29.13 - Interactive problem: alternating current

In this simulation, there is a circuit that contains a signal generator and a resistor.

The emf of the signal generator causes a current to flow. Your task is to use an

oscilloscope to determine the maximum current Imax, the maximum potential

difference across the resistor ǻVmax and the signal frequency f.

You will see sinusoidal graphs in the oscilloscope similar to those you see in

illustrations in other sections of this chapter.

One probe from the oscilloscope is used to measure current. The current is the

same everywhere in the circuit. The two ends of the potential difference (“voltage”)

probe in this exercise are placed so it measures ǻV across the resistor.

The oscilloscope displays information in the form of two graphs. In the upper graph,

the current is displayed on the vertical axis and time is on the horizontal axis. In the

lower graph, ǻV is displayed on the vertical axis, and time once again is on the

horizontal axis.

Sliders allow you to change the scales of the graphs. The base unit is a single gray

square in the oscilloscope window. For example, if you set the current slider to

“8.00×10í^2 A,” then the height of a gray box represents 8.00×10í^2 amperes. You can set the scale for the time in an analogous fashion. All this

resembles the working of typical laboratory oscilloscopes.

You set the scale by trial and error. If you cannot see any signal, move the slider so the units are smaller until you can see the graph. If the

waves you see are so large that you cannot see their peaks, increase the scale until their peaks “shrink” into view.

When you launch the simulation, the signal generator is on and all the probes are attached. The peak of the wave in the current display is the

maximum current, Imax. The peak of the wave in the voltage display is ǻVmax.

You also have to analyze the wave to determine the frequency of the signal. To do so, identify two adjacent matching points in the wave (such

as two peaks) in order to identify a cycle. The time it takes to complete this cycle is the period of the wave, and the reciprocal of the period is

frequency. For instance, if the time between two peaks is 1.0×10í^3 seconds, then the frequency is 1/(1.0×10í^3 ), or 1,000 hertz (Hz).

A couple of tips may help here. Since the horizontal axis measures time, you can determine the period by counting how many gray boxes there

are between the corresponding points on the wave (be careful to note what the time scale is). It can be easier to use locations where the wave

crosses the horizontal axis, since more precise values can be noted there. But if you use those points, remember that matching points are

where the graph is in both cases going down, or in both cases going up. Two adjacent points where the wave crosses the horizontal axis, but in

opposite directions, represent only half a cycle.

When you have analyzed a signal and determined Imax,ǻVmax and the signal frequency, enter these values in the boxes in the simulation.

Press CHECK to see if you are correct.

To measure the second signal, press the Signal Two button on the signal generator. The signal generator will create a signal with a different

potential difference and frequency. You can also analyze this signal, enter your conclusions and check them by pressing CHECK.

29.14 - Mutual induction

Mutual induction: The induction of an emf in

one circuit by a changing current in another

circuit.

Currents create magnetic fields. A changing current in a circuit creates a changing

magnetic field. This changing magnetic field will induce an emf and a current in a

nearby second circuit, in a process called mutual induction.

At the right, you see an illustration of the steps involved in mutual induction:

1. A switch is closed in the left-hand circuit. This circuit contains a battery.


2. A current in the left-hand circuit starts to flow after the switch is closed, increasing
   toward its steady state value.


3. The increasing current on the left generates a changing magnetic field.


4. The changing magnetic field causes a changing magnetic flux through the circuit
   on the right, which induces an emf in this circuit.


5. The induced emf in the right-hand circuit causes a current there, lighting the light
   bulb for a brief time.


6. Once the current on the left reaches its steady state, the light bulb on the right goes out, because there is no longer a changing
   magnetic field to induce a current there.

An important point to remember is that currents are induced by achangingmagnetic field. The closing of the battery circuit starts a current that

in turn creates a magnetic field where before there was none. It is this change in magnetic field that induces the emf and the current in the

second circuit. When the current caused by the battery reaches a steady state, which occurs quite quickly, then its magnetic field will also be

Mutual induction

Changing current in one circuit creates

current in other

·Changing current causes changing

magnetic field

·Changing magnetic field induces

current

(^548) Copyright 2007 Kinetic Books Co. Chapter 29