Cracking the AP Physics 2 Exam, 2019 Edition by Princeton Review

Author:Princeton Review
Language: eng
Format: epub
Publisher: Random House Children's Books
Published: 2018-11-12T16:00:00+00:00

Ammeters

Ammeters are connected in series because they measure current, and current stays the same in series.

Voltmeters

Voltmeters are connected in parallel because they measure voltage, and voltage stays the same in parallel.

Solution. We consider the ammeter to be ideal; this means it has negligible resistance, so it doesn’t alter the current that it’s trying to measure. Similarly, we consider the voltmeter to have an extremely high resistance, so it draws negligible current away from the circuit.

Our first goal is to find the equivalent resistance in the circuit. The 600 Ω and 300 Ω resistors are in parallel; they’re equivalent to a single 200 Ω resistor. This is in series with the battery’s internal resistance, r, and R3. The overall equivalent resistance is therefore R = 50 Ω + 200 Ω + 150 Ω = 400 Ω , so the current supplied by the battery is I = V/R (2400 V)/(400 Ω) = 6 A. At the junction marked J, this current splits. Since R1 is twice R2, half as much current will flow through R1 as through R2; the current through R1 is I1 = 2 A, and the current through R2 is I2 = 4 A. The voltage drop across each of these resistors is I1R1 = I2R2 = 1,200 V (matching voltages verify the values of currents I1 and I2). Since the ammeter is in the branch that contains R2, it will read I2 = 4 A.

The voltmeter will read the voltage drop across R3, which is V3 = IR3 = (6 A) (150 Ω) = 900 V. So the potential at point b is 900 V lower than at point a.

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