Electric Current AP Physics 2 FRQ – Exam Style Questions etc.
Electric Current AP Physics 2 FRQ
Unit 11: Electric Circuits
Weightage : 15–18%
Exam Style Practice Questions , Electric Current AP Physics 2 FRQ
Question
You are given three 2-ohm resistors, some wire, a variable DC voltage supply, a voltmeter, and an ammeter.
(a) Draw a schematic diagram of a circuit that will produce an equivalent resistance of 3 ohms as well as measure the circuit current and source voltage.
(b) Which, if any, of these components are assumed to have zero internal resistance? Which, if any, of these components are assumed to have infinite resistance? Justif your choices.
(c) Trace out one complete loop in your circuit, and prove Kirchoff’s loop rule. Once you set up your rule, assume a setting of 6 volts on the power supply. Verify your rule numerically.
(d) A student wishes to measure the voltage and current in a simple circuit using small lightbulbs instead of commercially manufactured resistors.She finds that after a short while, the current in the ammeter is decreasing. How might she account for this?
Answer/Explanation
(a)
(b) Ammeters, wires, and power supplies are assumed to have zero internal resistance as they have no resistance indicated in the circuit diagram. If there were any resistance, unaccounted losses of voltage would occur in our circuit. When modeling actual resistance inside of components, an “internal” resistor is added to the circuit diagram. Voltmeters would create a parallel path if they did not have infinite resistance. To the extent they have less than infinite resistance, the addition of voltmeters to the circuit would effectively lower the resistance, thereby lowering the voltage reading.
(c) Kirchhoff’s loop rule states that the sum of voltages changing around the loop must be zero. As an example of this, start to the right of the power supply and trace a loop clockwise through the top parallel resistor and then the resistor to the right and back to our starting point. Tracing our changes in voltage:
+ V − ( I /2) R − IR = 0
For the entire circuit:
V = IR
6 = I (3)
I = 2 amps
Loop rule:
+6 − (1)(2) − (2)(2) = 6 − 2 − 4 = 0
(d) The lightbulbs get hot very quickly, which increases their electrical resistance. This, in turn, reduces the current measured by the ammeter.
Question
The circuit shown is built and the voltage source supplies voltage for 15 minutes before the battery is completely drained. Assume the voltage supplied by the battery is constant at 12 V until the battery is drained, after which the battery supplies o V.
(a) What is the equivalent resistance of the circuit?
(b) Two students are discussing the apparatus. Student 1 says, “If the 20 n resistor were not present, the overall resistance of the circuit would have been lower and the battery would have lasted longer.” Student 2 says, “If the 20 n resistor were not present, I think the power output would have been higher and the battery would have drained faster.”
(i) Use equations to show whether the overall resistance would have been lower without the 20Ω resistor present.
(ii) Use equations to determine whether the overall power output would have been higher.
(iii) Which student is correct about the battery life?
(c) The 20 Ω resistor is replaced with a capacitor.
(i) As soon as the circuit is connected, explain without using equations how the current drawn out of the battery compares between the original circuit and the circuit with
the capacitor.
(ii) After the capacitor has been connected for a long time, but before the battery is completely drained, how does the current drawn out of the battery compare between the
original circuit and the circuit with the capacitor?
Answer/Explanation
Ans:
(a) The resistors in parallel are combined first.
Req-1 = 40-1 + 60 -1
Req = 40
The total equivalent resistance is then Rtotal = 24 + 20 = 44 Ω.
(b) (i) Without the resistor, there would only be the parallel resistor combination.
R-1 total = 40-1 + 60 -1
R total = 40 Ω.
(ii) P =IV= \(\left ( \frac{V}{R} \right )V,\) so a higher resistance results in a lower power output for the same battecy.
(iii) A lower power output means that it will take more time for the energy to be used up, so student 1 is correct.
(c) (i) As soon as the circuit is connected with the capacitor, the capacitor acts as if it had no resistance. This is exactly like
(b, i). The lower resistance would cause a greater current to be drawn from the battecy.
(ii) After the capacitor is fully charged, no more current can flow in the branch of the circuit with the capacitor. The capacitor is in series with the battecy, so there will be no
current drawn from the battecy.