Action and use of circuit components

1. Overview

This topic explores how different electrical components can be used to control the flow of current and the distribution of voltage within a circuit. Understanding these principles is essential for designing practical electronic systems, such as light-sensitive switches, temperature alarms, and volume controls.

Key Definitions

• Potential Difference (p.d.): The energy transferred per unit charge between two points in a circuit, measured in Volts (V).
• Resistance: The opposition to the flow of electrical current, measured in Ohms (Ω).
• Potential Divider: A simple circuit which uses two or more resistors in series to split (divide) the voltage of a supply into smaller parts.
• Potentiometer: A type of variable resistor with three terminals used as a variable potential divider to provide a manually adjustable output voltage.
• LDR (Light Dependent Resistor): A component whose resistance decreases as the intensity of light falling on it increases.
• Thermistor: A component whose resistance decreases as its temperature increases (specifically Negative Temperature Coefficient or NTC thermistors).

Core Content

Relationship between Resistance and Potential Difference In a series circuit, the total voltage from the power supply is shared between the components. The way this voltage is shared depends directly on the resistance of each component.

• Rule: For a constant current flowing through a circuit, the potential difference across a conductor increases as its resistance increases.
• This is derived from Ohm’s Law: $V = I \times R$. If $I$ is constant, then $V$ is directly proportional to $R$ ($V \propto R$).
• If you have two resistors in series, the one with the higher resistance will take a larger "share" of the total voltage.

Worked Example: A 12V battery is connected to a 4Ω resistor and an 8Ω resistor in series.

• Because the 8Ω resistor has twice the resistance of the 4Ω resistor, it will receive twice the voltage.
• The 12V is split: 4V goes across the 4Ω resistor, and 8V goes across the 8Ω resistor.

Extended Content (Extended Curriculum Only)

The Variable Potential Divider A potential divider consists of two resistors in series. By changing the resistance of one of these resistors, we can change the output voltage ($V_{out}$) across it.

• Variable Resistors: Using a potentiometer allows you to manually adjust $V_{out}$ from 0V up to the full supply voltage.
• Sensors: We can replace a fixed resistor with a sensor like an LDR or a Thermistor.
  • Light Sensors: In an LDR circuit, as light intensity increases, LDR resistance decreases. This causes the p.d. across the LDR to decrease, while the p.d. across the fixed resistor increases.
  • Temperature Sensors: In a thermistor circuit, as temperature increases, thermistor resistance decreases, changing the p.d. shared across it.

The Potential Divider Equation To calculate the specific output voltage ($V_{out}$) across one resistor ($R_2$) in a pair, use:

$$V_{out} = \left( \frac{R_2}{R_1 + R_2} \right) \times V_{in}$$

• $V_{in}$ = Supply voltage
• $R_1$ = Resistance of the first component
• $R_2$ = Resistance of the component you are measuring the voltage across

Worked Example: A potential divider circuit has a $V_{in}$ of 10V. $R_1$ is a fixed 200Ω resistor and $R_2$ is an LDR. In bright light, the LDR resistance is 50Ω. Calculate $V_{out}$ across the LDR.

• $V_{out} = (50 / (200 + 50)) \times 10$
• $V_{out} = (50 / 250) \times 10 = 0.2 \times 10 = 2\text{V}$

Key Equations

• Ohm’s Law: $V = I \times R$
  • $V$: Potential Difference (V)
  • $I$: Current (A)
  • $R$: Resistance (Ω)
• Potential Divider Formula: $V_{out} = \frac{R_2}{R_{total}} \times V_{in}$
  • $R_{total} = R_1 + R_2$

Common Mistakes to Avoid

• ❌ Wrong: Decreasing a component's resistance decreases the current in a series circuit.
• ✓ Right: Decreasing resistance decreases the total resistance, which increases the current ($I = V/R$).
• ❌ Wrong: LDRs have low resistance when it is dark.
• ✓ Right: LDR resistance is high when dark and low when light (Remember: Light Up, Resistance Down).
• ❌ Wrong: In a potential divider, the larger resistance gets a smaller share of the voltage.
• ✓ Right: The larger the resistance, the larger the share of the total voltage it receives.
• ❌ Wrong: Assuming a change in one branch of a parallel circuit automatically changes the voltage in other branches.
• ✓ Right: In a standard parallel circuit with a constant voltage source, the p.d. across each branch remains equal to the source voltage regardless of changes in other branches.

Exam Tips

1. Identify the "Fixed" vs "Variable": When looking at a potential divider, identify which resistor is fixed and which is the sensor (LDR/Thermistor). If the sensor resistance goes down, its share of the voltage goes down.
2. Check the Output: Always check which component the $V_{out}$ (or the voltmeter) is connected across. If the voltmeter is across the fixed resistor, it will show a higher voltage when the sensor's resistance decreases.
3. Diode Orientation: If a diode is in the circuit, check the arrow direction. Current only flows in the direction of the arrow. If the battery is trying to push current "against" the arrow, the resistance is effectively infinite and no current flows.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A student builds a simple circuit to investigate the resistance of a thermistor. The circuit consists of a 6.0 V battery, an ammeter, the thermistor, and a fixed resistor all connected in series.

(a) State the purpose of the fixed resistor in this circuit. [1]

(b) As the temperature of the thermistor increases, describe how its resistance changes. [2]

(c) Explain how the reading on the ammeter will change as the temperature of the thermistor increases. [2]

Worked Solution:

(a)

1. The fixed resistor limits the current in the circuit. Prevents excessive current that could damage the thermistor or other components.

How to earn full marks:

• State that the resistor limits current to protect components.

(b)

1. As temperature increases, the resistance of the thermistor decreases. This is the defining property of a thermistor.
2. The decrease is non-linear. While not strictly required, mentioning the non-linear relationship shows deeper understanding.

How to earn full marks:

• State the resistance decreases.
• State that the decrease is non-linear.

(c)

1. As the thermistor's resistance decreases, the total resistance of the circuit decreases. Because the thermistor is in series with a fixed resistor.
2. Since the battery voltage is constant, and resistance decreases, the current increases according to Ohm's Law. $V = IR \implies I = V/R$. If $R$ decreases and $V$ is constant, $I$ must increase.

How to earn full marks:

• State that the total resistance decreases.
• State that the ammeter reading (current) increases.

Common Pitfall: Many students forget that the ammeter reading depends on the total resistance in the series circuit, not just the thermistor's resistance. They might only focus on the thermistor and not consider how the fixed resistor affects the overall current.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

A light-dependent resistor (LDR) is used in a circuit to automatically switch on a garden light when it gets dark. The circuit includes a 9.0V supply, the LDR, a fixed resistor, and a relay. The relay switches on the garden light when the voltage across it reaches a certain threshold.

(a) Define resistance. [1]

(b) Describe how the resistance of the LDR changes as the intensity of light falling on it increases. [2]

(c) Explain how the potential difference across the fixed resistor changes as the intensity of light falling on the LDR decreases. [3]

Worked Solution:

(a)

1. Resistance is the opposition to the flow of electric current. Or: The ratio of potential difference to current.

How to earn full marks:

• Correct definition of resistance.

(b)

1. As light intensity increases, the resistance of the LDR decreases. This is the key property of an LDR.
2. The decrease is non-linear. While not required, mentioning the non-linear relationship shows deeper understanding.

How to earn full marks:

• State the resistance decreases.
• State the decrease is non-linear.

(c)

1. As light intensity decreases, the resistance of the LDR increases. This is the inverse relationship between light and LDR resistance.
2. Since the LDR and the fixed resistor are in series, they form a potential divider. Recognising the potential divider arrangement is key.
3. As the LDR's resistance increases, it takes a larger share of the potential difference, so the potential difference across the fixed resistor decreases. Higher resistance component gets higher voltage in a series circuit.

How to earn full marks:

• State that the resistance of the LDR increases.
• State that the LDR and fixed resistor form a potential divider.
• State that the potential difference across the fixed resistor decreases.

Common Pitfall: Students often struggle to connect the change in LDR resistance to the voltage across the fixed resistor. Remember that in a series circuit, the component with the higher resistance will have a larger share of the total voltage.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A potential divider circuit is constructed using a 1500 $\Omega$ resistor and a variable resistor (potentiometer) connected in series with a 6.0 V battery. The potentiometer has a maximum resistance of 3000 $\Omega$. A voltmeter is connected across the 1500 $\Omega$ resistor.

(a) Draw a circuit diagram of the potential divider, including the battery, the 1500 $\Omega$ resistor, the potentiometer (with its symbol), and the voltmeter. [3]

(b) Calculate the minimum potential difference that can be measured by the voltmeter. [2]

(c) Calculate the maximum potential difference that can be measured by the voltmeter. [3]

Worked Solution:

(a)

1. Circuit diagram includes a battery.
   📊A circuit diagram showing a 6.0V battery connected in series with a 1500 Ohm resistor and a potentiometer (variable resistor) with a total resistance of 3000 Ohms. A voltmeter is connected in parallel across the 1500 Ohm resistor.

How to earn full marks:

• Correct battery symbol.
• Correct fixed resistor symbol.
• Correct potentiometer symbol.
• Voltmeter in parallel with 1500 $\Omega$ resistor.
• All components in series.

(b)

1. Minimum voltage across the 1500 $\Omega$ resistor occurs when the potentiometer's resistance is at its maximum (3000 $\Omega$). The voltage is divided proportionally to the resistance.
2. Use the potential divider formula: $V_1 = V_{total} \times \frac{R_1}{R_1 + R_2}$ Where $V_1$ is the voltage across the 1500 $\Omega$ resistor, $V_{total}$ is 6.0 V, $R_1$ is 1500 $\Omega$, and $R_2$ is 3000 $\Omega$.
3. Calculate: $V_1 = 6.0 \times \frac{1500}{1500 + 3000} = 6.0 \times \frac{1500}{4500} = 6.0 \times \frac{1}{3} = 2.0$ V This is the minimum voltage.

How to earn full marks:

• Recognize that maximum potentiometer resistance gives minimum voltmeter reading.
• Correct use of potential divider formula.
• Correct calculation, with answer: $\boxed{2.0 \text{ V}}$

(c)

1. Maximum voltage across the 1500 $\Omega$ resistor occurs when the potentiometer's resistance is at its minimum (0 $\Omega$). When the potentiometer is at 0 $\Omega$, it's effectively a short circuit.
2. Therefore, all of the voltage from the battery will be across the 1500 $\Omega$ resistor. Since there is no other resistance in the circuit.
3. The maximum potential difference is 6.0 V.

How to earn full marks:

• Recognize that minimum potentiometer resistance gives maximum voltmeter reading.
• State that all voltage is across the 1500 $\Omega$ resistor when the potentiometer is at 0 $\Omega$.
• Correct answer: $\boxed{6.0 \text{ V}}$

Common Pitfall: A common mistake is forgetting to use the potential divider formula correctly, especially when finding the minimum voltage. Students might also incorrectly assume that the maximum voltage across the fixed resistor is always less than the battery voltage, even when the potentiometer is at zero resistance.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A student is investigating the behavior of a negative temperature coefficient (NTC) thermistor in a circuit. The student connects the thermistor in series with a 560 $\Omega$ resistor and a 4.5 V battery. A voltmeter is connected across the thermistor. The student heats the thermistor in a water bath and measures the potential difference across it at different temperatures.

(a) Explain why the potential difference across the thermistor changes as its temperature changes. [3]

(b) At a certain temperature, the potential difference across the thermistor is measured to be 1.5 V. Calculate the current flowing through the circuit at this temperature. [3]

(c) Using your answer from (b), calculate the resistance of the thermistor at this temperature. [3]

Worked Solution:

(a)

1. The thermistor's resistance changes with temperature. This is the defining characteristic of a thermistor.
2. As temperature increases, the resistance of the NTC thermistor decreases. Specifically, an NTC thermistor's resistance decreases with increasing temperature.
3. Since the thermistor and the 560 $\Omega$ resistor are in series, they form a potential divider. The potential difference across the thermistor depends on its resistance relative to the 560 $\Omega$ resistor. The potential divider principle is the key to understanding this circuit.

How to earn full marks:

• State that the thermistor's resistance changes with temperature.
• State that the resistance of the thermistor decreases as temperature increases.
• State that the thermistor and resistor form a potential divider and the voltage depends on the ratio of their resistances.

(b)

1. The total potential difference across the circuit is 4.5 V, and the potential difference across the thermistor is 1.5 V. Given in the question.
2. Therefore, the potential difference across the 560 $\Omega$ resistor is $4.5 \text{ V} - 1.5 \text{ V} = 3.0 \text{ V}$ Kirchhoff's Voltage Law.
3. Using Ohm's Law, $V = IR$, the current flowing through the 560 $\Omega$ resistor is $I = \frac{V}{R} = \frac{3.0}{560} = 0.00536$ A. Since the components are in series, the current is the same throughout the circuit. $\boxed{0.0054 \text{ A}}$ (2 s.f.)

How to earn full marks:

• Correctly calculate the potential difference across the 560 $\Omega$ resistor.
• Correct use of Ohm's Law.
• Correct answer with unit.

(c)

1. Using Ohm's Law again, $V = IR$, the resistance of the thermistor is $R = \frac{V}{I}$ Applying Ohm's Law to the thermistor.
2. The potential difference across the thermistor is 1.5 V, and the current is 0.00536 A (from part (b)). Using the value calculated in the previous part.
3. Therefore, $R = \frac{1.5}{0.00536} = 280 \Omega$ The resistance of the thermistor. $\boxed{280 \text{ } \Omega}$ (2 s.f.)

How to earn full marks:

• Correct use of Ohm's Law.
• Correct substitution of values, using the current from part (b) (ECF allowed if part (b) was incorrect).
• Correct answer with unit.

Common Pitfall: Students often forget to subtract the thermistor's voltage from the battery voltage to find the voltage across the fixed resistor. Also, be careful to use the current value calculated in part (b) to find the thermistor's resistance in part (c); errors in (b) will carry over if you don't use your calculated value.