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
(12 points, suggested time 25 minutes)
A circuit consists of an ideal battery of emf Ɛ and four identical resistors \(R_{1}\) , \(R_{2}\), \(R_{3}\), and \(R_{4}\), each of resistance R, as shown in Figure 1.
(a) For parts (a)(i) and (a)(ii), express your answers in terms of numerical values, Ɛ, and R only.
i. Derive an expression for the current \(I_{1}\) in Resistor \(R_{1}\).
ii. Derive an expression for the current \(I_{3}\) in Resistor \(R_{3}\).
(b) The partially completed bar chart in Figure 2 shows a bar that represents the absolute value \(\left| \Delta V\right|\) of the potential difference across the ideal battery.
• In Figure 2, draw a bar to represent \(\left| \Delta V\right|\) across each resistor, relative to the emf Ɛ of the ideal battery.
• The height of each bar should be proportional to the value of \(\left| \Delta V\right|\) represented by that bar. If \(\left| \Delta V\right|\) is zero, write a “0” in that column
A student claims that the rate at which energy is dissipated (power) by the circuit can be expressed as \(p = \frac{3\varepsilon ^{2}}{5R}\)
(c) State whether the expression for P is correct or incorrect. Justify your answer by referring to the derivations from part (a) or the bar chart from part (b).
When the ideal battery is connected in the original circuit, the rate at which energy is dissipated by Resistor \(R_{1}\) is \(P_{original}\). The ideal battery is now replaced with a nonideal battery of emf Ɛ and internal resistance r to form the new circuit shown in Figure 3. The rate at which energy is dissipated by Resistor \(R_{1}\) in the new circuit is \(P_{new}\).
(d) Indicate whether \(P_{new}\) is greater than, less than, or equal to \(P_{original}\).
_____ \(P_{new}\)> \(P_{original}\)_____ \(P_{new}\) < \(P_{original}\) _____ \(P_{new}\) = \(P_{original}\)
Briefly justify your answer
▶️Answer/Explanation
1(a(i))Example Response
Determine the total resistance of the circuit.
The resistance of the right-most branch containing resistors connected in series:
\(R_{s} =\sum_{i}^{}R_{i}\)
\(R_{s} =R+R\)
\(R_{s} =2R\)
The resistance of parallel branches that contain resistors:
\(\frac{1}{R_{p}} = \sum_{i}^{}\frac{1}{R_{i}}\)
\(\frac{1}{R_{p}} = \frac{1}{R} + \frac{1}{2R} = \frac{3}{2R}\)
\(R_{p} =\frac{2R}{3}\)
The total resistance of the circuit:
\(R_{s} =\sum_{i}R_{i}\)
\(R_{total} =R+\frac{2R}{3} =\frac{5R}{3}\)
The total current in the circuit:
\(I =\frac{\Delta V}{R} = \frac{3\varepsilon }{5R}\)
1(a(ii))Example Response
1(b)Example Response
1(c)Example Solution
The equation is correct. The equation for P , which is power, can be written as (\(p= \frac{\left ( \Delta v \right )^{2}}{R}\). According to the bar chart in part (b), the potential difference across the battery is \(\varepsilon \). The total resistance of the circuit is \(\frac{5R}{3}\) , according to the derivation from part (a)(i). Therefore, (\(p= \frac{\left ( \Delta v \right )^{2}}{R}\) = \(\frac{\left ( \varepsilon \right )^{2}}{\left ( \frac{5R}{3} \right )}\) = \(\frac{3\varepsilon ^{2}}{5R}\)
1(d)Example Response
\(P_{new}\) < \(P_{original}\). Since the emf of the battery is the same in the new circuit and the total resistance of the new circuit is greater, the current in \(R_{1}\) is less in the new circuit. Therefore, \(P_{new}\) is less than \(P_{original}\)
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