I have a new 240VAC isolation transformer connected to mains. With the transformer switched off I measure 240AC from either output terminal to mains. Is this a fault with the isolation tranformer or is this to be expected?
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1\$\begingroup\$ "Phantom voltage'. It's a real voltage, but it VERY VERY weak. Since your meter is nearly infinite input impedance (probably 10Meg or so), it doesn't load down this phantom voltage. Put like 10k resistor across the terminals, I bet the voltage drops to zero. appliantology.org/blogs/entry/1058-what-is-ghost-voltage \$\endgroup\$– Kyle BNov 25, 2022 at 2:19
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\$\begingroup\$ thank you Kyle B, right on the money, when loaded with a 5k resistor the voltage drops to a few millivolts. Thought there must be some explanation for it. Thanks again... \$\endgroup\$– PeterNov 25, 2022 at 5:46
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2 Answers
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"Phantom voltage'.
It's a real voltage, but it's VERY, VERY weak. Since your meter has nearly infinite input impedance (probably 10 MΩ or so), it doesn't load down this phantom voltage. Put like a 10 kΩ resistor across the terminals, I bet the voltage drops to zero.
https://appliantology.org/blogs/entry/1058-what-is-ghost-voltage
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What you're doing is measuring the "open-circuit" voltage of a circuit with a high output resistance. When the isolation transformer is switched to its off state, it's equivalent output resistance goes high (open circuits are technically very high resistances). Since the input resistance of the multimeter is high and doesn't load your circuit, there is little to no current draw. Therefore the voltage you'll see is the 240V source from the original 240V mains across its transformer coupling with no voltage drop from the output resistance, because current needs to flow in order for a voltage drop to occur.
This voltage source + output resistance model equivalence is known as a Thevenin equivalent circuit, which basically indicates how good a circuit is as a voltage source. (A circuit with a low output resistance is a good voltage source, as there is low voltage drop across the output, and the load uses most of this voltage due to the principle of voltage division for series connected resistances).
This can be demonstrated in simulation. The multimeter's been given an input resistance of 1GOhm.
It's the same reason why "dead" batteries seem to have a voltage across them, but once you add a load, it will drop to near zero. Measuring the open-circuit voltage is only half the information you require when looking to evaluate whether its sourcing any voltage or current. To gather the other half, you'll need to draw current by adding a load resistance and see the voltage drop across the load. Alternatively, you could measure its short circuit current (not recommended in practice unless you have some idea that you know the output resistance is high).
