What is the relationship between driving point resistance and Thevenin resistance? Thevenin Theorem
A linear, active, resistive network which contains one or more voltage and current sources, can be replaced by a single voltage source $V_T$ (the Thevenin voltage) and a series resistance $R_T$ (the Thevenin resistance).
However, when the entire network has only a single voltage source $V_s$, then the rest of the network is passive; if current through this source is $I_s$ the driving point resistance is defined as,
$$
R_{\text{dp}} := \frac {V_s}{I_s}
$$
Now my simple question, can't we say that in this scenario,$R_{\text{dp}}$ is $R_T$ and $V_s$ is $V_T$?
Thanks.
 A: You have to define a pair of nodes (the "output") across which you have the Thevenin voltage when no extra components are connected across those two nodes.
That voltage is not necessarily $V_{\rm S}$.
Imagine a simple series circuit consisting of two unequal resistors, $R_1$ and $R_2$ and a voltage source.
You now have three possible outputs, across $R_1$, across $R_2$ and across the voltage source and so three possibilities for a Thevenin equivalent circuit.
A: If I understand you correctly, it looks like you are considering the trivial case where there is only one voltage source and only one resistor in series with that source, what you call the driving point resistance, such that $I_s$ is the short circuit current and $V_s$ is the open circuit voltage. If that's the case, then obviously $V_{s}=V_{T}$ and
$$R_{dp}=R_T=\frac{V_{s}}{I_s}$$
If that's not what you meant, please clarify your question.
Hope this helps.
A: For an a.c. circuit
(applies also for d.c. ideal circuits without any reactance $X$ impedance):
Since the actual circuit and the simplified Thevenin (i.e. one source with one ohmic resistor load in series, replacing the real part of the possible complex discrete impedance components and ohmic resistor components of the actual a.c. circuit e.g. coils, capacitors, resistors etc. and the algebraic sum of all the active sources)  are in any way equivalent, regarding the real power consumption in your circuit, then an actual circuit having one source and multiple impedance components and or ohmic resistors  (could be any arbitrary combination of in series or in parallel connection),
$$
|Z|=\sqrt{Z Z^{*}}=\sqrt{R^{2}+X^{2}} = R_{\text{dp}} = \frac {V_s}{I_s} = \frac {V_{Th}}{I_{Th}} = R_{\text{Th}}
$$
where $R$ is the net ohmic resistance and $X$ the net reactance of all passive circuit components for a single source, multiple mixed ohmic and reactive components circuit. $V_{s}$ and $I_{s}$ are the RMS values of source voltage and source current in the a.c. actual circuit.
Of course it depends also as previously described in a previous answer where you assign the output of your circuit since $R_{\text{Th}}$ will change if your assigned output is not in parallel with the source and at the end of  the actual circuit. Thevenin describes a power consumption equivalent circuit depending where you sample the power in the circuit. If the output is not as above described then obviously $R_{\text{dp}}$ ≠ $R_{\text{Th}}.$
Also, notice that in this case, $V_{s}$ and $I_{s}$ are not equal with ${V_{Th}}$ and ${I_{Th}}$.
