1
b)
Use Kirchoff’s second law to determine the p.d. V.
…………………………………………………………………………………………
……………....…………………………………………………………………………
…………
…………………………………………………………………………………………
(2)
Total [4]
2)a) Kirchoff’s first law is based on the conservation of an electrical quantity. State
the law and the quantity conserved.
…………………………………………………………………………………………
……
…………………………………………………………………………………………
……
…………………………………………………………………………………………
(2)
b)
Fig 2.1 shows a potential divider circuit.
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fig 2.1
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i)
Show that the voltmeter reading is 1.5V
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
Total [4]
3)
Fig 3.1 shows part of an electric circuit.
fig 3.1
a)
Name the component marked X
…………………………………………………………………………………………
(1)
b)
Determine the magnitude of the currents I1, I2 and I3.
I1 =………………………mA
I2 =………………………mA
I3 =………………………mA
(3)
Total [4]
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Lesson 16 notes - Resistor Combination Formulae
Objectives
Be able to:
Recall and use formula for the combined resistance of two or more resistors in
series
Recall and use formula for the combined resistance of two or more resistors in
parallel
Resistors in combinations
Remember that R = V / I where V is the pd across a component and I is the
current through it.
Resistors in series
E
I
V
A
T
R1
V
1
From Kirchoff’s first law:
Current constant around series circuit
R2
V
2
From Kirchoff’s second law:
E=V1 + V2
and VT = V1 + V2
Therefore IRT = IR1 + IR2
Cancelling I
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RT = R1 + R2
This is true for any number of resistors in series.
Resistors in parallel
E
I1
Loop 1
R1
I2
Loop 2
I3
R2
From Kirchoff’s first law:
I1 = I2 + I3
(1)
Applying Kirchoff’s scond law to loop 2:
V2 + V1 = E0
So, (I3 x R2) – (I2 x R1) = 0 V
Which leads to I3 x R2 = I2 x R1
This is saying that the 2 resistors have the same p.d. across them.
So we can write,
I1 = V/RT
I2 = V/R1
I3 = V/R2
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Enter these into (1):
V/RT=V/R1 + V/R2
Cancel V
1/RT = 1/R1 + 1/R2
This is true for any number of resistors in parallel.
So RT is the reciprocal of 1/RT
In AS level physics most problems encountered involve 2 resistors in parallel
and for this we can write.
RT = R1R2/(R1+R2)
Because
and so,
Therefore,
1/RT = 1/R1 + 1/R2
1/RT = R2/(R1xR2) + R1/(R2xR1)
1/RT = (R2 + R1)/(R2xR1)
RT = (R2xR1)/(R2 + R1) or RT = R1R2/(R1+R2)
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Lesson 16 questions – resistors in series and parallel
1
Calculate the total resistance of the following resistance combinations:
a)
4Ω
12Ω
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
(1)
b)
4Ω
12Ω
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
(2)
c)
3Ω
2Ω
6Ω
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
Total [6]
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2)
Calculate the total resistance between the points X and Y for the circuit shown
in fig. 1.1
fig1.1
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
Total [3]
3)
A circuit of a combination of resistors is shown in figure 3.1.
6V
3Ω
4Ω
6Ω
fig 3.1
Calculate:
a)
the combined resistance of the 3Ω resistor and the 6Ω resistor in parallel,
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
b)
the total resistance of the circuit,
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (1)
c)
the battery current,
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
d)
the power supplied to the 4Ω resistor.
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
Total [8]
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Lesson 17 notes – Internal resistance
Objectives
You will be able to:
Use the concept that E.M.F. is defined in terms of the energy transferred by a
source in driving unit charge round a complete circuit.
Use energy considerations to distinguish between E.M.F. and P.D.
Appreciate that sources of E.M.F. have internal resistance and understand the
simple consequences of internal resistance for external circuits.
What you should know
 E.M.F is the gaining of energy by a charge as it passes through a
supply
 Potential difference is the transferring of energy from the charge to a
resistor
 Ohm’s Law (V=IR)
 Voltage = Energy / Charge
 Current = Charge moved / time taken
 Kirchoff’s 1st Law (KCL)
 Kirchoff’s 2nd Law (KVL)
 Power = current x voltage (also P=I2R)
Internal Resistance
Current flows through all parts of the circuit, including the cells. Since the cells
themselves are made of material with resistance there will be some energy
lost in the cell. This will increase with current, so there is more loss at higher
currents. Consider the diagram below:
Here we represent a ‘real’ cell as two components in a circuit diagram:
an ‘ideal’ cell, with no internal resistance, and a small resistor, representing
the internal resistance.
Energy is conserved, so the energy transferred from chemical to electrical by
Electrical
energy to heat:
Internal
resistance r
Chemical energy to electrical
energy: Cell of emf E
CELL
Electrical energy to work and
heat: external load resistance R
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the emf of the cell must ‘pay’ for the energy transferred to other forms in the
circuit - this is the sum of the energy transferred in the cell itself and the
energy transferred by the external load.
Clearly the energy transferred to heat in the cell is not doing useful work and
so is in a sense ‘lost’.
Deriving an equation
There are three ways to arrive at the equation relating emf, terminal pd,
current and internal resistance.
‘lost
volts’
e.m.f
r
E
CELL
I
Terminal
Voltage
R
1.
Kirchhoff’s 2nd Law: As charge goes around the circuit the sum of e.m.f
s must equal the sum of voltage drops leading to:
E=IR+Ir
The terminal voltage is equal to I R so this can be rearranged to give:
V=E–Ir
and interpreted as terminal voltage = emf – ‘lost volts’
2.
Energy is conserved. Imagine a unit of charge, Q, moving around the
circuit:
QE=QIR+QIr
This leads to the same equations as in (1) above.
3.
Use Ohm’s law with E ‘driving’ current through the combined resistance
(R + r):
I = E / (R+r)
Multiplying throughout by (R+r) leads to the same equations and
conclusions as in (1).
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Practical effects of internal resistance
Lets us a car as an example. The headlamps are connected in parallel across
a twelve-volt battery. The starter motor is also in parallel controlled by the
ignition switch. Since the starter motor has a low resistance it demands a very
high current (say 60 A). The battery itself has a low internal resistance (say
0.01 Ω). The headlamps themselves draw a much lower current.
What happens when the engine is started? (switch to starter motor closed for
a short time).

sudden demand for more current

large lost volts (around 0.01 Ω  60 A = 6 V)

terminal voltage drops to 12 V – 6 V = 6 V

headlamps dim
When the engine fires, the starter motor switch is opened and the current
drops. The terminal voltage rises and the headlamps return to normal. It’s
better to turn the headlamps off when starting the car.
Voltage, Current and Resistance
V
I
The voltage of a supply decreases if it is required to supply more current
across the same load (R=V/I)
The cell p.d. decreases as the current increases. (the internal resistance stays
constant)
The graph is shown opposite.
E = IR + Ir becomes IR = E – Ir
Compare this to y = c +mx
The intercept with the y axis gives the supply e.m.f
The intercept with the x axis gives the maximum current that can be drawn
from the supply
And the gradient gives the internal resistance of the supply
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Power and internal resistance
Electrical power, P = IV = I2R
So power supplied by the cell,
IE = I2R + I2r
Output Power
=
power delivered to R
+
power wasted in the cell
due to its internal resistance
The power supplied to R =
I2R = E2R/(R + r) 2
And because of this the peak of the power verses resistance curve is at R = r,
Maximum power is delivered when the load resistance is equal to the internal
resistance. (When the load is matched to the source)
The graph plotting the Output power against Load resistance for the internal
resistance of 1Ω shows this the best. As the load approaches 1Ω, the output power is
at a maximum.
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Lesson 17 questions – Internal resistance
1)
Fig. 1.1 shows two soldered resistors connected across terminals of a cell.
fig 1.1
The cell has a resistance of 0.8Ω and e.m.f. 1.5V.
a)
Define e.m.f. in terms of energy transformed and electric charge.
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
b)
Suggest why a cell has internal resistance.
…………………………………………………………………………………………
(1)
c)
Calculate the total resistance R of the circuit in Fig. 1.1.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
d)
Hence calculate the current I in the circuit.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
e)i)
Write an equation for the power dissipated by a current-carrying resistor.
…………………………………………………………………………………………
…………………………………………………………………………………………
ii)
For the circuit in Fig. 1.1, calculate the ratio:
power dissipated by internal resistance__
power dissipated by total external resistance
ratio=……………(4)
Total [7]
2)
Fig. 2.1 shows a cell of e.m.f. E and internal resistance r connected to a
variable resistor.
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fig 2.1
fig 2.2
Fig. 2.2 shows the variation of the p.d. V across the terminals of the cell with the
current I drawn from the cell.
a)
Explain how fig 2.2 shows that the e.m.f. E is 1.4V
…………………………………………………………………………………………
……………………………………………………………………………………… (1)
b)i)
Use fig. 2.2 to determine the maximum possible current that can be drawn
from the cell.
…………………………………………………………………………………………
……………………………………………………………………………………… (1)
ii)
Calculate the internal resistance r of the cell.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
Total [5]
3)
Fig. 3.1 shows the I/V characteristics for a tungsten filament lamp.
fig 3.11
A 5.0Ω resistor and the tungsten filament lamp are connected in series to a d.c. power
supply of e.m.f. 24V. The current drawn from the power supply is 2.0 A.
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a)
Calculate the total power delivered by the supply.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
(2)
b)
Use fig 3.1 to determine the resistance of the filament lamp when the current
in it is 2.0 A.
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
c)
Calculate the total resistance of the series combination of the filament lamp
and the resistor.
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (1)
d)
Calculate the internal resistance of the supply.
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
Total [7]
4)
Fig 4.1 shows a car battery of e.m.f 12V and internal resistance 0.014Ω
connected to the starter motor of a car. When the car engine is being started, the car
battery provides a currnt of 160A to the starter motor.
fig 4.1
a)
Show that the p.d. across the internal resistance is about 2.2V.
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (1)
b)
Determine the terminal p.d. across the battery.
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (1)
Total [2]
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Lesson 18 notes – Potential dividers
Objectives
You will understand the use of a potential divider as a source of variable p.d.
You will be able to describe and explain the use of thermistors and lightdependent resistors in potential dividers to provide a potential difference
which is dependent on temperature and on light intensity respectively.
Thermistor Characteristics
Increased temp
Low temp
Ohmic-Conductor keeping temperature constant
Potential Dividers
Potential dividers are resistors connected in series across a voltage source
used to obtain a desired fraction of the voltage. An example is shown below:
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The potential divider formula
For an unloaded potential divider the current through each resistor is the
same so the voltage is proportional to the resistance. This means that the pd
across the pair of resistors is divided in the same ratio as the resistors
themselves:
i.e. V1 / V2 = I R1 / I R2
or
V1 / V2 = R1 / R2
If R1 >> R2 then V1 is more or less the supply voltage and if R1 << R2 then V1
is close to 0 V.
VS is an input to the potential divider and V1 is an output. The circuit itself
provides a way to tap off a voltage between 0 V and VS.
This can, of course be done continuously using a rheostat or potentiometer
shown below:
Rotary potentiometers are used as volume controls in hi-fi systems.
The potential divider equation can be derived by rearranging the ratios above
to give:
V1 = VS x (R1 / (R1+R2))
Lets do it now:
Use KVL: Vs = I(R1+R2) so I = Vs/(R1+R2)
Now: V1/V2 = IR1/IR2 = R1/R2
And V1=I x R1
So, V1= (Vs/(R1+R2)) x R1
Therefore: V1 = VS x (R1 / (R1+R2))
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Extension
The effect of the load on output
Connecting a load across R1 reduces the output voltage.
This is because the effective resistance in the lower arm of the potential
divider is now a parallel combination of R1 and R load (less than R1) so a
smaller fraction of the voltage is ‘tapped off’.
If Rload >> R1 then there is no significant effect on the output voltage.
Consider what happens when a lit bulb goes out when “shorted out” by a
piece of wire. It is because the low resistance of the wire in parallel reduced
the combination’s total resistance, compared to the rest of the circuit.
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Lesson 14 questions – Potential dividers
1)
A potential divider circuit based on a light-dependent resistor (LDR) is shown
in fig 1.1. The supply has negligible internal resistance.
fig 1.1
a)
The light intensity falling upon the LDR is increased. State:
i)
how the resistance of the LDR changes,
…………………………………………………………………………………………
….………………………………………………………………………………………
(1)
ii)
how the p.d. across the LDR changes.
…………………………………………………………………………………………
….………………………………………………………………………………………
(1)
b)
At a particular light intensity, the resistance of the LDR is 420Ω.
i)
Calculate the p.d. across the LDR.
…………………………………………………………………………………………
….………………………………………………………………………………………
…….……………………………………………………………………………………
……….…………………………………………………………………………… (3)
ii)
An electronic circuit of resistance 50Ω is connected between the terminals X
and Y.
1.
Show that the total resistance of the parallel combination of this
electronic circuit and the LDR is about 45Ω.
…………………………………………………………………………………………
….………………………………………………………………………………………
…….……………………………………………………………………………………
……….…………………………………………………………………………… (1)
2.
Calculate the p.d. across the electronic circuit.
…………………………………………………………………………………………
….………………………………………………………………………………………
…….……………………………………………………………………………………
……….…………………………………………………………………………… (2)
Total [8]
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2)
Fig 2.1 shows a potential divider circuit. The battery has negligible internal
resistance and the voltmeter has a very high resistance.
fig2.1
i)
Show that the voltmeter reads 1.5V.
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
ii)
An electric device rated at 1.5V, 0.1A is connected between the terminals X
and Y. The device has constant resistance. The voltmeter reading drops to a very low
value and the device fails to operate, even though the device itself is not faulty.
1.
Calculate the total resistance of the device and the 400Ω resistor in parallel.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (3)
2.
Calculate the p.d. across the device when it is connected between X and Y.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
3.
Why does the device fail to operate?
……………………………………………………………………………………… (1)
Total [8]
3)a) Fig 3.1 shows the circuit symbol for a component.
fig 3.1
Name the component and state how its resistance changes as the temperature of the
component is increased.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
(2)
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b)
Fig 3.2 shows a potential-divider circuit. The battery has a negligible internal
resistance and the voltmeter has a very high resistance.
fig 3.2
i)
At a particular temperature, the resistance of component X is 4.2kΩ. Calculate
the voltmeter reading.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
(3)
ii)
State how your answer to (b)(ii) changes when the temperature of the
component X is increased.
…………………………………………………………………………………………
…………………………………………………………………………………………
(1)
Total [6]
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RAB Plymstock School
Lesson 18 notes – Potential dividers
Objectives
(a) draw a simple potential divider circuit;
(b) explain how a potential divider circuit can be used to produce a variable p.d.;
(c)recall and use the potential divider equation Vout = Vin (R1 / R1 + R2 ).
Thermistor Characteristics
Increased temp
Low temp
Ohmic-Conductor keeping temperature constant
Potential Dividers
Potential dividers are resistors connected in series across a voltage source
used to obtain a desired fraction of the voltage. An example is shown below:
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The potential divider formula
For an unloaded potential divider the current through each resistor is the
same so the voltage is proportional to the resistance. This means that the pd
across the pair of resistors is divided in the same ratio as the resistors
themselves:
i.e. V1 / V2 = I R1 / I R2
or
V1 / V2 = R1 / R2
If R1 >> R2 then V1 is more or less the supply voltage and if R1 << R2 then V1
is close to 0 V.
VS is an input to the potential divider and V1 is an output. The circuit itself
provides a way to tap off a voltage between 0 V and VS.
This can, of course be done continuously using a rheostat or potentiometer
shown below:
Rotary potentiometers are used as volume controls in hi-fi systems.
The potential divider equation can be derived by rearranging the ratios above
to give:
V1 = VS x (R1 / (R1+R2))
Lets do it now:
Use KVL: Vs = I(R1+R2) so I = Vs/(R1+R2)
Now: V1/V2 = IR1/IR2 = R1/R2
And V1=I x R1
So, V1= (Vs/(R1+R2)) x R1
Therefore: V1 = VS x (R1 / (R1+R2))
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Extension
The effect of the load on output
Connecting a load across R1 reduces the output voltage.
This is because the effective resistance in the lower arm of the potential
divider is now a parallel combination of R1 and R load (less than R1) so a
smaller fraction of the voltage is ‘tapped off’.
If Rload >> R1 then there is no significant effect on the output voltage.
Consider what happens when a lit bulb goes out when “shorted out” by a
piece of wire. It is because the low resistance of the wire in parallel reduced
the combination’s total resistance, compared to the rest of the circuit.
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 18 questions – Potential dividers
1)
A potential divider circuit based on a light-dependent resistor (LDR) is shown
in fig 1.1. The supply has negligible internal resistance.
fig 1.1
a)
The light intensity falling upon the LDR is increased. State:
i)
how the resistance of the LDR changes,
…………………………………………………………………………………………
….………………………………………………………………………………… (1)
ii)
how the p.d. across the LDR changes.
…………………………………………………………………………………………
….………………………………………………………………………………… (1)
b)
At a particular light intensity, the resistance of the LDR is 420Ω.
i)
Calculate the p.d. across the LDR.
…………………………………………………………………………………………
….………………………………………………………………………………………
…….……………………………………………………………………………………
……….…………………………………………………………………………… (3)
ii)
An electronic circuit of resistance 50Ω is connected between the terminals X
and Y.
1.
Show that the total resistance of the parallel combination of this
electronic circuit and the LDR is about 45Ω.
…………………………………………………………………………………………
….………………………………………………………………………………………
…….……………………………………………………………………………………
……….………………………………………………………………………………
(1)
2.
Calculate the p.d. across the electronic circuit.
…………………………………………………………………………………………
….………………………………………………………………………………………
…….……………………………………………………………………………………
……….…………………………………………………………………………… (2)
Total [8]
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2)
Fig 2.1 shows a potential divider circuit. The battery has negligible internal
resistance and the voltmeter has a very high resistance.
fig2.1
i)
Show that the voltmeter reads 1.5V.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…… (2)
ii)
An electric device rated at 1.5V, 0.1A is connected between the terminals X
and Y. The device has constant resistance. The voltmeter reading drops to a very low
value and the device fails to operate, even though the device itself is not faulty.
1.
Calculate the total resistance of the device and the 400Ω resistor in
parallel.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
2.
Calculate the p.d. across the device when it is connected between X
and Y.
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
3.
Why does the device fail to operate?
……………………………………………………………………………………… (1)
Total [8]
3)a) Fig 3.1 shows the circuit symbol for a component.
fig 3.1
Name the component and state how its resistance changes as the temperature of the
component is increased.
…………………………………………………………………………………………
…………………………………………………………………………………………
………… (2)
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b)
Fig 3.2 shows a potential-divider circuit. The battery has a negligible internal
resistance and the voltmeter has a very high resistance.
fig 3.2
i)
At a particular temperature, the resistance of component X is 4.2kΩ. Calculate
the voltmeter reading.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
………………………………………………………………………………………(3)
ii)
State how your answer to (b)(ii) changes when the temperature of the
component X is increased.
…………………………………………………………………………………………
………………………………………………………………………………………(1)
Total [6]
©2011 science-spark.co.uk
RAB Plymstock School