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farkasditke73 Strictly speaking the emitted light is not laser so there's no need for a laser marking. An internal laser source is used to excite the phosphor: LEP, Laser Excited Phosphor. If it were laser the light wouldn't be white.

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farkasditke73 Impossible. It would require two hundred and fifty amperes on twelve volts. The resistance of the heating element should be forty-eight milliohms.

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farkasditke73 Willsgood gave the perfect answer. I may add the following: The AC voltage generated by the stray electric fields has a very high impedance, and so has the voltage (or frequency, for that matter) input of a DMM. As soon as you connect the lest leads to a voltage source, its lower impedance practically shunts the voltage generated by the electric fields so they won't affect your measurement.

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farkasditke73 For an astronomical telescope the diameter of the objective is equally important. The description says 90mm which is not bad for this price.

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Q: Can I connect both outputs to a single speaker (bridge) ?

Asked by furukawa on 2023-12-05 05:08:49

farkasditke73 Not directly as the two outputs are in phase. However, with a bit of DIY it may be possible. All you need to do is invert the signal sent to either input. Take a twin op-amp IC, set their amplification to 1 (connect the output through a small resistor to the inverting input (-), then feed one op-amp through the inverting input (+), the other through the non-inverting input (-). Short the amplifier's electrolytic capacitors at the output if there are any. Make sure that both outputs are on the same potential as they are now DC coupled to the loudspeakers. Alternatively, you can replace them with bipolar electrolytic caps but these are not cheap. A good result can't be guaranteed as at higher frequencies the phase shift of the two channels may not be exactly identical. It's worth a try but remember that double the output voltage means four times the output power.

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Q: solar will it mesure power from solar cells back to the grid ?

Asked by jbozelie on 2020-11-29 17:24:36

farkasditke73 It will measure the current flowing through the meter BUT it can't detect the "direction" of the energy flow, i.e. whether you are drawing current from, or feeding current back into, the grid. This means that energy fed back into the grid will be registered as consumption.

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