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.
The wire gauge you need is a function of several things.
- The acceptable voltage drop or power loss (that appears to be the
only thing considered in the website you linked). The voltage drop (and power loss) is proportional to wire length and inversely
proportional to the cross-sectional area of the wire- in other words
inversely proportional to the square of the wire diameter (assuming constant current).
- The acceptable temperature rise. This is a function of the number
of current-carrying wires bundled together, the environment (maximum ambient temperature
and air pressure or altitude, for example), the insulation type, the
wire type (some types of wire are plated to withstand higher
temperatures than bare copper without corroding).
- Regulatory requirements and other considerations- for example, the
wire may be rated for 200°C insulation, but you might not want the
wire to run that hot.
- Fusing- the fuse or circuit breaker should protect the wire in the
case of faults such as overload or short circuit.
Very short lengths of wire can depend on heat sinking through the ends (indeed, in a vacuum, that may be the main heat loss mechanism), but usually that's not taken into account.
Normally you'd run through a checklist such as the above to make sure ALL the requirements are satisfied simultaneously, so you might find that using PTFE insulated wire allows you to use AWG 18 wire, but because of the voltage drop limitation you'll have to use AWG 12 wire.
Best Answer
I'll assume that the requirement is really 1 mV/cm of voltage drop along the wire. Here's one way to get there:
1 mV/cm × 100 cm = 100 mV total drop required.
100 mV / 3 Ω = 33.3 mA of current needs to flow.
2 V cell / 33.3 mA = 60 Ω total resistance is required.
60 Ω total - 3 Ω wire = 57 Ω additional required.