CHAPTER FOUR Khrushchev's laws in AC circuits. Now this is going to be a short chapter, but I think it's significant enough to stand on its own and it will relieve some of the length Enos of the chapters on either side of it. So I thought I would let it stand out by itself. kerchief voltage law provides us with a check on our calculations. On the example in our previous chapter. To demonstrate kerchief Voltage Law in an AC circuit, we can look at the answers we derived for component voltage drops.
In the last circuit. kerchief voltage law tells us that the algebraic sum of the voltage drops across the resistor, the inductor and a capacitor should equal the applied voltage from the source even though this may not look true at first sight, a bit of complex number addition proves otherwise. So, the voltage drop across the resistor was given by 3.1472 plus j 19.177 volts, the inductor has a voltage drop of minus 18.797 plus j 30848 volts and the capacitors one 135.65 minus j 22.262. If you add up the real components, they do indeed add up to exactly 120 and the imaginary components or the J components do indeed add up to zero, which is equivalent to our supply voltage. That's providing a check on our calculations. So let's move our individual components to a parallel circuit now.
So we now have the hundred and 20 volt 60 hertz power supply, directly connected across all of the three components. The impedances of the components don't change, they're still the same and so I've written them in there. The fact that these components are connected in parallel instead of series has absolutely no effect on their individual impedances which should not be a surprise So long as the power supply is the same, which it is the inductive and capacitive reactance is will not have changed at all. With all of the component values expressed as impedances we can set up an analysis table and proceed as in the last example problem with a series circuit except this time following the rules of parallel circuits instead of the series circuits. Knowing that the voltage is shared equally by all the components in the parallel circuit we can transfer the finger for total voltage to all the components in the table.
It is equal to the supply voltage. We can now apply ohms law is equal to E all over Add vertically to each column to determine the current through each component, and again, the M designates milliamps. From here, there's two strategies for calculating the total current and total impedance. First, we could calculate the total impedance from all of the individual impedances in parallel. In other words, the total impedance would be given by one over the quantity one over Zed r plus one over set L plus one over said C, and then calculate the current by dividing the source voltage by the impedance. However, working through the parallel impedance equation with complex numbers is no easy.
Task two can be done with all the reciprocation and changing back and forth from rectangular to polar form. Again it can be done but it is very labor intensive. The second way to calculate the total current total impedance is to add up all the branch currents to arrive at a total current. Remembering that a total current in a parallel circuit AC or DC is equal to the sum of the branch currents according to current shops current law, then use ohms law to determine the total impedance for the total voltage and the total current. This ends chapter four