(A) Find the time constant of the circuit shown in Figure 32.11a.

(B) The switch in Figure 32.11a is 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.

Question

(A) Find the time constant of the circuit shown in Figure 32.11a.

(B) The switch in Figure 32.11a is 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.

Accepted Answer

The time constant is given by Equation 32.8:

[latex] au=\frac{L}{R} [/latex] (32.8).

[latex] au=\frac{L}{R}=\frac{30.0 imes 10^{-3} H }{6.00 \Omega}=5.00 ms [/latex].

(B) The switch in Figure 32.11a is closed at t = 0. Calculate the current in the circuit at t = 2.00 ms.

Using Equation 32.7 for the current as a function of time (with t and τ in milliseconds), we find that at t = 2.00 ms,

[latex] I=\frac{\varepsilon}{R}\left(1-e^{-t / au} ight) [/latex] (32.7).

[latex] I=\frac{\varepsilon}{R}\left(1-e^{-t / au} ight)=\frac{12.0 V }{6.00 \Omega}\left(1-e^{-0.400} ight)=0.659 A [/latex].

A plot of the current versus time for this circuit is given in Figure 32.11b.

(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 thus 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 12.0 V as the inductor tries to maintain the zero-current condition. (The left end of the inductor is at a higher electric potential than the right end.) As time passes, the emf across the inductor decreases and the current in the resistor (and hence the potential difference across it) increases, as shown in Figure 32.12. The sum of the two potential differences at all times is 12.0 V.

Question 32.3: (A) Find the time constant of the circuit shown in Figure 32...

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