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The voltage (potential difference) across a length of wire is 1.5 V - Edexcel - GCSE Physics - Question 8 - 2023 - Paper 2

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The voltage (potential difference) across a length of wire is 1.5 V. A charge of 0.042 C flows through the wire. Calculate the energy transferred. Use the equatio... show full transcript

Worked Solution & Example Answer:The voltage (potential difference) across a length of wire is 1.5 V - Edexcel - GCSE Physics - Question 8 - 2023 - Paper 2

Step 1

Calculate the energy transferred.

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Answer

To calculate the energy transferred, we can use the formula:

E=Q×VE = Q × V

Substituting the values:

E=0.042imes1.5E = 0.042 imes 1.5

Calculating gives:

E=0.063extJE = 0.063 ext{ J}

Thus, the energy transferred is 0.063 J.

Step 2

Add connecting wires, a voltmeter and an ammeter.

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Answer

The voltmeter should be connected in parallel with the iron wire to measure the voltage across it, while the ammeter should be connected in series with the iron wire to measure the current. The diagram should show the connections appropriately, ensuring that the symbols for both devices are included.

Step 3

Draw a straight line of best fit.

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Answer

A straight line of best fit should be drawn on Figure 21, ideally passing through the center of the data points to represent the relationship observed in the experiment.

Step 4

Use Figure 21 to estimate the resistance of a 100 cm length of the iron wire.

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Answer

By referencing Figure 21, find the corresponding resistance value for a 100 cm length of iron wire. Read the value at the 100 cm mark and note that it may be approximately in the range of the measured values represented in the graph.

Step 5

Calculate the resistance of this piece of wire.

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Answer

Using the equation:

V=I×RV = I × R

We can rearrange this to find resistance:

R=VIR = \frac{V}{I}

Substituting in the known values:

  • Voltage, V = 1.56 V
  • Current, I = 0.45 A

Calculating gives:

R=1.560.45=3.47extΩR = \frac{1.56}{0.45} = 3.47 ext{ Ω}

Therefore, the resistance of the piece of wire is approximately 3.47 Ω.

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