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Potential Dividers
Topic 3
Potential Dividers A potential divider is a resistor network that produces a fixed or variable potential (voltage). This potential is lower than the potential of the supply. The potential divider is probably the most often used circuit module. There are many instances of potential dividers being used in this book. These include the circuits on pp. 12, 14, 15, 17, and 18 of Topic 2. Essentially, a potential divider consists of two resistors connected in series.
CURRENT AND PD Assuming that no current flows out of the divider at the junction of R1 and R2, the same current, i, flows through both resistors. Ohm’s law applies to each resistor. For R2: i 5 vOUT =R2
Ohm’s Law also applies to the two resistors when connected in series: i 5 vIN =ðR1 1 R2 Þ
i
Combining these two equations: vOUT =R2 5 vIN =ðR1 1 R2 Þ
R1
vOUT 5 vIN 3 R2 =ðR1 1 R2 Þ VIN R2
VOUT
We use this equation to calculate vOUT, given vIN and the values of the two resistors. Example
FIGURE 3.1
Built on a breadboard it might look like this:
Calculate vOUT, given that vIN 5 9 V, R1 5 1.6 kΩ, and R2 5 2 kΩ. vOUT 5 9 3 2000=ð1600 1 2000Þ vOUT 5 5 V The output of the potential divider is 5 V.
9V R1 1k6 VIN R2 2k
FIGURE 3.2 The supply vIN comes from a battery or power pack on the left. The potential vOUT across R2 is measured by test leads on the right. Electronics: Circuits and Systems. r 2011 Owen Bishop. Published by Elsevier Ltd. All rights reserved.
VOUT =5V
0V FIGURE 3.3 When the source voltage is 9 V, the output voltage is 5 V.
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PART | 1
THE EFFECT OF THE LOAD
Circuits
Example
The foregoing calculation assumes that no current flows out of the output terminal. This is almost true when we use a digital meter to measure vOUT. We find that the measured value of vOUT is almost correct, allowing for possible tolerance errors in the resistor values. However, current is drawn from the output in most practical circuits. This is what happens: Current i passes through R1. At the output of the divider, a current iLOAD passes through the load while the remainder i2 passes on through R2. The diagram shows that R2 and the load (RLOAD) are resistances in parallel.
A load with resistance 3 kΩ is connected to the potential divider described in the previous example. What is the new value of vOUT? The combined resistance of R2 and the load is: R5
2000 3 3000 5 1200 Ω 2000 1 3000
In effect, the divider now consists of R1 in series with 1.2 kΩ, so: vOUT 5 9 3
1200 5 3:86 V 1600 1 1200
The output voltage is reduced to 3.86 V. i
In the example, the output voltage of the divider falls by 1.14 V when the load is connected. The amount of the fall depends on the resistance of the load.
R1 VIN
i LOAD i2
INCREASING THE PRECISION
R2
VOUT
RLOAD
FIGURE 3.4 When a load is connected to the divider, it is in parallel with R2.
The usual technique for preventing excessive drop in output voltage is to have a relatively large current flowing through the divider. Then the current drawn by the load has relatively less effect. As a rule-ofthumb, make the divider current about ten times the current taken by the load. Example
If R2 and the load are resistances in parallel, their combined resistance is less than the resistance of either of them separately. Therefore the pd across R2 is less when the load is connected. The measured value of vOUT is less than expected.
In the previous example, the current drawn by the 3 kΩ load is: iLOAD 5 5=3000 5 1:67 mA
9V
9V R1 1k6
R1 1k6 equivalent
VIN
to R2 2k
0V
VOUT
VIN
RLOAD 3k
1k2
0V R2 RLOAD
FIGURE 3.5 Connecting a load to the divider lowers the resistance across which vOUT appears.
ROUT = 3.86 V
Topic | 3
Potential Dividers
The current flowing through the divider must be at least ten times that amount, which is 16.7 mA. For a current of this size, R1 and R2 in series must total no more than R 5 vIN/0.0167 5 539 Ω. We have to specify two resistors that have the same ratio as in the example, but total no more than 539 Ω. Self Test a Select two E24 resistors that total 540 Ω and form a potential divider that has 5 V output with 9 V input, under no load. What is the output when a load of 2 kΩ is connected? b Design a potential divider using E24 resistors that gives an output of 3.3 V when the input is 6 V and there is no load. What is the output when a load of 1 kΩ is connected? Select the resistor values to minimize the voltage drop under this load.
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R1. We can adjust the temperature at which the circuit changes state by adjusting VR1.
MEASUREMENT ERRORS Unlike a digital testmeter, an analogue moving-coil testmeter needs appreciable current to drive it. Typically, the coil resistance is 200 kΩ when switched to the 10 V range. A cheap analogue meter may have a coil resistance of only 20 kΩ. If a potential divider or other network of high-value resistors is tested with an analogue meter, serious errors may occur. This is because the coil of the meter acts as a load on the network. Current is drawn by the meter and this results in a voltage drop at the point where voltage is being measured.
Unfortunately, using resistor values totalling only 539 Ω means that the divider takes a current of over 25 mA. In a battery powered device we might not be able to afford to waste so much current.
R1 100k 10 V R2 56k
VOUT
R METER 20k
There are more calculations to try on the Companion website.
USING POTENTIAL DIVIDERS WITH SENSORS In an electronic system, the output signal from the sensor may be a change in its resistance. For example, when the sensor is a light dependent resistor (LDR), as in the circuit on p. 14, variations in the amount of light falling on the LDR result in changes in its resistance. Because of this, the LDR is said to be a resistive sensor. Thermistors, photodiodes, and strain gauges are other examples of resistive sensors. In the circuit on p. 14, the LDR resistor is part of a potential divide so the potential at the junction of R1 and R2 varies according to the amount of light falling on the LDR (R2). The transistor Q1 is switched on and off, so activating or deactivating the audible warning device. Another example is shown on p. 15. In this case the sensor is a thermistor. Together with the variable resistor, VR1, it forms a potential divider with its output at junction A. The output voltage at that junction depends on the setting of VR1 and the temperature of
FIGURE 3.6 An analogue meter may give low voltage readings on a high-resistance network.
Example In the diagram, the actual value of vOUT is: 10 3 56 000=156 000 5 3:59 V But the meter has a coil resistance of 20 kΩ, so R2 has 20 kΩ in parallel with it. Their combined resistance is: R5
56 000 3 20 000 5 14:7 kΩ 56 000 1 20 000
The value of vOUT, as read with the meter connected to the divider, is: 10 3 14 700=11 470 5 1:28 V
The meter reading is 2.31 V too low.
The example demonstrates that a moving-coil meter is far from accurate when used to measure voltages in a network of high resistances. Using a digital meter gives no such problems, as it has an input
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PART | 1
resistance of many megohms and takes virtually no current from the network.
If there is no circuit connected to the cell, as in the previous diagram, no current flows through the cell. So the current through the internal resistance is zero. Using the Ohm’s Law equation:
VARIABLE POTENTIAL DIVIDERS
V 5 IR 5 0 3 R 5 0
The diagrams show three ways of building a potential divider with variable output voltage:
With no current flowing there is zero pd across the internal resistance. The total pd between the terminals of the cell is 1.5 1 0 5 1.5 V. pd between the terminals 5 emf of cell
R1 VIN
VIN
VR1
VR1
VOUT (A)
Circuits
pd and Volts VOUT
(B)
The unit of pd (symbol, V) and emf (symbol, E) is the volt (symbol, V). Spot the difference? (p. 349)
R1 VIN
VR1
Current and Amps
VOUT
R2
The unit of current (symbol, I ) is the amp (symbol, A).
(C) FIGURE 3.7
When calculating vOUT, we think of the variable resistor as two resistors joined at the point of contact of the wiper.
Let the internal resistance of the cell be Rint. If the cell is connected to an external circuit, with resistance Rext, a current flows through this and also through the internal resistance. These two resistances are in series and the circuit below is a potential divider. Potential = (1.5–iRint) V
INTERNAL RESISTANCE When analysing circuits we have so far ignored the fact that the source itself has resistance. Often this is so small that it can be ignored, but there are times when it can not. To understand the effect of the internal resistance of a cell (and also the meaning of the term emf), think of the cell as two components, connected in series.
current = i A R int source of emf
pd = iRint
Rext
pd = 1.5 V
cell 0V FIGURE 3.9 When a current flows there is a fall of potential (or voltage drop) across the internal resistance.
+1.5 V
cell
0V
+1.5 V internal resistance
pd = 0V
source + of emf cell
pd = 1.5 V
The fall in potential across the internal resistance (sometimes called the ‘lost volts’) means that the: pd between the terminals , emf of cell
0V
FIGURE 3.8 A cell can be thought of as a source of emf (the chemical energy of the cell) in series with a resistance (the internal resistance of the cell).
Taking this circuit as a potential divider, it divides the emf and its ‘output’ is at the positive terminal of the cell or other source. The equation for a potential divider is: vOUT 5 vIN 3 R2 =ðR1 1 R2 Þ
Topic | 3
Potential Dividers
29
In the case of the circuit above this becomes: terminal pd 5 emf 3 Rext =ðRint 1 Rext Þ
If Rint is small compared with Rext, as is the case with batteries and other power supplies, the terminal potential is only slightly less than the emf. The voltage drop can be ignored. If the source has high internal resistance, such as the crystal in a piezo-electric microphone, and it is connected to an external circuit of low resistance, most of the emf from the crystal will be lost in its internal resistance. To obtain the maximum signal from the microphone the external circuit must have high resistance. An amplifier with an operational amplifier (p. 87) as its input would be suitable for this.
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6
ACTIVITIES 1 Set up several potential divider circuits, with different input voltages and using various pairs of resistors. Calculate the expected output voltage and check the result with a digital multimeter. Use the resistor colour code (see Supplement A) to select suitable resistors from the E24 series. If possible, use resistors with 1% tolerance. 2 Extend Activity 1 by experimenting with loads of various resistances. 3 Measure the internal resistance of an alkaline cell. Connect a digital testmeter across the terminals of the cell to measure its emf of the cell with no load attached (the current flowing to a digital meter is virtually zero). Then measure the pd of the cell when a low-value resistor is connected across it. A 10 Ω, 0.5 W resistor can be used. Warning! it will become very hot! Take the reading quickly and then disconnect. Calculate the internal resistance using the equation opposite.
QUESTIONS ON POTENTIAL DIVIDERS 1 A load of 300 Ω is connected in parallel with the lower resistor (R2) of a potential divider. The resistance of R2 is 1.2 kΩ. What is the combined resistance of R2 and the load? 2 A potential divider consists of two resistors in series. R1 is 4.5 kΩ and R2 is 3.3 kΩ. The potential applied to the divider is 12 V. What is the combined resistance of R1 and R2? How much current flows
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10
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12
13
14
15
through the divider with no load connected? What is the potential across R2? A potential divider is built from two resistors of 36 Ω and 24 Ω. The supply potential is 8 V. What two output potentials are obtainable from this divider? State Ohm’s Law and use it to prove that the combined resistance of several resistors in series is equal to their sum. Use Ohm’s law to obtain a formula for calculating the combined resistance of three resistances connected in parallel. Three resistors, 47 Ω, 120 Ω, and 91 Ω, are connected in series with each other and with an unknown fourth resistor. When a pd of 17 V is applied across the 4-resistor chain, a current of 50 mA flows through it. What is the pd across each resistor? What is the resistance of the fourth resistor? Suggest practical uses for each of the three types of variable divider. A potential divider consists of a 390 Ω resistor and a 470 Ω resistor. The load is 2.5 kΩ, connected across the 470 Ω resistor. What is the output potential of the divider when the input is 10 V? Design a potential divider, using resistors of the E24 series, to produce a 4V output from a 10 V input. It must be able to supply up to 1.8 mA to the load without a serious drop in voltage. Design a potential divider, using E24 series resistors, in which the output potential is approximately 45% of the input. A variable potential divider produces an output ranging from 6.75 V to 9 V when the input potential is 9 V. What values and types of resistor are being used? For each of the variable dividers shown in Figure 3.4, describe in words the range of output voltages obtainable. Design a divider to provide a variable output of from 0 V to 5 V, given a constant input of 16 V. The resistance of the variable resistor is 100 kΩ. A battery has an emf of 6 V. The pd across its terminals falls to 5.88 V when the battery is connected to a 120 Ω resistor. What is its internal resistance? A cell has an emf of 1.2 V and an internal resistance of 0.5 Ω. What is the pd across its terminals when it is connected to a 22 Ω resistor?
Potential Dividers A potential divider is a resistor network that produces a fixed or variable potential (voltage). This potential is lower than the potential of the supply. The potential divider is probably the most often used circuit module. There are many instances of potential dividers being used in this book. These include the circuits on pp. 12, 14, 15, 17, and 18 of Topic 2. Essentially, a potential divider consists of two resistors connected in series.
CURRENT AND PD Assuming that no current flows out of the divider at the junction of R1 and R2, the same current, i, flows through both resistors. Ohm’s law applies to each resistor. For R2: i 5 vOUT =R2
Ohm’s Law also applies to the two resistors when connected in series: i 5 vIN =ðR1 1 R2 Þ
i
Combining these two equations: vOUT =R2 5 vIN =ðR1 1 R2 Þ
R1
vOUT 5 vIN 3 R2 =ðR1 1 R2 Þ VIN R2
VOUT
We use this equation to calculate vOUT, given vIN and the values of the two resistors. Example
FIGURE 3.1
Built on a breadboard it might look like this:
Calculate vOUT, given that vIN 5 9 V, R1 5 1.6 kΩ, and R2 5 2 kΩ. vOUT 5 9 3 2000=ð1600 1 2000Þ vOUT 5 5 V The output of the potential divider is 5 V.
9V R1 1k6 VIN R2 2k
FIGURE 3.2 The supply vIN comes from a battery or power pack on the left. The potential vOUT across R2 is measured by test leads on the right. Electronics: Circuits and Systems. r 2011 Owen Bishop. Published by Elsevier Ltd. All rights reserved.
VOUT =5V
0V FIGURE 3.3 When the source voltage is 9 V, the output voltage is 5 V.
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26
PART | 1
THE EFFECT OF THE LOAD
Circuits
Example
The foregoing calculation assumes that no current flows out of the output terminal. This is almost true when we use a digital meter to measure vOUT. We find that the measured value of vOUT is almost correct, allowing for possible tolerance errors in the resistor values. However, current is drawn from the output in most practical circuits. This is what happens: Current i passes through R1. At the output of the divider, a current iLOAD passes through the load while the remainder i2 passes on through R2. The diagram shows that R2 and the load (RLOAD) are resistances in parallel.
A load with resistance 3 kΩ is connected to the potential divider described in the previous example. What is the new value of vOUT? The combined resistance of R2 and the load is: R5
2000 3 3000 5 1200 Ω 2000 1 3000
In effect, the divider now consists of R1 in series with 1.2 kΩ, so: vOUT 5 9 3
1200 5 3:86 V 1600 1 1200
The output voltage is reduced to 3.86 V. i
In the example, the output voltage of the divider falls by 1.14 V when the load is connected. The amount of the fall depends on the resistance of the load.
R1 VIN
i LOAD i2
INCREASING THE PRECISION
R2
VOUT
RLOAD
FIGURE 3.4 When a load is connected to the divider, it is in parallel with R2.
The usual technique for preventing excessive drop in output voltage is to have a relatively large current flowing through the divider. Then the current drawn by the load has relatively less effect. As a rule-ofthumb, make the divider current about ten times the current taken by the load. Example
If R2 and the load are resistances in parallel, their combined resistance is less than the resistance of either of them separately. Therefore the pd across R2 is less when the load is connected. The measured value of vOUT is less than expected.
In the previous example, the current drawn by the 3 kΩ load is: iLOAD 5 5=3000 5 1:67 mA
9V
9V R1 1k6
R1 1k6 equivalent
VIN
to R2 2k
0V
VOUT
VIN
RLOAD 3k
1k2
0V R2 RLOAD
FIGURE 3.5 Connecting a load to the divider lowers the resistance across which vOUT appears.
ROUT = 3.86 V
Topic | 3
Potential Dividers
The current flowing through the divider must be at least ten times that amount, which is 16.7 mA. For a current of this size, R1 and R2 in series must total no more than R 5 vIN/0.0167 5 539 Ω. We have to specify two resistors that have the same ratio as in the example, but total no more than 539 Ω. Self Test a Select two E24 resistors that total 540 Ω and form a potential divider that has 5 V output with 9 V input, under no load. What is the output when a load of 2 kΩ is connected? b Design a potential divider using E24 resistors that gives an output of 3.3 V when the input is 6 V and there is no load. What is the output when a load of 1 kΩ is connected? Select the resistor values to minimize the voltage drop under this load.
27
R1. We can adjust the temperature at which the circuit changes state by adjusting VR1.
MEASUREMENT ERRORS Unlike a digital testmeter, an analogue moving-coil testmeter needs appreciable current to drive it. Typically, the coil resistance is 200 kΩ when switched to the 10 V range. A cheap analogue meter may have a coil resistance of only 20 kΩ. If a potential divider or other network of high-value resistors is tested with an analogue meter, serious errors may occur. This is because the coil of the meter acts as a load on the network. Current is drawn by the meter and this results in a voltage drop at the point where voltage is being measured.
Unfortunately, using resistor values totalling only 539 Ω means that the divider takes a current of over 25 mA. In a battery powered device we might not be able to afford to waste so much current.
R1 100k 10 V R2 56k
VOUT
R METER 20k
There are more calculations to try on the Companion website.
USING POTENTIAL DIVIDERS WITH SENSORS In an electronic system, the output signal from the sensor may be a change in its resistance. For example, when the sensor is a light dependent resistor (LDR), as in the circuit on p. 14, variations in the amount of light falling on the LDR result in changes in its resistance. Because of this, the LDR is said to be a resistive sensor. Thermistors, photodiodes, and strain gauges are other examples of resistive sensors. In the circuit on p. 14, the LDR resistor is part of a potential divide so the potential at the junction of R1 and R2 varies according to the amount of light falling on the LDR (R2). The transistor Q1 is switched on and off, so activating or deactivating the audible warning device. Another example is shown on p. 15. In this case the sensor is a thermistor. Together with the variable resistor, VR1, it forms a potential divider with its output at junction A. The output voltage at that junction depends on the setting of VR1 and the temperature of
FIGURE 3.6 An analogue meter may give low voltage readings on a high-resistance network.
Example In the diagram, the actual value of vOUT is: 10 3 56 000=156 000 5 3:59 V But the meter has a coil resistance of 20 kΩ, so R2 has 20 kΩ in parallel with it. Their combined resistance is: R5
56 000 3 20 000 5 14:7 kΩ 56 000 1 20 000
The value of vOUT, as read with the meter connected to the divider, is: 10 3 14 700=11 470 5 1:28 V
The meter reading is 2.31 V too low.
The example demonstrates that a moving-coil meter is far from accurate when used to measure voltages in a network of high resistances. Using a digital meter gives no such problems, as it has an input
28
PART | 1
resistance of many megohms and takes virtually no current from the network.
If there is no circuit connected to the cell, as in the previous diagram, no current flows through the cell. So the current through the internal resistance is zero. Using the Ohm’s Law equation:
VARIABLE POTENTIAL DIVIDERS
V 5 IR 5 0 3 R 5 0
The diagrams show three ways of building a potential divider with variable output voltage:
With no current flowing there is zero pd across the internal resistance. The total pd between the terminals of the cell is 1.5 1 0 5 1.5 V. pd between the terminals 5 emf of cell
R1 VIN
VIN
VR1
VR1
VOUT (A)
Circuits
pd and Volts VOUT
(B)
The unit of pd (symbol, V) and emf (symbol, E) is the volt (symbol, V). Spot the difference? (p. 349)
R1 VIN
VR1
Current and Amps
VOUT
R2
The unit of current (symbol, I ) is the amp (symbol, A).
(C) FIGURE 3.7
When calculating vOUT, we think of the variable resistor as two resistors joined at the point of contact of the wiper.
Let the internal resistance of the cell be Rint. If the cell is connected to an external circuit, with resistance Rext, a current flows through this and also through the internal resistance. These two resistances are in series and the circuit below is a potential divider. Potential = (1.5–iRint) V
INTERNAL RESISTANCE When analysing circuits we have so far ignored the fact that the source itself has resistance. Often this is so small that it can be ignored, but there are times when it can not. To understand the effect of the internal resistance of a cell (and also the meaning of the term emf), think of the cell as two components, connected in series.
current = i A R int source of emf
pd = iRint
Rext
pd = 1.5 V
cell 0V FIGURE 3.9 When a current flows there is a fall of potential (or voltage drop) across the internal resistance.
+1.5 V
cell
0V
+1.5 V internal resistance
pd = 0V
source + of emf cell
pd = 1.5 V
The fall in potential across the internal resistance (sometimes called the ‘lost volts’) means that the: pd between the terminals , emf of cell
0V
FIGURE 3.8 A cell can be thought of as a source of emf (the chemical energy of the cell) in series with a resistance (the internal resistance of the cell).
Taking this circuit as a potential divider, it divides the emf and its ‘output’ is at the positive terminal of the cell or other source. The equation for a potential divider is: vOUT 5 vIN 3 R2 =ðR1 1 R2 Þ
Topic | 3
Potential Dividers
29
In the case of the circuit above this becomes: terminal pd 5 emf 3 Rext =ðRint 1 Rext Þ
If Rint is small compared with Rext, as is the case with batteries and other power supplies, the terminal potential is only slightly less than the emf. The voltage drop can be ignored. If the source has high internal resistance, such as the crystal in a piezo-electric microphone, and it is connected to an external circuit of low resistance, most of the emf from the crystal will be lost in its internal resistance. To obtain the maximum signal from the microphone the external circuit must have high resistance. An amplifier with an operational amplifier (p. 87) as its input would be suitable for this.
3
4
5
6
ACTIVITIES 1 Set up several potential divider circuits, with different input voltages and using various pairs of resistors. Calculate the expected output voltage and check the result with a digital multimeter. Use the resistor colour code (see Supplement A) to select suitable resistors from the E24 series. If possible, use resistors with 1% tolerance. 2 Extend Activity 1 by experimenting with loads of various resistances. 3 Measure the internal resistance of an alkaline cell. Connect a digital testmeter across the terminals of the cell to measure its emf of the cell with no load attached (the current flowing to a digital meter is virtually zero). Then measure the pd of the cell when a low-value resistor is connected across it. A 10 Ω, 0.5 W resistor can be used. Warning! it will become very hot! Take the reading quickly and then disconnect. Calculate the internal resistance using the equation opposite.
QUESTIONS ON POTENTIAL DIVIDERS 1 A load of 300 Ω is connected in parallel with the lower resistor (R2) of a potential divider. The resistance of R2 is 1.2 kΩ. What is the combined resistance of R2 and the load? 2 A potential divider consists of two resistors in series. R1 is 4.5 kΩ and R2 is 3.3 kΩ. The potential applied to the divider is 12 V. What is the combined resistance of R1 and R2? How much current flows
7 8
9
10
11
12
13
14
15
through the divider with no load connected? What is the potential across R2? A potential divider is built from two resistors of 36 Ω and 24 Ω. The supply potential is 8 V. What two output potentials are obtainable from this divider? State Ohm’s Law and use it to prove that the combined resistance of several resistors in series is equal to their sum. Use Ohm’s law to obtain a formula for calculating the combined resistance of three resistances connected in parallel. Three resistors, 47 Ω, 120 Ω, and 91 Ω, are connected in series with each other and with an unknown fourth resistor. When a pd of 17 V is applied across the 4-resistor chain, a current of 50 mA flows through it. What is the pd across each resistor? What is the resistance of the fourth resistor? Suggest practical uses for each of the three types of variable divider. A potential divider consists of a 390 Ω resistor and a 470 Ω resistor. The load is 2.5 kΩ, connected across the 470 Ω resistor. What is the output potential of the divider when the input is 10 V? Design a potential divider, using resistors of the E24 series, to produce a 4V output from a 10 V input. It must be able to supply up to 1.8 mA to the load without a serious drop in voltage. Design a potential divider, using E24 series resistors, in which the output potential is approximately 45% of the input. A variable potential divider produces an output ranging from 6.75 V to 9 V when the input potential is 9 V. What values and types of resistor are being used? For each of the variable dividers shown in Figure 3.4, describe in words the range of output voltages obtainable. Design a divider to provide a variable output of from 0 V to 5 V, given a constant input of 16 V. The resistance of the variable resistor is 100 kΩ. A battery has an emf of 6 V. The pd across its terminals falls to 5.88 V when the battery is connected to a 120 Ω resistor. What is its internal resistance? A cell has an emf of 1.2 V and an internal resistance of 0.5 Ω. What is the pd across its terminals when it is connected to a 22 Ω resistor?