In this experiment, you will investigate how a variable resistor (rheostat) can be used to control the brightness of a lamp. You connects a 6.0 V battery, a...
Question 1 Report
In this experiment, you will investigate how a variable resistor (rheostat) can be used to control the brightness of a lamp. You connects a 6.0 V battery, a rheostat and a lamp in series. A voltmeter is connected across the lamp. You turns the rheostat slider to different positions and records the voltmeter reading Vlamp each time. You also notes the brightness of the lamp. The circuit is shown in Fig. 28.1.
| Rheostat position | Vlamp / V | Brightness |
|---|---|---|
| Maximum resistance | 1.8 | very dim |
| 3/4 resistance | 2.5 | dim |
| 1/2 resistance | 3.4 | medium |
| 1/4 resistance | 4.6 | bright |
| Zero resistance | 5.8 | very bright |
(a) Record the readings in Table 28.1. [1]
(b) Describe the pattern shown by the results. [1]
(c) Explain why the voltage across the lamp changes when the rheostat is adjusted. [2]
(d) State why the voltage across the lamp does not reach 6.0 V even when the rheostat is at zero resistance. [1]
(e) Calculate the voltage across the rheostat when the lamp voltage is 3.4 V. Show your working. [1]
(f) Describe how the voltmeter is connected in the circuit to measure Vlamp. [1]
(g) State one advantage of using a rheostat rather than replacing fixed resistors to vary the brightness. [1]
(h) Plan how you would modify this experiment to investigate how the current through the lamp changes as the rheostat resistance increases. State the extra measurement you would take. [2]
Answer Details
(a) All readings are correctly recorded in Table 28.1. [1]
(b) As the rheostat resistance decreases, the voltage across the lamp increases and the lamp gets brighter. [1]
(c) The rheostat and lamp are connected in series, so they share the total battery voltage of 6.0 V. [1] When the rheostat resistance is large, a larger fraction of the supply voltage is dropped across the rheostat (since \( V = IR \) and the rheostat has a larger R), leaving a smaller voltage across the lamp. When the rheostat resistance is reduced, less voltage is dropped across it and more is available for the lamp, making it brighter. [1]
(d) Even when the rheostat is set to zero resistance, the lamp voltage (5.8 V) does not quite reach 6.0 V because the lamp filament itself has resistance, the connecting wires have a small resistance, and the battery has internal resistance. These all cause a small voltage drop that is not measured by the voltmeter across the lamp. [1]
(e)
\[ V_{\text{rheostat}} = V_{\text{battery}} - V_{\text{lamp}} = 6.0 - 3.4 = 2.6 \; \text{V} \][1]
(f) The voltmeter is connected in parallel with (across) the lamp, so that it measures the potential difference across the lamp only. [1]
(g) A rheostat allows the brightness to be adjusted continuously and smoothly without needing to break the circuit or swap components. Fixed resistors would require disconnecting and reconnecting for each resistance value. [1]
(h) Add an ammeter in series with the circuit to measure the current through the lamp at each rheostat setting. [1] Record the ammeter reading for each rheostat position and plot a graph of current against rheostat resistance to observe how the current changes. The current should decrease as the rheostat resistance increases. [1]
Why this matters: This circuit is a voltage divider. The total voltage is shared between the rheostat and the lamp in proportion to their resistances. Increasing the rheostat resistance increases its share of the voltage and decreases the lamp's share. The brightness of a lamp depends on the electrical power it receives: \( P = \frac{V^2}{R_{\text{lamp}}} \). A higher voltage across the lamp means more power dissipated and therefore more light output. Dimmer switches in homes work on a similar voltage-division principle (though modern ones use electronic switching rather than a simple rheostat).
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