University:
University of Michigan
Course:
MATH 116 | Calculus II
Academic year:

2024
Views:

158

Pages:

2
Author:

banenetta0l

Question 1 Answer the following T/F questions based on the behavior of the circuit in the above Figure 5: A transient current flows for a short time after the RELAY is switched to either the N.O. ("normally open") or N.C. ("normally closed") position. The transient currents that flow in this experiment decay exponentially to zero. The capacitor charge 𝑄 Q exponentially decays to zero when the RELAY is thrown to position N.O. When the current is positive, the capacitor charge 𝑄 Q is decreasing. When the current is positive, the measured VOLTAGE IN 2 will be negative. When the capacitor charge 𝑄 Q is positive, the measured VOLTAGE IN 1 is positive. The quantity 𝑡 1 / 2 = 𝑅 ⋅ 𝐶 ln ⁡ ( 2 ) t 1/2 =R⋅Cln(2) is called the half-life of an exponential decay, where 𝑅 ⋅ 𝐶 R⋅C is the time constant in an RC circuit. The current in a discharging RC circuit drops by half whenever 𝑡 t increases by 𝑡 1 / 2 t 1/2 . For a circuit with 𝑅 = 2 𝑘 Ω R=2kΩ and 𝐶 = 3 𝜇 F C=3μF, if the current is 6 mA 6mA at 𝑡 = 5 ms t=5ms, at what time (in ms) will the current be 3 mA 3mA? Answer True. The capacitor is initially uncharged, and the voltage on a capacitor cannot change instantaneously. Therefore, voltage before the switch throws a voltage after, and the initial current is 𝑉 𝑠 / 𝑅 Vs/R. True. At any point in time, the instantaneous current is ( 𝑉 𝑠 − 𝑉 𝑐 ) / 𝑅 (Vs−Vc)/R, so as capacitor voltage increases, the current decreases until 𝑉 𝑐 = 𝑉 𝑠 Vc=Vs and then 𝐼 = 0 I=0. True. The charge decays exponentially until it reaches a steady-state value when 𝑉 𝑐 = 𝑉 𝑠 Vc=Vs. Then, 𝑄 = 𝐶 𝑉 𝑠 Q=CVs. False. With the direction/polarity shown in the diagram, a positive 𝐼 I results in increasing capacitor charge. False. The capacitor voltage will be positive, and the voltage across the resistor will be positive at the "top" of the resistor; therefore, IN 2 is positive. False. VOLTAGE OUT 1 and VOLTAGE IN 1 are equal and always zero volts. True. The equation for the current is: 𝐼 ( 𝑡 ) = ( 𝑉 𝑠 / 𝑅 ) exp ⁡ ( − 𝑡 / 𝑅 𝐶 ) I(t)= exp(−t/RC) (Vs/R) You are given 𝑅 R and 𝐶 C, so use the data ( 𝐼 = 6 mA I=6mA at 𝑡 = 5 ms t=5ms) to determine that 𝑉 𝑠 / 𝑅 = 10.6119 mA Vs/R=10.6119mA. Then solve for the time when 𝐼 = 3 mA I=3mA: 𝑡 = 1.993 ms . t=1.993ms.