   Voltage Dividers
Introduction

In earlier work you will have met the idea that the current flowing through each of several resistors connected in series with a power supply is the same, even if their individual resistances are different.

By applying Ohm's Law to the type of circuit shown above, we can show that different resistances connected in series will have different p.d.s across them and that the sum of the individual p.d.s gives the voltage of the supply connected to the circuit. In this unit we will see how two resistors in series with a power supply can be treated as a voltage divider and can be used as the basis for environmental sensors. (Note that some texts refer to such circuits as potential dividers.)

Voltage divider formula
Alter the value of the variable resistor in Fig.1 to confirm the laws governing the
current
The rate of flow of charge past any specific point in a circuit. The base unit of current is the Ampere.
current
and p.d.s in this
series circuit
Components are connected in series when the same electrical charges pass through both.
series circuit
.

Click on the figure below to interact with the model.  Figure 1.  Resistors in series.

Complete the following statements for the circuit in Fig.1.

• When the variable resistor in Fig.1 has a value of to 100 Ω the p.d. across the variable resistor is  V. The total resistance in the circuit is  Ω, so the variable resistor now accounts for per cent of the total resistance in the circuit.
• When the resistors in Fig.1 have the same value, the supply
voltage
The voltage across a component is the electrical energy transferred by 1 coulomb of charge passing through the component.
voltage
is divided equally between the two resistors.

Complete the following statements for Fig.1.

• When the variable resistor has a resistance of 50 Ω the p.d. across the variable resistor is  V, the supply voltage for the circuit is still  V, and the total circuit resistance is now  Ω. The p.d. across the variable resistor accounts for per cent of the p.d. supplied to the circuit.
• When the resistors in Fig.1 are in the ratio two to one, the supply voltage is divided in the ratio two-thirds to one-third.

Complete the following statements for Fig.1.

• When the variable resistor has a value of 200 Ω the total resistance in the circuit is  Ω and the variable resistor accounts for per cent of this total. The p.d. across the variable resistor is  V and the supply voltage for the circuit is still  V. This means that the p.d. across the variable resistor now accounts for per cent of the p.d. supplied to the circuit.
• When the resistor values are in the ratio two to one, the p.d.s across them are two-thirds and one-third of the supply voltage. The larger resistor has the larger share of the total supply voltage.

In each of these examples you should have noticed that the p.d. across the variable resistor is the same fraction of the total p.d. as its
resistance
The opposition to the flow of current provided by a circuit is called resistance. Resistance is measured in units called Ohms.
resistance
is of the total resistance.

Resistors with values of 90 kΩ and 10 kΩ are connected in series with a 12 V battery. What percentage of the supply voltage is across the 10 kΩ resistor?
• Resistors with values of 22 kΩ and 47 kΩ are connected in series with a 12 V battery. Which resistor will have the larger p.d. across its terminals?
• Figure 2. Resistors and their p.d.s. The circuit of Fig.2 is similar to that of Fig.1 except that some labels have been added. The fraction of the total p.d. across the variable resistor V1 is the same as the fraction that the variable resistor is of the total resistance.   We also know that the total resistance of two resistors in series is equal to the sum of their individual values.   We can combine the two equations.   And rearranging them. This
voltage divider
A voltage divider is a circuit involving two resistors in series. The voltage across each resistor is determined by the value of the resistors and the supply voltage, as given by the equation: These circuits are often used in sensors.
voltage divider
equation can also be expressed as:

We can use this voltage divider equation to calculate the p.d. across the 10 kΩ resistor V10k, in Fig.3.

 Figure 3. Using the voltage divider formula.  We use the voltage divider formula.   We then substitute the values from above.   We finally simplify to find the value of V10k. If the p.d. across the 10 kΩ resistor in Fig.3 is 6 V, what is the p.d. across the 5 kΩ resistor?
•  V   (to the nearest whole number)
• We can also use the voltage dividers equation to show that the p.d. across the 5 kΩ resistor is 3 V. This acts as a cross-check for the validity of the voltage divider equation. We use the voltage divider formula again.   We then substitute the values from above.   We finally simplify to find the value of V5k.  This is as we expect.

Set the variable resistor in Fig.4 below to 10 kΩ and note that the 9 V p.d. from the supply is shared equally between the two 10 kΩ resistors.

Click on the figure below to interact with the model.  Figure 4.  A fixed and variable resistor in series with a 9 V supply.

When the resistors of Fig.4 are of equal value, the p.d. of the supply is shared equally between the two. Alter the value of the variable resistor in Fig.4 and observe how the supply voltage is shared between the two resistors.

Complete the following statements.

• When the variable resistor in Fig.4 has a higher value than the fixed resistor, the p.d. across it is than half the supply voltage.

When the variable resistor has a lower value than the fixed resistor, the p.d. across it is than half the supply voltage.
• A convenient way of remembering how the voltage divider behaves is to think that the larger resistance 'grabs' the bigger share of the supply voltage.

Understanding the formula
The voltage at the midpoint of the voltage divider in Fig.5 has been labelled Vmp.

Click on the figure below to interact with the model.  Figure 5.  A voltage divider with the midpoint voltage labelled.

When the variable resistor in Fig.5 is set to 15 kΩ what is the p.d. across it?
•  V   (to 2 d.p.)
• When the variable resistor in Fig.5 is set to 15 kΩ the p.d. across the 10 kΩ fixed resistor is
•  V   (to 2 d.p.)
• When the variable resistor in Fig.5 is set to 15 kΩ the midpoint of the voltage divider is 3.60 V above the 0 V terminal of the power supply and 5.40 V below the 9 V terminal. The midpoint is said to be at a voltage 3.60 V (Vmp = 3.60 V). These voltages are relative to the 0 V terminal of the power supply, so the 0 V terminal of the power supply is used as a reference point.

Complete the following statements.

• When the value of the variable resistor in Fig.5 is increased the p.d. across it . Consequently the p.d. across the fixed resistor .
• The voltage at the midpoint of the voltage divider in Fig.5 reduces when the resistance of the variable resistor increases.

In Fig.6 the positions of the variable and fixed resistors have been interchanged. The voltage at the midpoint of the voltage divider has again been labelled Vmp.

Click on the figure below to interact with the model.  Figure 6.  A voltage divider, with the lower resistor variable.

When the variable resistor in Fig.6 is set to 15 kΩ, the p.d. across the variable resistor is
• Select the correct values in the following statements to summarize the p.d.s and supply voltage in Fig.6.

• When the variable resistor is set to 15 kΩ, the midpoint of the voltage divider is  V above the 0 V terminal of the power supply and  V below the 9 V terminal.

The voltage at the midpoint is  V.
• Increase the value of the variable resistor in Fig.6 and complete the following statements.

• When the value of the variable resistor is increased, the p.d. across it . Consequently the p.d. across the fixed resistor .
• If the fixed and variable resistors are arranged as shown in Fig.6, the voltage at the midpoint of the voltage divider increases when the resistance of the variable resistor increases.

The circuits of Fig.5 and 6 are very similar and can be easily confused. When considering how the voltage at the midpoint of a voltage divider changes as one of the resistance values alters, you must:

• consider how the p.d. across that resistor changes;

• consider how the p.d. across the resistor connected to the 0 V terminal of the supply changes;

• determine the effect on the voltage at the midpoint.

The behaviour of each of these voltage dividers is summarized below.

Circuit Effect of increasing the variable resistor
The midpoint voltage reduces when the value of the variable resistor increases.
The midpoint voltage increases when the value of the variable resistor increases.

Sensors
Your senses (such as touch, sight, and sound) are the inputs to your brain. In much the same way, sensors can be inputs in electronic systems. Sensors respond to changes in external conditions (such as heat or light). They do this because they are semiconductors and the resistance of a semiconductor changes when certain conditions change.
Click on the figure below to interact with the model.  Figure 7.  Two common sensors.

Fig.7 shows two common devices based on semiconductors. When these are used in circuits, moving the sliders simulates altering the surrounding conditions The resistance of the
thermistor
A thermistor is an electronic component whose resistance changes when its temperature alters. The resistance of a 'negative temperature coefficient' (n.t.c.) thermistor reduces as the temperature increases.
thermistor
depends upon its temperature, while the resistance of the
LDR
The resistance of a light dependent resistor reduces as the light intensity increases. This feature makes LDRs ideal for use in light sensing circuits.
LDR
is determined by the quantity of light hitting it.
When the temperature surrounding the thermistor increases, the resistance of the thermistor …
• When the quantity of light entering the LDR decreases, the resistance of the LDR …
• When the temperature of the thermistor is as low as possible, the resistance of the thermistor is at its maximum value. As the temperature increases, the resistance of the thermistor decreases and when the temperature is at its maximum, the resistance of the thermistor is at its minimum value.

When the light shining on the LDR is at its brightest, the resistance of the LDR is at its minimum value. Decreasing the amount of light reaching the LDR increases the resistance of the LDR.

Temperature-sensing circuit
The thermistor can be used in the type of voltage divider circuit shown in Fig.8 to make a temperature-sensing circuit.

Click on the figure below to interact with the model.  Figure 8.   Thermistor in a voltage divider.

Complete the following sentences.

• When the temperature of the thermistor in Fig.8 increases, its resistance . Consequently the p.d. across the thermistor and so the p.d. across the fixed resistor .
• The reading on the voltmeter in Fig.8 increases as the temperature increases and falls as the temperature decreases.

Placing a voltmeter across the fixed resistance in a circuit such as Fig.8 produces a voltage readout which changes with temperature. This relationship is exploited in commercial devices which use additional electronics to provide a direct readout of the temperature.

Light-sensing circuits
An LDR can be used in a voltage divider such as shown in Fig.9 to make a light-sensing circuit. Use the slider to alter the quantity of light entering the LDR and observe the changes in the voltmeter reading.

Click on the figure below to interact with the model.  Figure 9.  Voltage divider used as a light level indicator.

Complete the following sentences to explain the operation of the light-sensing circuit in Fig.9.

• When the quantity of light entering the LDR increases, its resistance . Consequently the p.d. across the LDR and so the p.d. across the fixed resistor .

The reading on the voltmeter as the light level increases and falls as the light level .
• In commercial applications additional circuits convert the voltage from the voltage divider circuit into a light-level reading so that the readout on the lightmeter's display indicates the light level entering the LDR.

A potential problem
At first sight the circuit in Fig.10 might be considered suitable for a system that switches on a lamp automatically in the dark. When darkness falls, the resistance of the LDR increases and so we would expect the p.d. across the LDR to increase. This should make the voltage at the midpoint of the voltage divider high enough to light the lamp.

Click on the figure below to interact with the model.  Figure 10.  A possible circuit for automatic lighting.

Does reducing the light level in the circuit of Fig.10 cause the lamp to light?
• The operation of a voltage divider relies on the fact the current through both resistors in a series circuit is the same. In Fig.10 the lamp is connected to the midpoint of the voltage divider and so not all the current passing through the top resistor passes through the lower resistor. In fact, the resistance of the lamp is much lower than the LDR so almost all the current in the circuit passes through the lamp regardless of conditions surrounding the LDR. However, this is still not sufficient to light the lamp due to the current limiting effect of the 10 kΩ resistor.

In this situation electronic engineers would say that the lamp is 'crippling' the operation of the voltage divider.

Fig.11 shows how electronic engineers can overcome this loading effect. Decrease the level of light entering the LDR and note what happens to the lamp now.

Click on the figure below to interact with the model.  Figure 11.  Automatic light level adjustment.

Does reducing the light level in the circuit of Fig.11 cause the lamp to light?
• The additional component (a
transistor
A transistor is a semiconductor device that is used extensively in amplifier and switching circuits.
transistor
) ensures that sufficient current to operate the lamp in Fig.11 now comes directly from the supply. The current in both resistors in the voltage divider is almost the same, so the voltage divider switches on the current to the lamp.

You will learn more about these types of circuits in specialist electronics courses.

Summary

When resistors are connected in series with a power supply, the fraction of the total p.d. across any specific resistor is the same as the fraction that its resistance is of the circuit's total resistance.

The voltage across one resistor in a voltage divider can be calculated using the equation: Voltage dividers incorporating LDRs or thermistors can be used to produce circuits giving output voltages which depend on the environment in which the sensor is placed.

Exercises
1. In Fig.12 two resistors are connected in series with a 6 V power supply. The power supply has negligible
internal resistance
All batteries or power supplies have internal resistance. This resistance has the effect of reducing the output p.d. as the current supplied increases.
internal resistance
. Calculate the effective value of the two resistors when connected in series.
•  Ω   (to the nearest whole number)
• Figure 12. 2. Calculate the current in each of the resistors in Fig.12.
•  µA   (to the nearest whole number)
• 3. Calculate the p.d. across the 47 kΩ in Fig.12.
•  V   (to 2 d.p.)
• 4. Calculate the p.d. across the 22 kΩ resistor in Fig.12.
•  V   (to 2 d.p.)
• 5. A pupil studying voltage dividers sets up the circuit shown in Fig.13. When the switch is open the voltmeter shows a reading of 3 V. What is the voltage across the 75 Ω resistor?
•  V   (to the nearest whole number)
• 6. When the switch in Fig.13 is open the voltmeter shows a reading of 3 V. What is the current in the 75 Ω resistor?
•  mA.   (to the nearest whole number)
• Figure 13. 7. Calculate the value of resistor R in Fig.13.
•  Ω   (to 1 d.p.)
• 8. The student now closes the switch in Fig.13. Calculate the effective resistance of the resistors connected to the 9 V battery.
•  Ω   (to the nearest whole number)
• 9. Calculate the p.d. across the 75 Ω resistor in Fig.13.
•  V   (to 1 d.p.)
• 10. Calculate the current in the 25 Ω resistor in Fig.13 when the switch is closed.
•  mA   (to the nearest whole number)
• 11. When writing up her findings of the experiment shown in Fig.13, the student states: 'When I connect the 25 Ω load resistor across the lower resistor in the voltage divider, the voltage at the midpoint reduces.' Is this statement true or false.
• Figure 14. 12. Resistors P, Q, and R , shown in Fig.14, are connected in series with a cell with
EMF
EMF is the common abbreviation for the electromotive force – a measurement of a battery's capacity to provide energy to the charges.
EMF
E, which has no internal resistance. In trying to prove that the total resistance (Rtot) of this combination is P + Q + R, a pupil makes the statements listed below. Classify these statements as true or false.
•  The current through each resistor is identical. False True The sum of the p.d.s across P, Q, and R equals E. False True The heat generated by each resistor is the same even for different values of P, Q, and R. False True
• 13. A light dependent resistor (LDR) is connected in series with a 10 kΩ resistor and a 1.5 V battery of negligible internal resistance. If the LDR's resistance in daylight and darkness is as shown in Fig.15. Calculate the voltage reading in daylight.
•  V   (to 2 d.p.)
• Figure 15. Conditions LDR Resistance Daylight 500 Ω Darkness 90 kΩ

14. Calculate the voltmeter reading in Fig.15 when the LDR is covered.
•  V   (to 2 d.p.)
• Figure 16. 15. In Fig.16, X, Y, and Z are identical 10 kΩ resistors and V1, V2, and V3 are identical ideal voltmeters. A pupil is asked to connect another resistor in parallel with X but before doing so he makes the following predictions. Classify the prediction as true or false.
•  The reading on V1 will not change. False True The reading on V2 will reduce. False True The calculated value of V1 − (V2 + V3) will be greater than the p.d. across resistor Y. False True
• Well done!
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