A radio club builds a simple receiver for a school event. Fig. 1 shows a battery, switch, resistor and buzzer in series. When the switch is closed, charge f...
Question 1 Report
A radio club builds a simple receiver for a school event. Fig. 1 shows a battery, switch, resistor and buzzer in series. When the switch is closed, charge flows through the buzzer and it produces sound waves in the air. The resistor is included to limit the current so the buzzer is not damaged. The club uses a 9.0 V battery and measures a current of 0.15 A.
(a) What form of energy is transferred by the buzzer to the surroundings? [1]
(b) Calculate the total resistance of the circuit. [2]
(c) Which action stops the current without removing the battery? [1]
(d) Give one reason for using a resistor in series with the buzzer. [1]
The diagram shows a student testing a filament lamp. An ammeter is placed in series and a voltmeter is connected across the lamp. The student slowly increases the potential difference, allowing the filament temperature to rise before recording each reading. Unlike a fixed resistor, a filament lamp does not have a constant resistance because its metal filament gets hotter as current increases.
(a) What happens to the filament temperature as current increases? [1]
(b) Describe the change in resistance of the filament as it gets hotter. [1]
(c) Calculate the resistance when the voltmeter reads 6.0 V and the ammeter reads 0.40 A. [2]
(d) Which meter must be connected in parallel? [1]
Answer Details
Receiver circuit
(a) The buzzer transfers energy to the surroundings as sound energy. [1]
(b) \[R=\frac{V}{I}=\frac{9.0\ \mathrm{V}}{0.15\ \mathrm{A}}=60\ \Omega\]
The total resistance is 60 \(\Omega\). [2]
(c) Open the switch. This breaks the conducting path, stopping the current without removing the battery. [1]
(d) A resistor limits current and prevents the buzzer being damaged. [1]
Filament lamp
(a) As current increases, the filament temperature increases. [1]
(b) The resistance increases as the metal filament gets hotter. [1]
(c) \[R=\frac{V}{I}=\frac{6.0\ \mathrm{V}}{0.40\ \mathrm{A}}=15\ \Omega\]
The resistance is 15 \(\Omega\). [2]
(d) The voltmeter must be connected in parallel, because it measures potential difference across the lamp. [1]
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