4.3.1 Circuit diagrams and circuit components (3)
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1.
A student designs a circuit to illuminate an LED using a battery and a diode. The circuit diagram is shown below.
(a) Explain the function of the diode in this circuit.
(b) Calculate the power dissipated by the LED. Assume the LED has a forward voltage drop of 2V and a forward current of 20mA.
(c) State one advantage and one disadvantage of using a diode in this circuit compared to directly connecting the LED to the battery.
(a) The diode acts as a one-way valve for electrical current. It allows current to flow easily in one direction (forward bias) but blocks current flow in the opposite direction (reverse bias). In this circuit, the diode ensures that current only flows through the LED when the battery is connected in the correct polarity. This prevents the LED from being damaged by reverse voltage.
(b) Power (P) can be calculated using the formula P = V * I. In this case, V = 2V and I = 0.02A (20mA = 0.02A). Therefore, P = 2V * 0.02A = 0.04W. The power dissipated by the LED is 0.04 Watts.
(c)
- Advantage: The diode protects the LED from damage due to reverse polarity connection. Without the diode, reversing the battery could permanently damage the LED.
- Disadvantage: The diode introduces a voltage drop of approximately 0.7V (for silicon diodes). This reduces the amount of voltage available to the LED, potentially making it dimmer than if it were connected directly to the battery.
2.
Question 1
A circuit diagram is shown below. 
(Replace with an actual circuit diagram showing a cell, two resistors in series, an ammeter and a voltmeter.)
The ammeter reads 0.25 A and the voltmeter reads 1.5 V. Calculate the value of each resistor.
Solution:
- Series Circuit: In a series circuit, the current is the same through all components. Therefore, the 0.25 A current is the same through both resistors.
- Ohm's Law: We can use Ohm's Law (V = IR) to calculate the resistance of each resistor.
- Resistor 1: R1 = V / I = 1.5 V / 0.25 A = 6 Ω
- Resistor 2: R2 = V / I = 1.5 V / 0.25 A = 6 Ω
Answer: Each resistor has a value of 6 Ω.
3.
Question 3
A transformer is used to step down a voltage from the mains supply to a lower voltage for use in a household appliance. The primary coil has 250 turns and the secondary coil has 125 turns. The current in the primary coil is 2 A. Calculate the current in the secondary coil.
Solution:
- Transformer Equation: The relationship between the number of turns in the primary and secondary coils and the currents is given by: Vprimary / Vsecondary = Nprimary / Nsecondary = Isecondary / Iprimary.
- Calculate Voltage Ratio: Vprimary / Vsecondary = 250 / 125 = 2.
- Calculate Secondary Current: Isecondary / Iprimary = 2. Therefore, Isecondary = 2 * Iprimary = 2 * 2 A = 4 A.
Answer: The current in the secondary coil is 4 A.