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A Resistor [note spelling] when jumpered-out has a theoretical resistance of zero. The voltage across same is zero.

To "jumper-out" is presuming a wire across the resistor: Theoretically this "ideal" wire has zero resistance, however, in practice a "thin" copper wire of (say) 10mm length - e.g., about the length of a 1/8th-watt resistor - would have a resistance around the milliohm.

Depending on the current passing through a wire (per V=IR) the voltage might be nothing or a mega-volt, as a function of the current and cross-section of the wire.

However, anything over, say, a small number of amps through a real-world "thin" wire would cause the wire to act as a fuse and break, so, in the case of a thin wire (0.25 mm say) we're looking at near zero volts across the resistor/wire combination at say 0.01 Amps and tending the supplied voltage at, say, 10 amps.

If, on the other hand, the resistor is no longer in the circuit (literally "jumpered out") then the theoretical resistance of the "gap" where the resistor should be is infinity (real-world: The order of megaohms). The makes the voltage across same just about the voltage supplied to either side of said resistor as if it were not there, which will depend on the rest of the circuit.

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Q: When a Resister is jumpered out what is the voltage across the resistor?
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If the resistance in the circuit is increased what will happen to the current and voltage?

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