IGCSE Physics: Circuit Components & Potential Dividers

4.3 Electric circuits

4.3.3 Action and use of circuit components

Core

1. Know that the p.d. across an electrical conductor increases as its resistance increases for a constant current

For a conductor with a constant current, the potential difference (\(V\)) across it is given by Ohm's Law:

\[ V = IR \]

If the resistance (\(R\)) of the conductor increases while the current (\(I\)) remains constant, the potential difference (\(V\)) across the conductor will increase.

Supplement (Extended)

2. Describe the action of a variable potential divider

A potential divider is a circuit that provides a variable voltage output based on the values of two resistors in series. The output voltage depends on the ratio of the resistances.

If the value of \( R_1 \) or \( R_2 \) changes, the output voltage will change:

If \( R_1 \) increases with \( R_2 \) unchanged, \( V_1 \) increases and \( V_2 \) decreases.
If \( R_2 \) increases with \( R_1 \) unchanged, \( V_2 \) increases and \( V_1 \) decreases.

The ratio of the voltages across the two resistors is given by:

\[ \frac{R_1}{R_2} = \frac{V_1}{V_2} \]

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

For two resistors \( R_1 \) and \( R_2 \) in series, the potential difference across each resistor can be calculated using the ratio:

\[ \frac{R_1}{R_2} = \frac{V_1}{V_2} \]

For example, if \( R_1 = 20 \, \Omega \) and \( R_2 = 80 \, \Omega \), the ratio is \( 1:4 \). If the supply voltage is \( V \), the potential differences are:

\[ V_1 = 1 \times \frac{V}{5} \quad \text{and} \quad V_2 = 4 \times \frac{V}{5} \]

Light-Dependent Resistors (LDRs) and Thermistors

The resistance of a light-dependent resistor (LDR) decreases with increasing light levels. It is commonly used in circuits that respond to changes in light, such as security lights.

The resistance of a thermistor decreases significantly with increasing temperature. It is often used in temperature-sensitive circuits, such as fire alarms.

Relays

A relay is an electromagnetic switch used to control high-power devices with a low-power signal. For example, a relay can be used to start a washing machine motor when the water reaches the correct temperature.

When a small current flows through the relay's electromagnet, it attracts a switch, closing the circuit and turning on the appliance.

Sample Questions

Question 1: In a potential divider circuit, \( R_1 = 10 \, \Omega \) and \( R_2 = 50 \, \Omega \). The supply voltage is \( 12 \, V \). Calculate the potential difference across \( R_2 \).

Answer:

The ratio of the resistors is:

\[ \frac{R_1}{R_2} = \frac{10}{50} = \frac{1}{5} \]

The total number of shares is \( 1 + 5 = 6 \).

The potential difference across \( R_2 \) is:

\[ V_2 = 5 \times \frac{12}{6} = 10 \, V \]

Fire Alarm Circuit

Figure above shows a circuit that acts as a fire alarm. When the temperature of the thermistor rises, its resistance falls. The thermistor and fixed resistor \( R \) form a potential divider, so the p.d. between S and T rises, and enough current flows into the relay to switch on the bell.

Light-Sensitive Circuit

Figure above shows a circuit that acts as a warning when too much light enters an automated photographic laboratory. A light-emitting diode (LED) on the control panel outside the laboratory lights up to show the warning.

When operating correctly in the dark, the resistance of the LDR is high. The p.d. between V and W is low, so no current flows through the LED.

If light enters the laboratory, the resistance of the LDR falls. The LDR and fixed resistor \( R \) form a potential divider, so the p.d. between V and W rises, and enough current flows through the LED for it to light up and give a warning.

Extended Questions

Question 2: In the circuit shown in Figure below, lamp 1 has a resistance of \( R_1 = 6 \, \Omega \) and lamp 2 has a resistance of \( R_2 = 3 \, \Omega \). The circuit is connected to a 9 V cell. Calculate:

The combined resistance of the circuit.
The current through the circuit.
The voltage across each lamp.

Answer:

1. Combined resistance:

\[ R = R_1 + R_2 = 6 \, \Omega + 3 \, \Omega = 9 \, \Omega \]

2. Current through the circuit:

\[ I = \frac{V}{R} = \frac{9 \, \text{V}}{9 \, \Omega} = 1.0 \, \text{A} \]

3. Voltage across each lamp:

\[ V_1 = I \times R_1 = 1.0 \, \text{A} \times 6 \, \Omega = 6.0 \, \text{V} \] \[ V_2 = I \times R_2 = 1.0 \, \text{A} \times 3 \, \Omega = 3.0 \, \text{V} \]

The voltages across the lamps add up to the e.m.f. of the cell:

\[ V_{\text{cell}} = V_1 + V_2 = 6.0 \, \text{V} + 3.0 \, \text{V} = 9 \, \text{V} \]

Question 3: In the circuit shown in Figure (a), the p.d. across the lamp is equal to the e.m.f. across the cell (that is, 9 V).

In Figure (b), two lamps are connected in series with a 9 V cell. Both lamps have a resistance of 1 Ω. Calculate:

The current through the circuit.
The voltage across each lamp.

Answer:

a) Current through the circuit:

\[ I = \frac{V}{R} = \frac{9 \, \text{V}}{2 \, \Omega} = 4.5 \, \text{A} \]

b) Voltage across each lamp:

\[ V_1 = I \times R_1 = 4.5 \, \text{A} \times 1 \, \Omega = 4.5 \, \text{V} \] \[ V_2 = I \times R_2 = 4.5 \, \text{A} \times 1 \, \Omega = 4.5 \, \text{V} \]

The voltages across the lamps add up to the e.m.f. of the cell:

\[ V_{\text{cell}} = V_1 + V_2 = 4.5 \, \text{V} + 4.5 \, \text{V} = 9 \, \text{V} \]

Question 4: A potential divider circuit is required to produce an output voltage of 8 V across a resistor, \( R_1 \), of 600 Ω. The supply voltage is 12 V. What is the required value of the series resistor, \( R_2 \)?

Answer:

Step 1: Sketch the circuit diagram and label the known values:

\[ R_1 = 600 \, \Omega, \quad V_1 = 8 \, \text{V}, \quad V_{\text{supply}} = 12 \, \text{V} \]

Step 2: Apply the potential divider equation:

\[ \frac{R_1}{R_2} = \frac{V_1}{V_2} \]

Step 3: Substitute the known values:

\[ \frac{600 \, \Omega}{R_2} = \frac{8 \, \text{V}}{4 \, \text{V}} = 2 \]

Step 4: Solve for \( R_2 \):

\[ R_2 = \frac{600 \, \Omega}{2} = 300 \, \Omega \]

Final Answer: The series resistor \( R_2 \) needs a value of 300 Ω.

Question 5: Refer to Figure below.

a) Work out \( V_1 \) and \( V_{\text{in}} \) in Figure (a).

b) Work out \( V_1 \) and \( R_1 \) in Figure (b).

Answer:

a) Figure (a):

The potential divider rule is given by:

\[ V_x = \frac{R_x}{R_1 + R_2} \cdot V_{\text{in}} \]

For \( V_2 = 7.2 \, \text{V} \), let’s first find \( V_{\text{in}} \):

\[ V_{\text{in}} = \frac{V_2 \cdot (R_1 + R_2)}{R_2} \] Substituting the values: \[ V_{\text{in}} = \frac{7.2 \cdot (215 + 860)}{860} = \frac{7.2 \cdot 1075}{860} = 9.0 \, \text{V} \]

Now calculate \( V_1 \):

\[ V_1 = \frac{R_1}{R_1 + R_2} \cdot V_{\text{in}} \] Substituting the values: \[ V_1 = \frac{215}{215 + 860} \cdot 9.0 = \frac{215}{1075} \cdot 9.0 = 1.8 \, \text{V} \]

Thus, the results are:

\[ V_1 = 1.8 \, \text{V}, \quad V_{\text{in}} = 9.0 \, \text{V} \]

b) Figure (b):

We are given \( V_2 = 8.0 \, \text{V}, R_2 = 80 \, \Omega, \) and \( V_{\text{in}} = 36 \, \text{V}. \)

First, find \( R_1 \) using the potential divider rule:

\[ V_2 = \frac{R_2}{R_1 + R_2} \cdot V_{\text{in}} \] Rearranging for \( R_1 \): \[ R_1 = \frac{R_2 \cdot V_{\text{in}}}{V_2} - R_2 \] Substituting the values: \[ R_1 = \frac{80 \cdot 36}{8.0} - 80 = \frac{2880}{8.0} - 80 = 360 - 80 = 280 \, \Omega \]

Next, calculate \( V_1 \):

\[ V_1 = \frac{R_1}{R_1 + R_2} \cdot V_{\text{in}} \] Substituting the values: \[ V_1 = \frac{280}{280 + 80} \cdot 36 = \frac{280}{360} \cdot 36 = 28 \, \text{V} \]

Thus, the results are:

\[ V_1 = 28 \, \text{V}, \quad R_1 = 280 \, \Omega \]

4.3 Electric circuits

Core

Supplement (Extended)

Light-Dependent Resistors (LDRs) and Thermistors

Relays

Sample Questions

Fire Alarm Circuit

Light-Sensitive Circuit

Extended Questions

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