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Given a sinusoidal voltage source attached to a load, like the following,
I want to get both the resistive ## R_o ## and the reactive ## X_o ## components of the output impedance ## R + jX_o ##. Measuring the open circuit voltage will give ##V##. If I connect a test resistance ##R_1## at the output, and measure the magnitude of the resulting voltage across it ##V_o1## with an oscilloscope, and then do the same again with another test resistance ##R_2## and its corresponding voltage magnitude ##V_o2##, then using the voltage division equations, I can obtain values for ##Ro## and ##Xo^2##, and to walk through the Math (note the ##\Rightarrow [X]## at the end of some of the lines are just to label the equations for easy reference):
Testing Resistance ##R_1##:
## V_{o1} = V * \frac {R_1} {R_1 + R_o + jX_o} \Rightarrow [1]##
## \Rightarrow R_1 + R_o + jX_o = R_1 * \frac {V} {V_o1} ##
(taking magnitude of both sides)
## \Rightarrow \sqrt {(R_1 + R_o)^2 + X_o^2} = R_1 * \frac {V} {V_o1}##
(squaring both sides)
## \Rightarrow (R_1 + R_o)^2 + X_o^2 = (R_1 * \frac {V} {V_o1})^2 \Rightarrow [2]##
Likewise with ##R_2##:
## (R_2 + R_o)^2 + X_o^2 = (R_2 * \frac {V} {V_o1})^2 \Rightarrow [3]##
Now subtracting both equations and getting ride of the ##X_o^2## term,
## (R_1 + R_o)^2  (R_2 + R_o)^2 = (R_1 * \frac {V} {V_o1})^2  (R_2 * \frac {V} {V_o1})^2 \Rightarrow [4]##
(expanding the two expressions on the LHS)
## R_1^2 + 2R_1R_o + R_o^2  R_2^2  2R_2R_o  R_o^2 = (R_1 * \frac {V} {V_o1})^2  (R_2 * \frac {V} {V_o1})^2 ##
(rewriting the ##R_1^2## and ##R_2^2## terms as a difference of squares and factoring ##2R_o## from the other terms)
## (R_1  R_2)(R_1 + R_2) + 2R_o(R_1  R_2) = (R_1 * \frac {V} {V_o1})^2  (R_2 * \frac {V} {V_o1})^2 ##
(dividing out ##(R_1  R_2)## and ##2##)
## \frac {R_1 + R_2} {2} + R_o = \frac {1} {2(R_1  R_2)} * (R_1 * \frac {V} {V_o1})^2  (R_2 * \frac {V} {V_o1})^2 ##
(finally solving for ##R_o##)
## R_o = \frac {1} {2(R_1  R_2)} * (R_1 * \frac {V} {V_o1})^2  (R_2 * \frac {V} {V_o1})^2  \frac {R_1 + R_2} {2} \Rightarrow[5]##
And then from either of the two original voltage division equations ##[2]## or ##[3]##, you can solve for ##X_o^2## to get the following:
## X_o^2 = (R_{1 or 2} * \frac {V} {V_o1})^2  (R_{1 or 2} + R_o)^2 \Rightarrow [6]##
Now for ##X_o##, we can get the magnitude by taking the square root of ##[6]## but the sign could go either way. My thinking was, alright, calculate the current by using ## \frac {V_{1or2}} {R_{1or2}}## and then the phase of the current will indicate whether the output impedance is capacitive or inductive, but wait, we don't know the phase relative to the source, and there is no way of directly measuring both the ##Vo## and ##V## at the same time on a scope and seeing the phase difference with just the output terminals. And likewise to measure short circuit current and then divide the open circuit voltage by that doesn't work either, since they have to be measured at the same time, which is impossible. So, how can I determine the nature of the reactance of the output impedance?
I want to get both the resistive ## R_o ## and the reactive ## X_o ## components of the output impedance ## R + jX_o ##. Measuring the open circuit voltage will give ##V##. If I connect a test resistance ##R_1## at the output, and measure the magnitude of the resulting voltage across it ##V_o1## with an oscilloscope, and then do the same again with another test resistance ##R_2## and its corresponding voltage magnitude ##V_o2##, then using the voltage division equations, I can obtain values for ##Ro## and ##Xo^2##, and to walk through the Math (note the ##\Rightarrow [X]## at the end of some of the lines are just to label the equations for easy reference):
Testing Resistance ##R_1##:
## V_{o1} = V * \frac {R_1} {R_1 + R_o + jX_o} \Rightarrow [1]##
## \Rightarrow R_1 + R_o + jX_o = R_1 * \frac {V} {V_o1} ##
(taking magnitude of both sides)
## \Rightarrow \sqrt {(R_1 + R_o)^2 + X_o^2} = R_1 * \frac {V} {V_o1}##
(squaring both sides)
## \Rightarrow (R_1 + R_o)^2 + X_o^2 = (R_1 * \frac {V} {V_o1})^2 \Rightarrow [2]##
Likewise with ##R_2##:
## (R_2 + R_o)^2 + X_o^2 = (R_2 * \frac {V} {V_o1})^2 \Rightarrow [3]##
Now subtracting both equations and getting ride of the ##X_o^2## term,
## (R_1 + R_o)^2  (R_2 + R_o)^2 = (R_1 * \frac {V} {V_o1})^2  (R_2 * \frac {V} {V_o1})^2 \Rightarrow [4]##
(expanding the two expressions on the LHS)
## R_1^2 + 2R_1R_o + R_o^2  R_2^2  2R_2R_o  R_o^2 = (R_1 * \frac {V} {V_o1})^2  (R_2 * \frac {V} {V_o1})^2 ##
(rewriting the ##R_1^2## and ##R_2^2## terms as a difference of squares and factoring ##2R_o## from the other terms)
## (R_1  R_2)(R_1 + R_2) + 2R_o(R_1  R_2) = (R_1 * \frac {V} {V_o1})^2  (R_2 * \frac {V} {V_o1})^2 ##
(dividing out ##(R_1  R_2)## and ##2##)
## \frac {R_1 + R_2} {2} + R_o = \frac {1} {2(R_1  R_2)} * (R_1 * \frac {V} {V_o1})^2  (R_2 * \frac {V} {V_o1})^2 ##
(finally solving for ##R_o##)
## R_o = \frac {1} {2(R_1  R_2)} * (R_1 * \frac {V} {V_o1})^2  (R_2 * \frac {V} {V_o1})^2  \frac {R_1 + R_2} {2} \Rightarrow[5]##
And then from either of the two original voltage division equations ##[2]## or ##[3]##, you can solve for ##X_o^2## to get the following:
## X_o^2 = (R_{1 or 2} * \frac {V} {V_o1})^2  (R_{1 or 2} + R_o)^2 \Rightarrow [6]##
Now for ##X_o##, we can get the magnitude by taking the square root of ##[6]## but the sign could go either way. My thinking was, alright, calculate the current by using ## \frac {V_{1or2}} {R_{1or2}}## and then the phase of the current will indicate whether the output impedance is capacitive or inductive, but wait, we don't know the phase relative to the source, and there is no way of directly measuring both the ##Vo## and ##V## at the same time on a scope and seeing the phase difference with just the output terminals. And likewise to measure short circuit current and then divide the open circuit voltage by that doesn't work either, since they have to be measured at the same time, which is impossible. So, how can I determine the nature of the reactance of the output impedance?
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