9.3. Resistance and resistivity
A subsection of Physics, 9702, through 9. Electricity
Listing 10 of 234 questions
State Ohm’s law. A battery of electromotive force (e.m.f.) 6.2 V and negligible internal resistance is connected in a circuit to a uniform resistance wire, a voltmeter, a fixed resistor and a switch, as shown in . V 6.2 V wire of resistance 18 Ω S The resistance wire has resistance 18 Ω, length 0.94 m and cross‑sectional area 7.2 × 10–8 m2. The slider S is positioned half‑way along the length of the wire. Calculate the resistivity ρ of the material of the resistance wire. ρ = Ω m The switch is open. State the reading on the voltmeter. voltmeter reading = V The switch is now closed. State whether there is an increase, decrease or no change to: ● the current in the battery ● the voltmeter reading. The switch remains closed. The slider S is moved along the resistance wire so that the voltmeter reading is 3.1 V. On , draw a cross (×) on the resistance wire to show a possible new position of the slider. The circuit in is altered by changing the battery for one of a different e.m.f. The switch is open. A student records the following data for the resistance wire: current in the wire 0.93 A mean drift speed of charge carriers 1.3 × 10–3 m s–1 number density of charge carriers 9.0 × 1028 m–3. Determine the charge q of a charge carrier in the wire suggested by this data. q = C With reference to the value of q, explain why the data recorded by the student cannot be correct.
9702_w23_qp_23
THEORY
2023
Paper 2, Variant 3
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State what is meant by an electric current. A metal wire has length L and cross-sectional area A, as shown in . L A I I is the current in the wire, n is the number of free electrons per unit volume in the wire, v is the average drift speed of a free electron and e is the charge on an electron. State, in terms of A, e, L and n, an expression for the total charge of the free electrons in the wire. Use your answer in to show that the current I is given by the equation I = nAve. A metal wire in a circuit is damaged. The resistivity of the metal is unchanged but the cross- sectional area of the wire is reduced over a length of 3.0 mm, as shown in . d 0.69 d damaged length cross-section Y cross-section X current 0.50 A 3.0 mm The wire has diameter d at cross-section X and diameter 0.69 d at cross-section Y. The current in the wire is 0.50 A. Determine the ratio average drift speed of free electrons at cross-section Y average drift speed of free electrons at cross-section X . ratio = The main part of the wire with cross-section X has a resistance per unit length of 1.7 × 10–2 Ω m–1. For the damaged length of the wire, calculate 1. the resistance per unit length, resistance per unit length = Ω m–1 2. the power dissipated. power = W The diameter of the damaged length of the wire is further decreased. Assume that the current in the wire remains constant. State and explain qualitatively the change, if any, to the power dissipated in the damaged length of the wire.
9702_w17_qp_22
THEORY
2017
Paper 2, Variant 2
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Questions Discovered
234