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Most antennas have a standing wave along there length and are therefore effective impedance transformers. The feed impedance of a dipole might be of the order of 73Ω, at its ends, the impedance will be at least 2kΩ, if not higher.

Solving for voltage at the antenna ends, we will have:

$P=\frac{V^2}{R}\Rightarrow V_{rms}=\sqrt{P\cdot R}$

Peak voltage is indeed $\sqrt{2}\cdot V_{rms}$, but the voltages at the ends of a dipole antenna are balanced with respect to ground so: $V_{peak}=\frac{\sqrt{2}}{2}V_{rms}$, only half the value.

Assuming a transmitter power of 1kW: $V_{peak}=\frac{\sqrt{2}}{2}\sqrt{P\cdot R}=\sqrt{\frac{P\cdot R}{2}}=\sqrt{\frac{10^3 \cdot \not 2 \cdot 10^3}{\not 2}}=10^\frac{6}{2}=10^3=1kV$

If the antenna is loaded with a coil, the impedance will be transformed up to an even higher value. This is how a Tesla coil works. The resulting corona effect might be quite dramatic as shown in the picture below (1kW on 80m in short W4JRW dual-band dipole @ HB9DWU).

Most antennas have a standing wave along there length and are therefore effective impedance transformers. The feed impedance of a dipole might be of the order of 73Ω, at its ends, the impedance will be at least 2kΩ, if not higher.

Solving for voltage at the antenna ends, we will have:

$P=\frac{V^2}{R}\Rightarrow V_{rms}=\sqrt{P\cdot R}$

Peak voltage is indeed $\sqrt{2}\cdot V_{rms}$, but the voltages at the ends of a dipole antenna are balanced with respect to ground so: $V_{peak}=\frac{\sqrt{2}}{2}V_{rms}$, only half that value.

Assuming a transmitter power of 1kW: $V_{peak}=\frac{\sqrt{2}}{2}\sqrt{P\cdot R}=\sqrt{\frac{P\cdot R}{2}}=\sqrt{\frac{10^3 \cdot \not 2 \cdot 10^3}{\not 2}}=10^\frac{6}{2}=10^3=1kV$

If the antenna is loaded with a coil, the impedance will be transformed up to an even higher value. This is how a Tesla coil works. The resulting corona effect might be quite dramatic as shown in the picture below (1kW on 80m in short W4JRW dual-band dipole @ HB9DWU).

Most antennas have a standing wave along there length and are therefore effective impedance transformers. The feed impedance of a dipole might be of the order of 73Ω, at its ends, the impedance will be at least 2kΩ, if not higher.

and solving for voltage at the antenna ends, we will have:

$P=\frac{V^2}{R}\Rightarrow V_{rms}=\sqrt{P\cdot R}$

Peak voltage is indeed $\sqrt{2}\cdot V_{rms}$, but the voltages at the ends of a dipole antenna are balanced with respect to ground so: $V_{peak}=\frac{\sqrt{2}}{2}V_{rms}$.

Assuming a transmitter power of 1kW: $V_{peak}=\frac{\sqrt{2}}{2}\sqrt{P\cdot R}=\sqrt{\frac{P\cdot R}{2}}=\sqrt{\frac{10^3 \cdot \not 2 \cdot 10^3}{\not 2}}=10^\frac{6}{2}=10^3=1kV$

If the antenna is loaded with a coil, the impedance will be transformed up to an even higher value. This is how a Tesla coil works. The end effect might be quite dramatic as shown in the picture below.

Most antennas have a standing wave along there length and are therefore effective impedance transformers. The feed impedance of a dipole might be of the order of 73Ω, at its ends, the impedance will be at least 2kΩ, if not higher.

Solving for voltage at the antenna ends, we will have:

$P=\frac{V^2}{R}\Rightarrow V_{rms}=\sqrt{P\cdot R}$

Peak voltage is indeed $\sqrt{2}\cdot V_{rms}$, but the voltages at the ends of a dipole antenna are balanced with respect to ground so: $V_{peak}=\frac{\sqrt{2}}{2}V_{rms}$, only half the value.

Assuming a transmitter power of 1kW: $V_{peak}=\frac{\sqrt{2}}{2}\sqrt{P\cdot R}=\sqrt{\frac{P\cdot R}{2}}=\sqrt{\frac{10^3 \cdot \not 2 \cdot 10^3}{\not 2}}=10^\frac{6}{2}=10^3=1kV$

If the antenna is loaded with a coil, the impedance will be transformed up to an even higher value. This is how a Tesla coil works. The resulting corona effect might be quite dramatic as shown in the picture below (1kW on 80m in short W4JRW dual-band dipole @ HB9DWU).

Most antennas have a standing wave along there length and are therefore effective impedance transformers. The feed impedance of a dipole might be of the order of 73Ω, at its ends, the impedance will be at least 2kΩ, if not higher.

and solving for voltage at the antenna ends, we will have:

$P=\frac{V^2}{R}\Rightarrow V_{rms}=\sqrt{P\cdot R}$

Peak voltage is indeed $\sqrt{2}\cdot V_{rms}$, but the voltages at the ends of a dipole antenna are balanced with respect to ground so: $V_{peak}=\frac{\sqrt{2}}{2}V_{rms}$.

Assuming a transmitter power of 1kW: $V_{peak}=\frac{\sqrt{2}}{2}\sqrt{P\cdot R}=\sqrt{\frac{P\cdot R}{2}}=\sqrt{\frac{10^3 \cdot \not 2 \cdot 10^3}{\not 2}}=10^\frac{6}{2}=10^3=1kV$

Most antennas have a standing wave along there length and are therefore effective impedance transformers. The feed impedance of a dipole might be of the order of 73Ω, at its ends, the impedance will be at least 2kΩ, if not higher.

and solving for voltage at the antenna ends, we will have:

$P=\frac{V^2}{R}\Rightarrow V_{rms}=\sqrt{P\cdot R}$

Peak voltage is indeed $\sqrt{2}\cdot V_{rms}$, but the voltages at the ends of a dipole antenna are balanced with respect to ground so: $V_{peak}=\frac{\sqrt{2}}{2}V_{rms}$.

Assuming a transmitter power of 1kW: $V_{peak}=\frac{\sqrt{2}}{2}\sqrt{P\cdot R}=\sqrt{\frac{P\cdot R}{2}}=\sqrt{\frac{10^3 \cdot \not 2 \cdot 10^3}{\not 2}}=10^\frac{6}{2}=10^3=1kV$

If the antenna is loaded with a coil, the impedance will be transformed up to an even higher value. This is how a Tesla coil works. The end effect might be quite dramatic as shown in the picture below.