Tuesday, 9 October 2007

Why is it that in arterioles, the resistance is highest and so the steepest drop in BP occurs? But shouldn't BP increase with increased resistance?

I presume you're looking at the graph of the arterial tree's pressures that looks something like this:
You ask a very good question. The answer is basically that it depends whether you are measuring the pressure before or after the obstruction (resistance).

Consider a pipe in which water is flowing. Ohm's law applies:

Flow = (Pressure difference) / resistance*

In other words, there will be flow in the pipe as long as there is a pressure difference between the two ends, but it'll be offset by resistance. The higher the resistance, the slower the flow. So far so good. Now let's manipulate the equation to show how the pressure difference changes:

Pressure difference = Flow x resistance

So, if you keep flow constant (which you must do in the human body, since the 'pipe' is basically a loop that starts and ends with the heart; you can't have flow in one part but not the other), the pressure difference INCREASES when there is resistance.

So does the pressure increase too? Not so fast. Before the resistance, the pressure increases a little, but after the obstruction, the pressure is DECREASED. Thus, the resistance still causes a drop in the pressures as measured after the obstruction.

In other words, there is a drop in pressure after a resistor. However, there is a rise in pressure before the resistor. This accounts for the validity of the blood pressure equation:

BP = cardiac output x total peripheral resistance

So if you want to raise the blood pressure in the body (measured at, say, the aorta), you can increase the resistance of the arterioles.

*In conventional physics notation: Q = ΔP / R

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