Consider the circuit in the figure. Suppose the circuit elements have the follow
ID: 1434359 • Letter: C
Question
Consider the circuit in the figure. Suppose the circuit elements have the following values: epsilon = 12.5 V, R = 7.00 ohm, and L = 34.0 mH. (A) Find the time constant of the circuit. (B) Switch S_2 is at position a_1 and switch S_1 is thrown closed at t = 0. Calculate the current in the circuit at t = 2.00 ms. (C) Compare the potential difference across the resistor with that across the inductor. (A) Find the time constant of the circuit. Conceptualize You should understand the behavior of the circuit in the figure from the discussion in this section. Categorize We evaluate the results using equations developed in this section, so this example is a substitution problem. Evaluate the time constant from the following equation: (B) Switch S_2 is at position a_1 and switch S_1 is thrown closed at t = 0. Calculate the current in the circuit at t = 2.00 ms. Evaluate the current at t = 2.00 ms from the following equation: (C) Compare the potential difference across the resistor with that across the inductor. At the instant the switch is closed, there is no current and therefore no potential difference across the resistor. At this instant, the battery voltage appears entirely across the inductor in the form of a back emf of 12.5 V as the inductor tries to maintain the zero-current condition. (The top end of the inductor in the figure is at a higher electric potential than the bottom end.) As time passes, the emf across the inductor decreases and the current in the resistor (and hence the voltage across it) increases as shown in the figure below. The sum of the two voltages at all times is 12.5 V. (a) Calculate the current in the circuit after the switch has been closed for one time constant. A (b) Calculate the voltage across the resistor after one time constant has elapsed since the switch has been closed. V When switch S_1 is thrown closed, the current increases and an emf that opposes the increasing current is induced in the inductor. When the switch S_2 is thrown to position b, the battery is no longer part of the circuit and the current decreases. An RL circuit. When switch S_1 is in position a_1 the battery is in the circuit.Explanation / Answer
A) time constant = L /R = (34 x 10^-3 ) / (7 ) = 4.86 x 10^-3 s
T = 4.86 ms
B) I = Imax [ 1 - e^(-t /T)]
Imax = e / R = 12.5/ 7 = 1.79 A
I = 1.79 [ 1 - e^(-2/4.86)]
I =0.602 A
C) V = iR = 0.602 x 7 = 4.22 Volt
MasterIT:
a) I = 1.79 [ 1 - e^(-T/T)] = 1.13 A
b) V = 1.13 x 7 = 7.92 Volt