Chapter 25 — Current, Resistance & Electromotive Force

15 points · 3 extra credit

Chapter 25 — Current, Resistance & Electromotive Force

Core: 15 points · Extra credit: 3 points. Show your reasoning; partial credit is given.

Worked solutions are available for every problem.
  1. 1. (2 pts ••) A copper wire of diameter 2.05 mm2.05\,\text{mm} carries a steady current of 8.0 A8.0\,\text{A}. Copper has n=8.5×1028n=8.5\times10^{28} free electrons per cubic metre.

    (a) Find the current density JJ in the wire.

    (b) Find the drift speed of the conduction electrons.

    (c) How many electrons pass through a cross-section of the wire in one minute?

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  2. 2. (2 pts ••) A heating element is made from nichrome wire (ρ=1.10×10−6 Ω ⁣⋅ ⁣m\rho=1.10\times10^{-6}\,\Omega\!\cdot\!\text{m}) of length 3.5 m3.5\,\text{m} and diameter 0.50 mm0.50\,\text{mm}.

    (a) Find the resistance of the element at room temperature.

    (b) A second element uses the same material but with twice the length and half the diameter. By what factor does its resistance change relative to part (a)? Give the new resistance.

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  3. 3. (2 pts ••) A copper coil has a resistance of 25.0 Ω25.0\,\Omega at 20 ∘C20\,{}^\circ\text{C}. The temperature coefficient of resistivity for copper is α=3.9×10−3 K−1\alpha=3.9\times10^{-3}\,\text{K}^{-1}, and over this range you may treat resistance as varying linearly with temperature.

    (a) Find the resistance of the coil when it heats up to 85 ∘C85\,{}^\circ\text{C}.

    (b) At what temperature would the coil's resistance read 30.0 Ω30.0\,\Omega?

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  4. 4. (1 pt •) A resistor is connected across an ideal 12 V12\,\text{V} source and obeys Ohm's law with R=470 ΩR=470\,\Omega.

    (a) Find the current through the resistor.

    (b) Find the power it dissipates.

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  5. 5. (2 pts ••) A single AA cell has an emf of 1.50 V1.50\,\text{V} and internal resistance r=0.35 Ωr=0.35\,\Omega. It is connected to an external resistor R=4.5 ΩR=4.5\,\Omega.

    (a) Find the current in the circuit.

    (b) Find the terminal voltage of the cell.

    (c) Find the power dissipated in the internal resistance and the power delivered to RR, and confirm they sum to the total power the emf supplies.

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  6. 6. (2 pts ••) A household space heater is rated 1500 W1500\,\text{W} when operated at 120 V120\,\text{V}.

    (a) Find the operating resistance of the heater and the current it draws.

    (b) If the heater runs 2.52.5 hours per day for 3030 days and electricity costs \0.14$ per kilowatt-hour, find the total energy used (in kWh) and the cost.

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  7. 7. (2 pts ••) In a cylindrical conductor of radius R=1.2 mmR=1.2\,\text{mm}, the current is not uniform: the current density points along the axis and has magnitude J(r)=J0 (r/R)J(r)=J_0\,(r/R), where rr is the distance from the axis and J0=1.5×106 A/m2J_0=1.5\times10^{6}\,\text{A/m}^2.

    (a) By integrating over the cross-section, find the total current carried by the conductor. (Hint: use a thin annular ring of radius rr and thickness drdr.)

    (b) Find the average current density I/(πR2)I/(\pi R^2) and compare it to J0J_0.

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  8. 8. (2 pts ••) A battery of emf E\mathcal{E} and internal resistance r=0.50 Ωr=0.50\,\Omega drives an external resistor RR. When R=3.0 ΩR=3.0\,\Omega the current is 2.4 A2.4\,\text{A}.

    (a) Find the emf E\mathcal{E}.

    (b) The external resistor is then removed and the battery's terminals are short-circuited (external resistance →0\to 0). Find the short-circuit current.

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  9. Extra Credit (3 pts •••)

    A solid conductor is shaped as a truncated cone (frustum) of length LL. Its circular cross-section has radius aa at the left face and grows linearly to radius b>ab>a at the right face. The material has uniform resistivity ρ\rho, and current flows along the axis. Assume the current density is uniform across each cross-section (a good approximation when the taper is gentle), so each thin disc acts as a resistor in series.

    (a) Let xx measure distance from the left face. Write the radius ϱ(x)\varrho(x) of the cross-section as a function of xx, and hence the resistance dRdR of a thin disc of thickness dxdx.

    (b) Integrate to show that the total resistance is

    R=ρLπab.R=\frac{\rho L}{\pi a b}.

    (c) Evaluate RR for a copper frustum (ρ=1.7×10−8 Ω ⁣⋅ ⁣m\rho=1.7\times10^{-8}\,\Omega\!\cdot\!\text{m}) with L=0.15 mL=0.15\,\text{m}, a=1.0 mma=1.0\,\text{mm}, and b=3.0 mmb=3.0\,\text{mm}. Check that your formula reduces to ρL/(πa2)\rho L/(\pi a^2) when b→ab\to a (a uniform cylinder).

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