Electric Power AP Physics 2 FRQ – Exam Style Questions etc.
Electric Power AP Physics 2 FRQ
Unit 11: Electric Circuits
Weightage : 15–18%
Exam Style Practice Questions ,Electric Power AP Physics 2 FRQ
Question
A certain light bulb is designed to dissipate 6 watts when it is connected to a 12-volt source.
a. Calculate the resistance of the light bulb.
b. If the light bulb functions as designed and is lit continuously for 30 days, how much energy is used?
Be sure to indicate the units in your answer.
The 6-watt, 12-volt bulb is connected in a circuit with a 1,500-watt, 120-volt toaster; an adjustable resistor; and a 120-volt power supply. The circuit is designed such that the bulb and the toaster operate at the given values and, if the light bulb fails, the toaster will still function at these values.
c. On the diagram below, draw in wires connecting the components shown to make a complete circuit that will function as described above.
d. Determine the value of the adjustable resistor that must be used in order for the circuit to work as designed.
e. If the resistance of the adjustable resistor is increased, what will happen to the following?
i. The brightness of the bulb. Briefly explain your reasoning.
ii. The power dissipated by the toaster. Briefly explain your reasoning.
Answer/Explanation
Ans:
a. P = V2/R gives R = 24 Ω
b. E = Pt where t = (30 days)(24 h/day)(3600 sec/h) gives E = 1.6 × 107 J
c. 
The bulb, needing only 12 V must have a resistor in series with it and the toaster, requiring 120 V must be connected directly to the power supply.
d. The current through the bulb is I = P/V = 0.5 A, which is also the current in the resistor, which must have 108 V across it to provide the light bulb only 12 V. R = V/I = (108 V)/(0.5 A) = 216 Ω
e. i. If the resistance of the resistor is increased, the current through the branch will decrease, decreasing the brightness of the bulb.
ii. Since the toaster operates in its own parallel branch, nothing will change for the toaster.
Question
Two lightbulbs, one rated 30 W at 120 V and another rated 40 W at 120 V, are arranged in two different circuits.
a. The two bulbs are first connected in parallel to a 120 V source.
i. Determine the resistance of the bulb rated 30 W and the current in it when it is connected in this circuit.
ii. Determine the resistance of the bulb rated 40 W and the current in it when it is connected in this circuit.
b. The bulbs are now connected in series with each other and a 120 V source.
i. Determine the resistance of the bulb rated 30 W and the current in it when it is connected in this circuit.
ii. Determine the resistance of the bulb rated 40 W and the current in it when it is connected in this circuit.
c. In the spaces below, number the bulbs in each situation described, in order of their brightness.
(1= brightest, 4 = dimmest)
____30 W bulb in the parallel circuit
____40 W bulb in the parallel circuit
____30 W bulb in the series circuit
____40 W bulb in the series circuit
d. Calculate the total power dissipated by the two bulbs in each of the following cases.
i. The parallel circuit
ii. The series circuit
Answer/Explanation
Ans:
a.(i) P = V2/R gives R = 480 Ω and V = IR gives I = 0.25 A
ii. P = V2/R gives R = 360 Ω and V = IR gives I = 0.33 A
b. i./ii. The resistances are unchanged = 480 Ω and 360 Ω. The total resistance in series is 480 Ω + 360 Ω = 840 Ω making the total current I = V/R = 0.14 A which is the same value for both resistors in series
c. The bulbs are brightest in parallel, where they provide their labeled values of 40 W and 30 W. In series, it is the larger resistor (the 30 W bulb) that glows brighter with a larger potential difference across it in series.
This gives the order from top to bottom as 2 1 3 4
d. i. In parallel, they each operate at their rated voltage so they each provide their rated power and PT = 30 W + 40 W = 70 W
ii. In series PT = VT2/RT = 17 W