Recall and use the equation for two resistors used as a potential divider R_1 / R_2 = V_1 / V_2

4.3.3 Action and Use of Circuit Components

Learning Objective (Cambridge IGCSE 0625)

• Recall and use the potential‑divider relationship \(\displaystyle\frac{R_1}{R_2}= \frac{V_1}{V_2}\) – the two resistors must be **in series** with the supply.
• Apply Ohm’s law (\(V = IR\)) to understand that, for a constant current, the voltage across a conductor increases when its resistance increases.
• Interpret circuit diagrams containing fixed and variable resistors, and describe how a variable potential divider (potentiometer) works.

1. What is a Potential Divider?

A potential (or voltage) divider is a simple series circuit that produces a lower output voltage from a higher input voltage by using two resistors. The output voltage is taken from the junction between the resistors and is usually denoted \(V_{\text{out}}\) (which, for a basic divider, is the same as \(V_2\)).

2. Derivation of the Divider Formula

1. Consider a supply voltage \(V_{\text{in}}\) connected to two resistors \(R_1\) and \(R_2\) that are **in series**.
2. The same current flows through both resistors: \[ I=\frac{V_{\text{in}}}{R_1+R_2} \]
3. Ohm’s law gives the voltage across each resistor: \[ V_1 = I R_1,\qquad V_2 = I R_2 \]
4. Dividing the two expressions eliminates the current: \[ \frac{V_1}{V_2}= \frac{I R_1}{I R_2}= \frac{R_1}{R_2} \]
5. Re‑arranging gives the form required by the syllabus: \[ \boxed{\frac{R_1}{R_2}= \frac{V_1}{V_2}} \]

3. Alternate Forms of the Divider Equation

• Output voltage (across the lower resistor, \(R_2\)): \[ V_{\text{out}} = V_2 = V_{\text{in}}\;\frac{R_2}{R_1+R_2} \]
• Input voltage: \[ V_{\text{in}} = V_1 + V_2 \]
• Using Ohm’s law, for a given current \(I\) the voltage across a resistor is \(V = IR\). Hence, if the resistance of a conductor is increased while the current stays the same, the voltage across it also increases – a core IGCSE concept.

4. Fixed vs. Variable Potential Dividers

4.1 Fixed‑Resistor Divider

Both resistors have fixed values. For a given supply voltage the output voltage is fixed.

4.2 Variable‑Resistor Divider (Potentiometer)

A potentiometer is a three‑terminal variable resistor. When used as a divider the middle terminal (the wiper) provides the output voltage. Turning the knob changes the ratio \(\displaystyle\frac{R_1}{R_2}\) and therefore varies \(V_{\text{out}}\) continuously, while the total resistance of the potentiometer remains constant.

Real‑world example: The volume control on a portable radio is a potentiometer. Adjusting the knob changes the proportion of the battery voltage that reaches the amplifier, allowing the user to set the loudness.

Transition to Diagrams

In the diagrams below the voltage taken from the junction is called the output voltage (\(V_{\text{out}}\)). For a simple two‑resistor divider, \(V_{\text{out}} = V_2\).

5. Using the Divider Equation

The relation \(\displaystyle\frac{R_1}{R_2}= \frac{V_1}{V_2}\) (or equivalently \(V_{\text{out}} = V_{\text{in}} \frac{R_2}{R_1+R_2}\)) allows you to determine any one of the four quantities (\(R_1\), \(R_2\), \(V_1\), \(V_2\)) when the other three are known.

6. Worked Examples

Example 1 – Output voltage of a fixed divider

Given: \(V_{\text{in}} = 12\ \text{V}\), \(R_1 = 2\ \text{k}\Omega\), \(R_2 = 3\ \text{k}\Omega\). Find \(V_{\text{out}}\).

1. Calculate the ratio \(\displaystyle\frac{R_1}{R_2}= \frac{2}{3}\).
2. Use the output‑voltage form: \[ V_{\text{out}} = 12\;\frac{3}{2+3}=12\;\frac{3}{5}=7.2\ \text{V} \]

Example 2 – Determining a resistor value

Given: \(V_{\text{in}} = 9\ \text{V}\), desired \(V_{\text{out}} = 3\ \text{V}\), \(R_2 = 1\ \text{k}\Omega\). Find \(R_1\).

1. Find the voltage across \(R_1\): \(V_1 = V_{\text{in}} - V_{\text{out}} = 9 - 3 = 6\ \text{V}\).
2. Apply the ratio form: \[ \frac{R_1}{1\ \text{k}\Omega}= \frac{6}{3}=2 \;\Rightarrow\; R_1 = 2\ \text{k}\Omega. \]

Example 3 – Potentiometer as a variable divider

Given: A 5 kΩ potentiometer is connected across a 12 V battery. The wiper is set so that the resistance from the top terminal to the wiper is \(R_1 = 1.5\ \text{k}\Omega\). Find \(V_{\text{out}}\).

1. Then \(R_2 = 5 - 1.5 = 3.5\ \text{k}\Omega\).
2. Output voltage: \[ V_{\text{out}} = 12\;\frac{3.5}{1.5+3.5}=12\;\frac{3.5}{5}=8.4\ \text{V}. \]

Example 4 – Loading the Divider

Given: The divider of Example 1 (\(R_1 = 2\ \text{k}\Omega\), \(R_2 = 3\ \text{k}\Omega\), \(V_{\text{in}} = 12\ \text{V}\)) now has a 1 kΩ load resistor \(R_L\) connected across the output terminals.

1. The load is in parallel with \(R_2\); the effective resistance at the lower leg becomes \[ R_{2,\text{eq}} = \frac{R_2\,R_L}{R_2+R_L}= \frac{3\ \text{k}\Omega \times 1\ \text{k}\Omega}{3\ \text{k}\Omega+1\ \text{k}\Omega}=0.75\ \text{k}\Omega. \]
2. Now use the output‑voltage form with \(R_{2,\text{eq}}\): \[ V_{\text{out, loaded}} = 12\;\frac{0.75}{2+0.75}=12\;\frac{0.75}{2.75}=3.27\ \text{V}. \]
3. Thus the presence of the load reduces the output from 7.2 V to about 3.3 V.

Example 5 – Using a Voltmeter

When measuring \(V_{\text{out}}\) with a voltmeter, the meter’s internal resistance \(R_{\text{vm}}\) should be much larger than \(R_2\) (ideally \(\;R_{\text{vm}} \gg R_2\)) so that the parallel combination does not noticeably change the divider ratio. For a typical digital voltmeter with \(R_{\text{vm}} \approx 10\ \text{M}\Omega\) and \(R_2\) in the kilohm range, the loading effect is negligible.

7. Common Pitfalls

• Loading the divider: Connecting a load (or a voltmeter with insufficient internal resistance) across the output changes the effective resistance of the lower leg, so the calculated voltage is no longer correct unless the load is included in the analysis.
• Label confusion: \(V_1\) is the voltage across \(R_1\); \(V_2\) (or \(V_{\text{out}}\)) is the voltage across \(R_2\). Keep the notation consistent throughout.
• Series requirement: The divider formula only works when \(R_1\) and \(R_2\) are connected **in series** with the supply. A parallel arrangement does not give a simple voltage division.
• Variable divider misunderstanding: Turning a potentiometer changes both \(R_1\) and \(R_2\) simultaneously; it is the **ratio** \(\displaystyle\frac{R_1}{R_2}\) that determines the output, not the absolute values.

8. Practice Questions

1. A 15 V supply is connected to a potential divider made of \(R_1 = 4\ \text{k}\Omega\) and \(R_2 = 6\ \text{k}\Omega\). Calculate the voltage across each resistor.
2. You need an output of 5 V from a 12 V battery using a divider. If \(R_2 = 2\ \text{k}\Omega\), what should \(R_1\) be?
3. A divider provides \(V_{\text{out}} = 2\ \text{V}\) across \(R_2 = 500\ \Omega\). The total supply voltage is 10 V. Find \(R_1\).
4. Explain why adding a 1 kΩ load across \(V_{\text{out}}\) in Question 1 changes the output voltage. (No calculation required.)
5. A 10 kΩ potentiometer is powered from a 9 V source. If the wiper is positioned so that the resistance from the top terminal to the wiper is 3 kΩ, what is the output voltage at the wiper?

9. Summary Table