www.nortonkit.com 18 अक्तूबर 2013
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Resistance and Reactance: [Series RC Circuits] [Series RL Circuits] [Parallel RC Circuits] [Parallel RL Circuits] [Series LC Circuits] [Series RLC Circuits] [Parallel LC Circuits] [Parallel RLC Circuits]
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Resistors and AC

When we apply an ac voltage to a resistor as shown to the right, current will flow through that resistor. This much seems logical and reasonable. But how much current? Can we apply Ohm's Law as we did with DC? How can we even specify the ac voltage, since it is constantly changing? To answer these questions, we must first define how we will specify the ac voltage applied to the resistor.

In most cases, it makes a lot of sense to describe ac voltage and current in terms of a "dc equivalent," such that the actual ac power delivered to the load does exactly the same amount of work as the same value of dc voltage and current applied to the same load. To do this, we need to find some sort of "average ac power" over the entire cycle. Unfortunately, the actual average voltage of the applied ac is zero. The same is true of the alternating current flowing back and forth through the circuit. Yet we know that real power is used, because light bulbs turn on, clocks run, electric motors work, etc. How do we resolve this?

### The RMS Values

The answer is to identify the power dissipated by the resistor, in terms of either the ac voltage by itself, or the current by itself. This is easy enough; we already know that we can use either of two expressions for this:

P = I²R = E²/R

Since squaring a number always results in a positive result (or zero, and we are not using imaginary numbers here), we should be able to find a real average of the squared value. Then we can take the square root of that average to get the effective current or voltage.

If we plot a unit sine wave and its square, we will get the graph shown to the right. Here, the sine wave (in red) varies in the range ±1, while its square (in blue) is a smaller sinusoidal waveform that varies from 0 to 1. Mathematically:

sin²(x) = ½ - [cos(2x)]/2

Since the average value of any sine (or cosine) wave by itself is zero, the average value of the above expression is simply 1/2. This is the average of the squared sine wave. Therefore we must take its square root to get the effective value, which is 1/ = 0.707. This factor gives us the square root of the mean (or average) of the squared value of the sine wave. Therefore, the effective value of the waveform is also known as the root-mean-square, or rms value.

The peak value of the actual sine wave can be any value; it is a constant multiplier which is squared and then has the square root taken of its squared value. As a constant, it can be left out of the averaging process. It is simply used in the final calculation. Thus:

Irms = Ipeak/ = Ip × 0.707
Erms = Epeak/ = Ep × 0.707

The above expressions apply specifically to sinusoidal waveforms. Other waveforms may have a different relationship between peak and rms values, and must be analyzed separately.

### Applying Ohm's Law

Now that we know what the rms value of a voltage or current is and how we got it, we can look at the relationship between voltage and current in a resistance. In the graph to the right, we have arbitrarily assigned the red curve to voltage and the blue one to current. We note here that the current increases or decreases directly with the applied voltage at any instant in time. Thus, Ohm's Law applies to this circuit directly, and is accurate for peak voltage and current, instantaneous voltage and current, and most importantly, for rms voltage and current. In all cases, E = IR. And if we use rms values, P = IE = I²R = E²/R.

Since voltage and current change in step with each other, they are said to be in phase with each other. This is always true for any resistance or resistive circuit.