For the ideal resistor, the voltage across is proportional to the current through and thus, their ratio is the constant \$R\$:
$$\frac{v_R}{i_R} = R $$
For the ideal (semiconductor) diode, we have
$$i_D = I_S(e^{\frac{v_D}{nV_T}}-1)$$
Inverting yields
$$v_D = nV_T\ln (1 + \frac{i_D}{I_S}) $$
thus, the diode voltage is not proportional to the diode current, i.e., the ratio of the voltage and current is not a constant.
$$\frac{v_D}{i_D} = \frac{nV_T}{i_D}\ln (1 + \frac{i_D}{I_S}) \ne R$$
Now, the small-signal or dynamic resistance is just
$$\frac{dv_D}{di_D} = \frac{nV_T}{I_S + i_D} \approx \frac{nV_T}{i_D} $$
how is it different from the normal resistance
As shown above, the diode static resistance (ratio of the diode voltage and current) differs from and is, in fact, larger than the diode dynamic resistance by the factor of \$\ln (1 + \frac{i_D}{I_S})\$
$$\frac{v_D}{i_D} = \frac{dv_D}{di_D} \ln (1 + \frac{i_D}{I_S})$$
which is to say that, in typical operating ranges, the diode dynamic resistance is much smaller than then diode static resistance.
Does the power dissipation relation, \$P=I^2r\$ hold in case of dynamic
resistances?
The instantaneous power associated with the diode is
$$p_D = v_D i_D = nV_Ti_D\ln (1 + \frac{i_D}{I_S}) \ne i_D^2\frac{nV_T}{i_D} = nV_Ti_D $$
Since the power associated with a circuit element is always the product of the voltage across and current through, one would not use the dynamic resistance but, rather the static resistance.
SUBSTITUTION THEOREM:
If the voltage across and the current through any branch of a dc bilateral network are known, this branch can be replaced by any combination of elements that will maintain the same voltage across and current through the chosen branch.
As long as terminal voltage and current is same, accordance with substitution theorem, you can substitute whatever in the branch. Here is an example that demonstrate how it works.
THÉVENIN’S THEOREM:
Any two-terminal dc network can be replaced by an equivalent circuit consisting solely of a voltage source and a series resistor.
You have asked an important question indeed. Thévenin’s equivalent circuit has a series resistor but in the second circuit diagram I have only used a source and in the third I have used both (more possible). Both are accordance with substitution theorem.It means one can replace a branch with any combination of elements which is not true for Thévenin’s theorem.For the marked branch in the main circuit if you use Thévenin’s theorem you will get Vth or Eth = 0V and Rth =3 Ohm. This is because Thévenin’s theorem doesn’t care about rest of the network or the load resistance but substitution theorem does. Without the whole circuit substitution theorem is not applicable but in Thévenin equivalent circuit the load resistance may vary.
Best Answer
You are right, if we assume that the input is driven by an ideal voltage source and that we have ideal switches then the circuit can not be analyzed using our standard definitions of circuit elements. Connecting two voltage sources with different voltages in parallel results in an invalid circuit because it violates our definition of what "parallel" means.
At the instant after the switch closes the voltage on the capacitor must be exactly what it was at the instant before the switch closed. An instantaneous change in voltage across a capacitor would require an infinite current, which once again makes the circuit invalid and impossible to analyze.
Power is most certainly consumed in such a circuit, but it is consumed in the resistances of the real-world switches and real-world voltage sources rather than in the capacitors.