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

Objective

Recall and use the equation for two resistors in series acting as a potential divider:

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

Where V₁ is the voltage across resistor R₁ and V₂ across R₂.

Water‑Flow Analogy ⚡️

Think of the circuit as a pipe system. The total voltage (V_in) is like the water pressure at the source. Each resistor is a valve that restricts flow. The voltage drop across a resistor is proportional to how much it restricts the flow.

If you have two valves (R₁ and R₂) in series, the pressure drop across each valve follows the same ratio as their resistances:

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

Just like a higher restriction valve drops more pressure, a higher resistance drops more voltage.

Step‑by‑Step Example 🧪

Suppose we have a 12 V supply and two resistors:

• R₁ = 2 kΩ
• R₂ = 3 kΩ

We want to find the voltage across each resistor.

1. Calculate the ratio: R₁/R₂ = 2 kΩ / 3 kΩ = 2/3.
2. Let V₁ be the voltage across R₁ and V₂ across R₂. We know V₁ + V₂ = 12 V.
3. Using the divider equation: V₁ / V₂ = 2/3.
4. Let V₂ = x. Then V₁ = (2/3)x.
5. Substitute into the total voltage: (2/3)x + x = 12 V → (5/3)x = 12 V → x = 12 V × (3/5) = 7.2 V.
6. Therefore, V₂ = 7.2 V and V₁ = 12 V – 7.2 V = 4.8 V.

Check: 4.8 V / 7.2 V = 2/3, which matches the resistance ratio.

Quick Reference Table 📊

Exam Tips for IGCSE Physics 0625

• Always write the full equation before simplifying: \$\frac{R1}{R2} = \frac{V1}{V2}\$.
• Check units – resistances in Ω, voltages in V.
• When given a total voltage, remember that V₁ + V₂ = V_in.
• Use a quick ratio method: V₁ = V_in × R₁ / (R₁ + R₂) and similarly for V₂.
• Always double‑check that the sum of your calculated voltages equals the supply voltage.
• For multiple choice, look for the answer that satisfies the ratio condition.
• Practice with different resistor values to become comfortable with the ratio concept.