The correct answer is: A. less than 1 Ohm.
The capacitive reactance of a capacitor is given by the formula:
$X_C = \frac{1}{2\pi fC}$
where
$f$ is the frequency in Hz and $C$ is the capacitance in farads.In this case, $f = 1000$ Hz and $C = 0.1 \mu F = 10^{-6}$ F. Substituting these values into the formula, we get:
$X_C = \frac{1}{2\pi (1000 \text{ Hz})(10^{-6} \text{ F})} = 1590 \Omega$
However, this is just the theoretical value. In practice, the capacitive reactance of a capacitor will be less than this value due to the presence of parasitic resistance. The parasitic resistance is due to the resistance of the wires used to connect the capacitor to the circuit, as well as the resistance of the dielectric material used to make the capacitor.
The parasitic resistance will cause the actual capacitive reactance to be lower than the theoretical value. The amount by which the actual capacitive reactance is lower than the theoretical value
will depend on the specific capacitor and the circuit in which it is used.In general, the parasitic resistance will be more significant for capacitors with a higher capacitance and a lower frequency. This is because the parasitic resistance is proportional to the capacitance and inversely proportional to the frequency.
For example, a 100 nF capacitor operating at 100 Hz will have a parasitic resistance that is about 10 times greater than the parasitic resistance of a 10 nF capacitor operating at 1000 Hz.
Therefore, the correct answer to the question is: A. less than 1 Ohm.