Lightning strike is categorized as DC because of its electrostatic nature. The negative flash/stroke is found to have 200KA of current with a voltage close to 300KV (cloud to ground). But why is it referred to as a High frequency source, since it does not have a frequency component? In an electric circuit diagram will lightning be represented as a voltage source or current source?
[Physics] Why is Lightning referred to as High frequency source
electric-currentelectrostaticslightningvoltage
Related Solutions
(1) To address your first question: you have to treat the cloud and the earth below it as forming as a capacitor. There's a good popular description of this at http://micro.magnet.fsu.edu/electromag/java/lightning/. A capacitor is described by it's capacitance, and this is related to the voltage and charge by:
$$C = \frac{Q}{V}$$
where $Q$ is the electric charge and $V$ is the voltage difference across the capacitor. You can approximate the cloud and earth as a parallel plate capacitor, and the capacitance is given by:
$$C = \frac{\epsilon A}{d}$$
where $A$ is the area of the cloud base, $d$ is the spacing between the cloud base and the earth, and $\epsilon$ is the permittivity of air ($8.854 \times 10^{-12}C^2N^{-1}m^{-2}$). Combining the two equations and a quick rearrangement gives:
$$V = \frac{Qd}{\epsilon A}$$
This is obviously a gross simplification, but should give you a rough idea of the potential difference.
(2) As to your second question: as you say, positive lightning requires a higher voltage to get it started. Looking at the equation for the voltage, assuming the cloud stays the same the only way the voltage can be higher is if the charge is higher. Current is defined as charge per unit time, and if the duration of the lightening strike is roughly constant a positive lightning bolt has to transfer more charge in the same time and therefore has a higher current.
Draw the circuit using ideal circuit elements:
Now, the series current is:
$$I = \dfrac{\mathcal{E}}{R_{internal}+ R_{load}}$$
The voltage across the internal resistance is:
$$V = \mathcal{E} \dfrac{R_{internal}}{R_{internal}+ R_{load}} $$
The power dissipated by the internal resistance can by found three equivalent ways:
$$P = VI = \dfrac{V^2}{R_{internal}} = I^2 R_{internal} = \mathcal{E}^2 \dfrac{R_{internal}}{(R_{internal}+ R_{load})^2}$$
Clearly, setting $R_{load} = 0$ yields:
$$I = \dfrac{\mathcal{E}}{R_{internal} + 0} = \dfrac{\mathcal{E}}{R_{internal}}$$
$$V = \mathcal{E} \dfrac{R_{internal}}{R_{internal + 0}} = \mathcal{E}$$
$$P = \dfrac{\mathcal{E}^2}{R_{internal}}$$
Simply put, the entire emf appears across the internal resistance. Zero volts appears across the source plus internal resistance due to the short circuit.
Best Answer
Of course lightning "has a frequency component". If it didn't have non-zero frequencies, it could never start or would last indefinitely.
Lightning is a huge current pulse over a short time, usually several pulses over a few milliseconds. But more importantly, the current is started and stopped very abruptly, which by necessity means it has a broad spectrum containing significant high frequency content.
Spark gaps in general are infamous for being broad band frequency sources well into the radio range. Some early radio transmitters harnesses this by connecting a spark gap to a resonant circuit. Since the spark gap produced a wide spectrum, it also produced the frequency of the resonance. However, the other frequencies aren't sufficiently attenuated, so spark gap transmitters are specifically banned in the US (probably most countries).
I think your question really comes down to a unawareness of Fourier analisys. That would be a good search term to look up additional information.