Question 1Multiple choice · 1 markA* challenge
A wire of length 5 m has a diameter of 0.5 mm. The resistivity of the metal is 1.7 × 10−8 Ω m.
What is the resistance of the wire?
- A 0.11 Ω
- B 0.22 Ω
- C 0.43 Ω
- D 1.7 Ω
Show answer
Answer: C
Free B.5 Current & circuits questions for IB Physics (SL & HL), part of B The particulate nature of matter. Each question has the answer and the marking points an examiner would look for.
Generate a full B.5 Current & Circuits worksheet →All B The particulate nature of matter questionsA wire of length 5 m has a diameter of 0.5 mm. The resistivity of the metal is 1.7 × 10−8 Ω m.
What is the resistance of the wire?
Answer: C
A cell of emf ε and internal resistance r is connected to a variable resistor. Which statements are correct?
I. The terminal potential difference falls as the current increases.
II. Maximum power is delivered to the variable resistor when its resistance equals r.
III. The power dissipated in the cell is εI.
Answer: A
A battery of emf 6 V and internal resistance 1.5 Ω is connected to a resistor of 2 Ω.
Distinguish between emf and terminal potential difference.
Calculate the current and the terminal potential difference.
Calculate the efficiency of the battery in this circuit.
A second identical resistor is connected in parallel with the first. Determine the new power dissipated in the external circuit.
State the external resistance for maximum power transfer and calculate that power.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | definitions emf is the energy transferred to electrical energy per unit charge by the source ✓ terminal pd is the energy per unit charge transferred in the external circuit (emf minus “lost volts”) ✓ | 2 |
| (b) | 1.71 A; 3.43 V I = 62 + 1.5 = 1.71 A ✓ V = IR = 3.43 V ✓ | 2 |
| (c) | 57.1% efficiency = useful powertotal power = RR + r ✓ 57.1% ✓ | 2 |
| (d) | 5.76 W external resistance 1 Ω ✓ I = 2.40 A ✓ P = I2R = 5.76 W ✓ | 3 |
| (e) | 6.00 W R = r = 1.5 Ω ✓ P = ε24r = 6.00 W ✓ | 2 |
A student measures the terminal potential difference V across a cell for different currents I, by changing a variable resistor in series with the cell and an ammeter.
| I / A ± 0.01 | 0.25 | 0.50 | 0.75 | 1.00 | 1.25 |
|---|---|---|---|---|---|
| V / V ± 0.02 | 4.21 | 3.90 | 3.59 | 3.29 | 2.99 |
Write down the equation relating V, I, the emf ε and the internal resistance r.
Use the graph to determine ε and r.
Calculate the maximum power that the cell can deliver to an external resistor.
The student suggests that a larger range of currents would improve the experiment. Discuss one advantage and one disadvantage.
Explain why the switch should be opened between readings.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | V = ε − Ir V = ε − Ir ✓ | 1 |
| (b) | ε ≈ 4.51 V, r ≈ 1.22 Ω line of best fit drawn ✓ ε = y-intercept = 4.51 V ✓ r = −gradient ✓ r = 1.22 Ω (accept ±10%) ✓ | 4 |
| (c) | 4.16 W maximum when R = r, so P = ε24r ✓ 4.16 W ✓ | 2 |
| (d) | range vs heating a wider range spreads the points and reduces the uncertainty in the gradient ✓ larger currents heat the cell, changing r, or drain it so ε falls ✓ | 2 |
| (e) | avoid heating / discharge so the cell does not heat up or run down, keeping ε and r constant ✓ | 1 |
A student measures the terminal potential difference V of a cell for different currents I by changing a variable resistor. When I = 0.40 A, V = 1.18 V. When I = 1.00 A, V = 0.700 V.
Explain why V changes when the resistance of the variable resistor is changed.
Show that the internal resistance of the cell is about 0.80 Ω.
Determine the emf of the cell.
Calculate the maximum power that can be delivered to the variable resistor.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | pd across internal resistance changes the current changes ✓ so the pd across the internal resistance (Ir) changes ✓ | 2 |
| (b) | 0.800 Ω r = ΔV/ΔI ✓ = 0.480/0.60 = 0.800 Ω ✓ | 2 |
| (c) | 1.50 V ε = V + Ir ✓ = 1.50 V ✓ | 2 |
| (d) | 0.70 W maximum when R = r ✓ P = ε2/4r = 0.70 W ✓ | 2 |
A nichrome wire has length 1.5 m and diameter 0.40 mm. The resistivity of nichrome is 1.1 × 10−6 Ω m.
Calculate the resistance of the wire.
The wire is connected to a cell of emf 6.0 V and internal resistance 1.0 Ω. Calculate the current.
Calculate the terminal potential difference and explain why it is less than the emf.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 13 Ω A = π(0.20 × 10−3)2 = 1.26 × 10−7 m2 ✓ R = ρL/A = 13 Ω ✓ | 2 |
| (b) | 0.43 A I = ε/(R + r) ✓ = 6.0/14.1 = 0.43 A ✓ | 2 |
| (c) | 5.6 V V = 6.0 − 0.43 × 1.0 = 5.6 V ✓ energy is dissipated in the internal resistance ✓ | 2 |
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