7
The short answer:
\begin{equation}
\frac{V_{p-p}}{V_{rms}} = 2\sqrt{2}
\end{equation}
The long answer, or how to derive the above:
As noted on the Wikipedia page for root mean square, the RMS of a sine wave is equal to its amplitude divided by the square root of two. (You can also derive this by doing the integral over a sine wave yourself.)
\begin{...
4
A transmission line is linear, which means we can consider the voltage at any point to be the superposition of two other voltages.
Asking what they are "physically" may or may not be a useful question.
I could for example define:
$$ V(z) = Ve^{jkz} + V_{DS}(t) $$
where $V_{DS}(t)$ is a function corresponding to the normalized amplitude of Pink Floyd's ...
4
Living down-under and apparently also off-the-wall, hyper-enthusiast David L. Jones made following excellent digital multimeter (DMM) reviews:
EEVblog #91 – \$50 Multimeter Shootout: Extech EX330, Amprobe AM220, Elenco, Vichy VC99 & GS Pro-50,
EEVblog #99 – \$100 Multimeter Shootout: Extech, Amprobe, BK Precision, Ideal, UEi & Uni-T.
David actually ...
3
I wouldn't risk directly connecting a 9.6v battery to a radio that expects 5-6v.
The simplest solution would probably be the diodes mentioned in the comment to your question, but as the battery voltage dropped the voltage reaching the radio would drop as well, and as the battery got low the output voltage after the diodes might be too low to power the radio,...
3
800 volts is far more likely to burn you. Just doubling the voltage from 120 to 240 will quadruple the power heating your flesh.
This is just basic Ohm's Law: [P=E*I]
DC is one nasty customer:
It is easy to get complacent after spending a youth and a career working with docile, harmless 5-24 volts DC, or well-behaved 100 - 240 V AC voltages because of ...
3
The answer is "it depends".
It depends on the Antenna design, but lets look at an example:
A typical (license free / home / consumer) installation of WiFi is 100mW (not a 1000mW as your post indicates)
A typical antenna has a 50 Ohm impedance, and lets take an example:
For a 1/4 wave vertical antenna the voltage distribution has a voltage minimum at feed ...
2
I believe the Fluke 79 did not have inductance/capacitance measurement built in. This is an affordable feature on modern multimeters which is quite useful for the ham. For example, it can be used to measure inductance on self-wound toroids which are used in kit-built radios and for repairs when OEM parts are not available. Likewise, capacitors are often ...
2
The specification means $0.15\ \mu\text{V}$ at the input terminals of the receiver, so this voltage across a $50\ \Omega$ load.
Power in a resistor can be calculated with
$P={V^2}/{R}$
so $0.15\text{ }\mu\text{V}$ is a power of ${(0.15 \times 10^{-6})^2 } /{50} = 4.5 \times 10^{-16} \text{ W}$ (or ask google)
This is -123.5 dBm into the receiver.
For ...
2
The output impedance is not that important: the (lowest) allowed load impedance is important. And the assumption is right: from the output impedance it is possible to calculate the effect of load on the output voltage. But be aware of frequency deviation as a function of load variation, and also eventual stop of oscillation with serious load.
This is not a ...
1
First of all, a voltage of 800VDC can be very dangerous if you happen to be touching anything at ground potential. An 800V potential is REALLY eager to pass some current down to ground, compared to a 120V potential.
Also, because of Ohm's Law, if you were to touch something at a potential 6.66 higher (800V against 120V), you will get 6.66 times the current ...
1
You must also know the impedance of the receiver which is typically 50 ohms.
Convert the received signal from dBm to Watts.
Then the input voltage is sqrt ( P * R )
We usually work this the other way around:
Convert 0.15 uV into 50 Ohms to dBm which is -123.5 dBm.
If your signal is greather than that you are Ok.
Say your received signal is -100 dBm you have ...
1
This doesn't actually answer your question per se, but I think you have an important misconception: An antenna is not a resistor.
Of course you know it's not, but in particular, an antenna having a 50 Ω input impedance does not dissipate as much heat as a 50 Ω resistor would for the same input. The majority of the energy is radiated, not converted to heat — ...
1
For an ideal meter, sure, it's possible. However, most multimeters are not designed to work at RF. There are all sorts of problems: what effect does the input capacitance have on the measurement? What's the characteristic impedance of the leads? Most multimeters don't take these things into account because the only AC voltage they are designed to measure is ...
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