I read in sci.electronics.design that Wayne <

[email protected]>

wrote (in said:

The resistor is in series with the capacitor and the signal gen is

across one lead of the capacitor and one lead of the resistor. In other

word, if they were both resistors it would be a voltage divider. The

resistor or the capacitor can be ground. Forgive me for asking a dumb

question but why are V_c^2 + V_r^2 = V_s^2 to the power of 2?

Because the two voltages have a phase difference of 90 degrees (or, in

practice, very nearly; the capacitor won't be perfectly loss-free).

If this

was a voltage divider then you would have V?=Vs(R/R+Rc)?

In method 2 could you give me an example with the following figures?

Vs=10v p-p

Vr=4v p-p

Phase = 45deg (with respect Vs)

It's helpful to use a 'phasor diagram' to see what is going on. The

current I is in phase with the voltage across the resistor, and the

voltage across the capacitor is at 90 degrees to the current. So our

diagram is (*use Courier font*):

/|

/ |

/ |

/ |

Vs = 10 / |

/ | Vc =?

/ |

/ |

/ |

/ 45 |

/__________| ----------> Current I

Vr = 4 V

Now, I hope you can see that this is not, in theory, possible. The angle

at the apex of the triangle must be 45 degrees as well and the triangle

must be isosceles. So Vc must be 4 V as well, and then by Pythagoras

(which is the same as the squared values in Tim's equation), Vs =

sqrt(4^2 + 4^2) = sqrt(32) = 5.66 V.

One practical explanation, if your figures are correct, is that the

capacitor is very lossy, thus having a lot of resistance of its own.

This add to the horizontal side of the triangle (without altering the 4

V across the resistor). Another 3.07 V brings the figure right. The Vc

must also be 7.02 V, of course, because of the 45 degrees. We then get

by Pythagoras, Vs = sqrt(7.07^2 + 7.07^2) = sqrt(100) = 10.

It's a very poor capacitor if this is true. There are other

possibilities. What frequency are you working at and what are the

resistor and capacitor values? Have you actually measured 10 V at the

signal generator, or is that just what the output control says? If the

signal generator source impedance isn't very much smaller than the

impedance of your RC circuit, the Vs value will be way off, indeed it

may well be 5.66 V!