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

Recall and use the equation for two resistors used as a potential divider:

\$\frac{R1}{R2} = \frac{V1}{V2}\$

1. What is a Potential Divider?

A potential divider (or voltage divider) is a simple circuit that produces a lower output voltage from a higher input voltage by using two series resistors. The output voltage is taken from the junction between the resistors.

2. Derivation of the Divider Formula

1. Consider a series circuit with a supply voltage \$V{\text{in}}\$ and two resistors \$R1\$ and \$R_2\$.
2. The same current \$I\$ flows through both resistors because they are in series:

   \$I = \frac{V{\text{in}}}{R1 + R_2}\$

3. The voltage across each resistor is given by Ohm’s law:

   \$V1 = I R1,\qquad V2 = I R2\$

4. Dividing the two expressions eliminates the current:

   \$\frac{V1}{V2} = \frac{I R1}{I R2} = \frac{R1}{R2}\$

5. Re‑arranging gives the useful form:

   \$\frac{R1}{R2} = \frac{V1}{V2}\$

3. Using the Formula

The formula allows you to find any one of the four quantities (\$R1\$, \$R2\$, \$V1\$, \$V2\$) when the other three are known.

4. Example Problems

Example 1 – Find the output voltage

Given: \$V{\text{in}} = 12\ \text{V}\$, \$R1 = 2\ \text{k}\Omega\$, \$R2 = 3\ \text{k}\Omega\$. Find \$V2\$ (the voltage across \$R_2\$).

1. First find the ratio \$\dfrac{R1}{R2} = \dfrac{2}{3}\$.
2. Using the divider relation \$\dfrac{R1}{R2} = \dfrac{V1}{V2}\$, write \$\dfrac{2}{3} = \dfrac{V1}{V2}\$.
3. Since \$V1 + V2 = V{\text{in}}\$, substitute \$V1 = V{\text{in}} - V2\$:

   \$\frac{2}{3} = \frac{12 - V2}{V2}\$

4. Cross‑multiply and solve:

   \$2V2 = 3(12 - V2) \Rightarrow 2V2 = 36 - 3V2 \Rightarrow 5V_2 = 36\$

   \$V_2 = 7.2\ \text{V}\$

Example 2 – Determine a resistor value

Given: \$V{\text{in}} = 9\ \text{V}\$, desired \$V2 = 3\ \text{V}\$, and \$R2 = 1\ \text{k}\Omega\$. Find \$R1\$.

1. From the divider relation:

   \$\frac{R1}{R2} = \frac{V1}{V2}\$

2. But \$V1 = V{\text{in}} - V_2 = 9\ \text{V} - 3\ \text{V} = 6\ \text{V}\$.
3. Thus \$\dfrac{R_1}{1\ \text{k}\Omega} = \dfrac{6}{3} = 2\$.
4. Therefore \$R_1 = 2 \times 1\ \text{k}\Omega = 2\ \text{k}\Omega\$.

5. Common Pitfalls

• Assuming the divider works when a load is connected across \$V_2\$ – the load changes the effective resistance.
• Mixing up \$V1\$ (voltage across \$R1\$) with \$V{\text{out}}\$ (often taken as \$V2\$).
• Forgetting that the resistors must be in series; parallel connections do not form a divider.

6. Practice Questions

1. A 15 V supply is connected to a potential divider made of \$R1 = 4\ \text{k}\Omega\$ and \$R2 = 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 \$R2 = 2\ \text{k}\Omega\$, what should \$R1\$ be?
3. A divider provides \$V2 = 2\ \text{V}\$ across \$R2 = 500\ \Omega\$. The total supply voltage is 10 V. Find \$R_1\$.
4. Explain why adding a 1 kΩ load across \$V_2\$ in Example 1 changes the output voltage. (No calculation required.)

7. Summary Table