AP · AP 2 · free response
AP Physics 2: Algebra-Based · Question 10
AP Physics 2: Algebra-Based · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 12
- Topics
- 1
- Answer
- Complete
Three identical bulbs, each of resistance R, are connected to a battery of e.m.f. ε and negligible internal resistance. Bulb X is connected in series with a parallel combination of bulbs Y and Z. The brightness of a bulb is determined by the power it dissipates.
- (a)
Indicate Indicate which bulb or bulbs are brightest in the original circuit. No justification is required in this part.
1 mark - (b)
Derive Derive expressions for the power dissipated by bulb X and by bulb Y in the original circuit, in terms of ε and R.
4 marks - (c)
Determine Bulb Z now burns out, breaking its branch of the circuit. Determine the new power dissipated by bulb X, and state whether X becomes brighter, dimmer, or stays the same.
3 marks - (d)
Explain Explain, without using equations, why bulb X becomes dimmer while bulb Y becomes brighter when bulb Z burns out.
4 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Indicate Indicate which bulb or bulbs are brightest in the original circuit. No justification is required in this part.
Identify what the question is testing, organise the response into distinct mark-earning points, and make every conclusion traceable to a physical principle or to the evidence supplied.
- 1
1 point: bulb X
(b)
Derive Derive expressions for the power dissipated by bulb X and by bulb Y in the original circuit, in terms of ε and R.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
1 point: parallel pair has resistance R/2, so the total resistance is 3R/2
- 2
1 point: current through X is I = ε/(3R/2) = 2ε/(3R)
- 3
1 point: P_X = I²R = 4ε²/(9R)
- 4
1 point: current through Y is half of I, so P_Y = (ε/(3R))²R = ε²/(9R)
(c)
Determine Bulb Z now burns out, breaking its branch of the circuit. Determine the new power dissipated by bulb X, and state whether X becomes brighter, dimmer, or stays the same.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
1 point: with Z's branch broken, X and Y are in series and the total resistance is 2R
- 2
1 point: I = ε/(2R), so P_X = ε²/(4R)
- 3
1 point: ε²/(4R) is less than 4ε²/(9R), so X becomes dimmer
(d)
Explain Explain, without using equations, why bulb X becomes dimmer while bulb Y becomes brighter when bulb Z burns out.
State the outcome first, then link cause to effect with the relevant physical principle. Each link in the reasoning should be explicit enough to earn its own marking point.
- 1
1 point: removing one of two parallel branches raises the resistance of that section, and so raises the total resistance of the circuit
- 2
1 point: a higher total resistance means a smaller current from the battery, and bulb X carries the whole of that current, so X dissipates less power and dims
- 3
1 point: bulb Y previously carried only half of the current through X, because the current divided between two identical branches
- 4
1 point: now Y carries all of the current, and although that total current is smaller than before, it is larger than the half-share Y used to have — so Y brightens
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