AP® Physics C: E&M - Kirchhoff’s Loop Rule - MCQs
Question
The circuit shown has a battery of negligible internal resistance, resistors, and a switch. There are voltmeters, which measure the potential differences \(V_{1}\) and \(V_{2}\), and ammeters \(A_{1}\), \(A_{2}\), \(A_{3}\), which measure the currents Ii, h, and h- The switch is initially in the closed position. With the switch still closed, which of the following relationships are true? (Select two answers.)
(A) f 1 +12 – h = 0
(B) L1VB – \(V_{1}\) – \(V_{2}\) = 0
(C) \(V_{1}\) > \(V_{2}\)
(D) h =h
Answer/Explanation
Ans:
A and B—Applying Kirchhoff s junction rule to the node above the \(A_{2}\) ammeter we get:
\(\Sigma ^{I_{junction}}=I_{1}+I_{2}-I_{3}=0\)
Applying Kirchhoff s current rule to the loop on the right of the circuit containing the battery we get:
\(\Sigma ^{\Delta V_{loop}}=\Delta V_{B}-\Delta V_{1}-\Delta V_{2}=0\)
\(J_{2}\) < \(J_{3}\): Ammeter 3 is in the main circuit line and carries the total current. This current splits with half going through ammeter 1 and 2. \(U_{1}\) = \(U_{2}\) If you work out the math you will see that these voltages are the same.
Questions (a) and (b) refer to the following material.
The diagram shows a circuit that contains a battery with a potential difference of VB and negligible internal resistance; five resistors of identical resistance; three ammeters A1, A2, A3; and a voltmeter.
Question(a)
Which of the following correctly ranks the readings of the ammeters?
(A) A 1 = A2 = A3
(B) A1 = A2 > A3
(C) A1>A2 >A3
(D) A2 > A1 > A3
Answer/Explanation
Ans:D
The resistance of the parallel set of three resistors on the far right is \(\frac{2}{3}R.\) Thus the total resistance of the circuit to the right of the battery is \(\frac{5}{3}R.\) The loop on the left has a resistance of 2R. This means that the reading of A2 > A 1. The current that goes through A3 must be less than A2 because it is in a branching pathway. Answer choice D is the only option that meets these requirements.
Question(b)
What will be the reading of the voltmeter?
(A) \(\frac{1}{5}V_{B}\)
(B) \(\frac{1}{3}V_{B}\)
(C) \(\frac{2}{5}V_{B}\)
(D) \(\frac{3}{5}V_{B}\)
Answer/Explanation
Ans:C
Ammeter A2 is in the main line supplying the current to the right-hand side of the circuit. The total resistance of the circuit to the right of the battery is \(\frac{5}{3}R\) (as described in the answer to question 20). This gives a total current passing through ammeter A2 of:
\(I = \frac{\Delta V}{R}=\frac{V_{B}}{\frac{5}{3}R}= \frac{3V_{B}}{5R}\)
Using this current we can calculate the voltage drop through the resistor in the wire passing through ammeter A2:
\( \Delta V = IR = \left ( \frac{3V_{B}}{5R}\right )(R) =\frac{3V_{B}}{5}\)
This leaves the remaining voltage drop of \(\frac{2}{5}V_{B}\) that will be read by the voltmeter.