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A student connects the circuit shown - Edexcel - A-Level Physics - Question 9 - 2023 - Paper 1

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A student connects the circuit shown. The battery has negligible internal resistance. The student increases the resistance of the variable resistor from 0Ω to 40Ω. ... show full transcript

Worked Solution & Example Answer:A student connects the circuit shown - Edexcel - A-Level Physics - Question 9 - 2023 - Paper 1

Step 1

When variable resistor is 0Ω voltmeter reading –

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Answer

When the variable resistor is set to 0Ω, the entire voltage of the battery will be across the 10Ω resistor. Therefore, the voltmeter will read:

V=6.0VV = 6.0V

So the maximum reading on the voltmeter is 6.0 V.

Step 2

Use of principle of potential divider

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Answer

To find the voltmeter reading when the variable resistor is at a different value, we use the potential divider formula:

V=R2(VsourceR1+R2)V = R_2 \left( \frac{V_{source}}{R_1 + R_2} \right) where:

  • Vsource=6.0VV_{source} = 6.0V (the battery voltage)
  • R1=10ΩR_1 = 10Ω (fixed resistor)
  • R2R_2 is the resistance of the variable resistor.

As R2R_2 varies from 0Ω to 40Ω, we can calculate the voltmeter reading at R2=40ΩR_2 = 40Ω:

V=10Ω(6.0V10Ω+40Ω)=10Ω(6.0V50Ω)=1.2VV = 10Ω \left( \frac{6.0V}{10Ω + 40Ω} \right) = 10Ω \left( \frac{6.0V}{50Ω} \right) = 1.2V

Thus, the minimum reading on the voltmeter is 1.2 V.

Step 3

When variable resistor is 40Ω voltmeter reading –

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Answer

When the variable resistor is set to 40Ω, the voltmeter will read:

V=1.2VV = 1.2V

So the minimum reading on the voltmeter is 1.2 V.

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