Simple circuits

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Question 2
Medium

A student carries out an experiment to investigate the electrical resistance of an unknown component, Z.

a.

The student needs to measure a potential difference of up to 12 V12\text{ V}12 V across the component.

The student has a choice of two different voltmeters: Voltmeter P and Voltmeter Q.

The diagram below shows the initial readings on these voltmeters before they are connected to any circuit.

Two voltmeters, Voltmeter P and Voltmeter Q, shown before connection. Voltmeter P is an analogue meter with a curved scale from 0 to 5 V. Its needle points to 0.4 V when disconnected. Voltmeter Q is a digital meter with a screen displaying "0.0 V", and a selector dial set to the "0-20 V DC" range.

Which voltmeter is best for the student to use? Give two reasons.

Voltmeter: ‾\underline{\hspace{4cm}}​

Reason 1: ‾\underline{\hspace{12cm}}​

Reason 2: ‾\underline{\hspace{12cm}}​

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b.

The student connects the circuit and takes readings of potential difference and current for component Z. They plot these readings on the graph below.

A graph of Potential Difference in volts (V) on the y-axis versus Current in amperes (A) on the x-axis. The y-axis scale goes from 0 to 14 V with grid lines every 0.2 V. The x-axis scale goes from 0 to 1.0 A with grid lines every 0.02 A. Points are plotted at (0, 0), (0.20, 3.0), (0.40, 6.0), (0.60, 9.0), and (0.80, 12.5). A straight line of best fit passes through the origin and the first four points, but the final point at (0.80, 12.5) lies above this line of best fit.

Explain why the student thinks that component Z is a fixed resistor.

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c.

The student has correctly plotted the point at 0.80 A0.80\text{ A}0.80 A on the graph. Suggest one reason why this point is not on the straight line of best fit.

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d.

Suggest two ways the student can check if their results are reproducible.

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e.

In the linear region of the graph, the resistor has a resistance of 15.0 Ω15.0\ \Omega15.0 Ω.

Calculate the power dissipated by the resistor when the current is 0.4 A0.4\text{ A}0.4 A.

Use the equation: power=(current)2×resistance\text{power} = (\text{current})^2 \times \text{resistance}power=(current)2×resistance

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Simple circuits Questions

  1. GCSE
  2. /Physics
  3. /Simple circuits