IAN LANG ELECTRONICS

# Working Through LCR Problems

Recently a reader has e-mailed asking for help with problems pertaining to inductive and capacitive reactance and impedance and I must thank him for that because looking through the site I haven't gone into much detail about it, and it strikes me that if one person is having problems understanding it then so must others. To address this issue then, the following pages are all about reactance and impedance. We'll be looking later at the maths behind it but first a short lecture about capacitors, inductors, voltage and current.

As you may be aware, capacitors store charge in an electric field and inductors store charge in a magnetic field. You are doubtless further aware that there are two types of current, DC and AC.

When you use DC (direct current) you get a steady, constant voltage and the waveform of voltage against time looks like this:  Time  ( t )

Voltage ( V )

Where the red lines represent the divisions of a graph which could be in seconds, minutes, hours or days, Babylonian epochs or whatever period of time you are considering as long as all divisions are equal. For arguments sake let's take point P on the time scale on the X axis above (the one that goes across the page) as 10 seconds after point O above. We've measured a time period there of ten seconds. Point H, which is halfway further out from O and halfway further back from P is then of course five seconds, and halfway between  O and H would be 2.5 seconds, halfway between H and P is of course 7.5 seconds and so on. So far so easy.  Over the course of that ten seconds we have measured the voltage output of three brand-new AAA batteries. It's 4.8 V,  and that's what the black line represents.

The black line therefore represents the voltage waveform of the three batteries. Batteries always give out direct current (DC) due to the action of their chemistry producing a constant reaction. Not that it's technically a waveform, more of a straight line. You'll notice how it never varies throughout any point in the time period, and the net effect of putting a capacitor or inductor in a DC circuit is that the capacitor will charge up in time periods and the inductor will build a magnetic field and thus a time delay in the current reaching the rest of the circuit can be made. It's rare and pointless to use an inductor in DC circuits and so if you look into any sort of time-delays I'm 99.9 per cent sure you'll find a capacitor in there. You can try it for yourself. You need a capacitor (a biggish one) a couple of resistors, an NPN transistor and an LED.

You'll also need a power supply of  4-9 V and three AA batteries will do for this. In the test circuit I'm going to use a 9V battery simply because I've got a 9V box lying about doing nothing.

Here's a circuit: P

0

H And here's a short video:

The time taken to get the required voltage depends on the Farad rating of the capacitor and the ohm rating of the resistor through which it charges. At first, all the voltage will go to the capacitor and so will the current.  As the capacitor begins to charge, some will flow to the capacitor and some to the 22k resistor. The more the capacitor charges, the more flows to the 22k resistor until eventually you get enough voltage to light the LED. So here comes the first bit of maths to get you gently eased into the maths behind reactance and impedance.

The equation for finding the time constant for a capacitor to get to approximately 63 per cent of charge is:

# t = RC

Or in words, time = resistance multiplied by capacitance.

BUT-  there's a hidden bug here. You have to state the values in OHMS and Farads. Above we have 330k and 470uF and so we have to convert them. 1K = 1000 and so we have 330, 000 ohms. 1uF = 1/1,000,000 Farads and so the microfaradage is converted as 470/1,000,000  which is 0.00047 F

Our time here (in seconds)  to get 9V out would be  330, 000  times 0.00047 which equals 155.1

It will in fact take longer due to vagaries in the components, and in this case we only need about 2V to allow the LED to shine and so the time will be considerably shorter.

You can do the same with an inductor but it's rarely if ever done because magnetic fields can adversely affect other parts of a circuit and it's more difficult to make an accurate inductor than capacitor.

In this instance as we are using DC there is no component of frequency to trouble us. In AC there is. Over the page we look at AC waveforms and how capacitors and inductors influence them.

More > Go Back Tables

A converting table it may be handy to print out.