CIE IGCSE Physics (0625) Action and use of circuit components Study Notes - New Syllabus
CIE IGCSE Physics (0625) Topic 4.3.3 Action and use of circuit components Study Notes
Key Concepts:
Core
- Know that the p.d. across an electrical conductor increases as its resistance increases for a constant current
Supplement
- Describe the action of a variable potential divider
- Recall and use the equation for two resistors used as a potential divider R₁ / R₂ = V₁ / V₂
Potential difference across an electrical conductor
Potential difference across an electrical conductor :
If the current in a conductor is kept constant, then the potential difference (p.d.) across the conductor increases as its resistance increases.
Explanation:
- The relationship between voltage, current, and resistance is given by:
\( V = IR \)
- For a constant current \( I \), if the resistance \( R \) increases, the voltage \( V \) must increase proportionally.
Conclusion:
- At constant current:
↑ Resistance → ↑ Potential Difference (Voltage)
Example:
A constant current of \( 0.5 \, \text{A} \) flows through two different resistors: one of \( 4 \, \Omega \) and another of \( 10 \, \Omega \). Calculate the p.d. across each resistor and compare.
▶️ Answer/Explanation
Using the formula \( V = IR \):
For \( R = 4 \, \Omega \): \( V = 0.5 \times 4 = 2 \, \text{V} \)
For \( R = 10 \, \Omega \): \( V = 0.5 \times 10 = 5 \, \text{V} \)
Conclusion: The p.d. increases with resistance when current is kept constant.
\(\boxed{V \propto R \text{ when } I \text{ is constant}}\)
Example :
An electric heater and a light bulb are connected one at a time to a power supply maintaining a constant current of \( 2 \, \text{A} \). The heater has a resistance of \( 20 \, \Omega \), and the bulb has \( 5 \, \Omega \). Find the voltage across each device.
▶️ Answer/Explanation
Heater: \( V = 2 \times 20 = 40 \, \text{V} \)
Bulb: \( V = 2 \times 5 = 10 \, \text{V} \)
Observation: The heater (higher resistance) has a larger p.d. across it than the bulb.
\(\boxed{V_{\text{heater}} > V_{\text{bulb}}}\)
Action of a Variable Potential Divider
Action of a Variable Potential Divider:
A potential divider is a simple circuit using two or more resistors in series to produce a fraction of the input voltage across one of the resistors.
- A variable potential divider uses a variable resistor (or potentiometer) to continuously adjust the voltage output across one part of the resistor.
- It allows us to “tap off” a desired voltage from a constant input voltage by adjusting the resistance values.
- It is commonly used in volume controls, dimmer switches, and sensor circuits (e.g. LDR or thermistor combinations).
Working Principle:
- The total voltage is divided in the ratio of the resistances.
- If a resistor is changed (e.g., using a slider on a potentiometer), the voltage drop across each section of the resistor changes proportionally.
Formula for Potential Divider:
\(\dfrac{V_1}{V_2} = \dfrac{R_1}{R_2}\)
\(\Rightarrow V_1 = \left( \dfrac{R_1}{R_1 + R_2} \right) V_{\text{total}}\)
This formula applies when two resistors \( R_1 \) and \( R_2 \) are connected in series across a voltage source \( V_{\text{total}} \), and we want to calculate the voltage \( V_1 \) across \( R_1 \).
Example:
A 12 V battery is connected across two resistors in series: \( R_1 = 2 \, \Omega \) and \( R_2 = 4 \, \Omega \). Calculate the voltage across each resistor.
▶️ Answer/Explanation
Total resistance: \( R = R_1 + R_2 = 2 + 4 = 6 \, \Omega \)
Using the formula: \( V_1 = \left( \frac{R_1}{R_1 + R_2} \right) V = \left( \frac{2}{6} \right) \times 12 = 4 \, \text{V} \)
Then \( V_2 = 12 – 4 = 8 \, \text{V} \)
\(\boxed{V_1 = 4 \, \text{V}, \quad V_2 = 8 \, \text{V}}\)