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The diagram shows a circuit used to investigate the variation of current with potential difference for a filament lamp - Leaving Cert Physics - Question 4 - 2011

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The diagram shows a circuit used to investigate the variation of current with potential difference for a filament lamp. (i) Name the instrument X. What does it meas... show full transcript

Worked Solution & Example Answer:The diagram shows a circuit used to investigate the variation of current with potential difference for a filament lamp - Leaving Cert Physics - Question 4 - 2011

Step 1

Name the instrument X. What does it measure?

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Answer

The instrument X is a voltmeter. It measures the potential difference (voltage) across the filament lamp in volts.

Step 2

Name the component V. What does it do?

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Answer

The component V is a rheostat (or variable resistor). It adjusts the resistance in the circuit, thereby controlling the current flowing through the circuit and the filament lamp.

Step 3

Draw a graph, on graph paper, of the current against the potential difference.

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Answer

To plot the graph:

  1. Label the x-axis as 'Potential Difference (V)' and the y-axis as 'Current (A)'.
  2. Plot the points from the table:
    • (2, 0.9)
    • (4, 1.6)
    • (6, 2.1)
    • (7, 2.5)
    • (8, 2.8)
    • (9, 3.0)
  3. Draw a smooth curve through the points to represent the relationship. Make sure to correctly label the axes and ensure the points are reasonably spaced.

Step 4

What does your graph tell you about the variation of current with potential difference for a filament lamp?

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The graph indicates that the current increases with potential difference, but not in a linear fashion. This suggests that the filament lamp does not follow Ohm's Law at higher voltages, as the resistance increases with temperature. This is characteristic of non-ohmic materials.

Step 5

Using your graph, calculate the resistance of the lamp when the potential difference across the lamp is 5.5 V.

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To calculate the resistance (R) using Ohm's Law:
R=VIR = \frac{V}{I}
From the graph, at V = 5.5 V, estimate the current (I) to be approximately 2.1 A. Then, substituting the values gives: R=5.52.12.62ΩR = \frac{5.5}{2.1} \approx 2.62 \: \Omega
Thus, the resistance of the lamp at 5.5 V is approximately 2.62 ohms.

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