Kirchhoff’s Junction Rule AP Physics 2 FRQ – Exam Style Questions etc.
Kirchhoff’s Junction Rule AP Physics 2 FRQ
Unit 11: Electric Circuits
Weightage : 15–18%
Exam Style Practice Questions ,Kirchhoff’s Junction Rule AP Physics 2 FRQ
Question
Consider the following circuit:
(a) Determine the current that passes through \(R_{3}\) and \(R_{1}\).
(b) At what rate does \(R_{4}\) resistor dissipate energy?
(c) What is the voltage read by the voltmeter in the circuit?
(d) In order to minimize the impact of the voltmeter on the circuit, should the voltmeter have a very high or very low internal resistance?
Answer/Explanation
Ans:
(a) First, label some points in the circuit.
There are three branches each with their own current, \(fabc(I_{1})\), \(cdef(I_{2})\), \(cf(I_{3})\). By the Junction Rule: \(I_{3}=I_{1}-I_{2}\).
By the Loop Rule, the total change around any closed loop must be zero. You can then write equations for the following loops:
\(abcfa\): \(-I_{1}R_{3}-I_{1}R_{4}-(I_{1}-I_{2})R_{2}-(I_{1}-I_{2})R_{1}+V_{1}=0\)
\(cdefc\): \(V_{2}+(I_{1}-I_{2})R_{2}+(I_{1}-I_{2})R_{1}=0\)
Adding the two equations, you get
\(-I_{1}(35\Omega )+25V=0\)
\(25V=I_{1}(35\Omega )\)
\(I_{1}=\frac{5}{7}A\)
Plugging this into the cdefc equation, you get
\(15V+(I_{1})(15\Omega )-(I_{2})(15\Omega )=0\)
\(15V+\frac{5}{7}A(15\Omega )-(I_{2})(15)=0\)
\(1V+\frac{5}{7}V-(I_{2})(1\Omega )=0\)
\((I_{2})(1\Omega )=\frac{12}{7}V\)
\(I_{2}=\frac{12}{7}A\)
So the current through \(R_{3}\) is \(\frac{5}{7}\)A and \(R_{3}\) is \(\frac{12}{7}\)A.
(b) The rate at which \(R_{4}\) dissipates energy is given by \(P=IR^{2}=\frac{5}{7}A(20\Omega )^{2}=285.7W\)
(c) The voltmeter will read the sum of the voltage drops across \(R_{3}\) and \(R_{4}\)
\(V_{3+4}=V_{3}+V_{4}=I^{3}R^{3}+I^{4}R^{4}=(\frac{5}{7}A)(15\Omega )+(\frac{5}{7}A)(20\Omega )=25v\)
(d) The voltmeter is connected in parallel with the circuit. To minimize the impact of the voltmeter on the circuit, the voltmeter should have a very high resistance. Consider the following calculation of the total resistance upon addition of the voltmeter:
\(\frac{1}{R_{eff}}=\frac{1}{R_{+4}}+\frac{1}{R_{volmeter}}\)
If \(R_{volmeter}\) was very high, the \(R_{eff}\) would be very close in value to \(R_{3+4}\):
\(V_{R3}+V_{R4}=I_{1}R_{3}+I_{1}R_{4}=\frac{5}{7}A(15\Omega )+\frac{5}{7}A(20\Omega )=25V\)
Question
Three identical light bulbs are connected in the circuit shown above. The switch S is initially in the open position and then is closed at time t.
(a) Describe any changes that occur in the current through each bulb when the switch is closed. Justify your answers.
(b) Describe any changes that occur in the brightness of each bulb when the switch is closed. Justify your answers.
(c) When the switch is closed, does the power output of the battery increase, decrease, or remain the same? Justify your answer.
(d) When the switch is closed, the current in Bulb 1 changes. Explain why this change in current does not violate the law of conservation of charge.
Answer/Explanation
Ans:
Part (a)
Initially, the circuit is just Bulbs 1 and 3 in series. When Bulb 2 is added, the voltage from the battery is unchanged. Yet the total resistance of the circuit decreases, because an additional parallel path is added. Therefore, by V = IR with constant V, the total current in the circuit increases. Bulb 1 takes the total current, so Bulb 1’s current increases. For Bulb 1 only, the resistance is a property of the bulb and thus doesn’t change. So by V = IR with constant R, Bulb 1 takes an increased voltage, too. Then by Kirchoff’s loop rule, an increase voltage across Bulb 1 means a decreased voltage across Bulb 3. And for Bulb 3 only, by V = IR with constant R, Bulb 3’s current also decreases. (Obviously Bulb 2’s current increases from nothing to something.)
Part (b)
All bulbs have an unchanging resistance. Brightness depends on power, which is \(I^{2}R\). With constant R, a bigger current means more brightness; a smaller current means less brightness. So Bulb 1 gets brighter and Bulb 3 gets dimmer.
Part (c)
For the whole circuit, use power = \(V^{2}/R\). The voltage of the battery is unchanged because it’s still the same battery. The resistance of the circuit decreases because of the extra parallel path. Decreasing the denominator increases the entire value of the equation, so power increases.
Part (d)
Conservation of charge in circuits is expressed in Kirchoff’s junction rule—the current entering a junction equals the current leaving the junction. At any given moment of time, the junction rule holds. Now, when the switch is closed, more current flows from the battery than before. That’s not a violation of charge conservation, because the materials in the battery contain way more charged particles than are ever flowing through the wires. After the switch is closed, more current flows into the junction right before the switch than before, but more current also flows out of that junction than before. Charge conservation doesn’t mean that the same current must always flow in a circuit, it just says that whatever charge does flow in a circuit must flow along the wires.