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What happen to voltage and current when rheostat is adjusted from maximum to minimum resistance using both ammeter and voltmeter?
When a rheostat is adjusted from maximum to minimum resistance, the overall resistance in the circuit decreases. As a result, the current flowing through the circuit increases, which can be observed on the ammeter as a higher reading. Conversely, the voltage across the rheostat will decrease, as the voltage drop across a lower resistance is less, which can be monitored using the voltmeter.
What happen when ammeter connected in parallel?
an ideal ammeter has zero or negligible resistance when this is connected in series no effective resistance would be added in the circuit so that the value of curret that we get is exactly of the circuit only. but when the ammeter is connected in parllel as it has zero resistance , the resistor to which it is connected in parllel gets shorted and due to his the effective resistance of the circuit is changed and so the effective current ... due to this the w=value measured by the ammeter would be different (incresed due to dec. in effective resistance)
What happen when a ammeter is placed in parallel with the circuit?
If an ammeter is placed in parallel with a a load on a circuit, the circuit can short out as the ammeter takes the place of the load, flowing freely through the meter. Never measure across a resister or other electronic load with an ammeter. Remember, ammeters are used in series while voltmeters are used in parallel.
What will happen to the reading on the ammeter if one bulb blows out?
If one bulb in a series circuit blows out, the circuit becomes open, and the current stops flowing. As a result, the reading on the ammeter will drop to zero since there is no current passing through the circuit. In a parallel circuit, if one bulb blows out, the current may decrease slightly due to the change in total resistance, but the ammeter will still show a reading corresponding to the remaining bulbs in operation.
What happen if you change the position of ammeter in the circuit in parallel and series?
-- In a series circuit, no matter where you install the ammeter, it will always read the same current. -- In a parallel circuit, the ammeter may read a different current when it's moved to a different parallel branch.
What will happen to the ammeter reading if the resistance is increased?
The current decreases due to I=V/R. The ammeter reading will decrease as R is increased.
What would happen to circuit if there was no resistance?
a circuit with no resistance or zero resistance can be considered as open circuit in which the current is zero. without resistance the circuit just becomes open ()
What happen to the circuit if there is higher resistance?
In what sense.
When an AC 5A and dc Current of 5A applied simultaneously to the circuit then what it happen?
A DC ammeter will read zero
What will happen if a voltmeter has a low shunt resistance instead of having high shunt resistance?
The purpose of a voltmeter is to indicate the potential difference between two points in a circuit.When a voltmeter is connected across a circuit, it shunts the circuit. If the voltmeter has a low resistance,it will draw a substantial amount of current. This action lowers the effective resistance of the circuit andchanges the voltage reading.
What will happen in a circuit if the voltage does not change but the resistance in the circuit increases?
If the resistance increases, while the voltage stays the same, current will decrease. Current = voltage divided by resistance
What will happen if the voltmeter was inserted in place of an ammeter?
You would load the circuit, and it is likely it would not operate correctly. A volt meter is designed to have a very high resistance between the two probes; an ammeter is designed to have a very low resistance. For instance, say you have a 120 watt light bulb that runs on 120 volts (you would then draw ~1 amp of current). If you tried to measure this with a meter that has .1 ohm resistance on ammeter setting, and 1,000,000 ohms on volt meter: Error due to loading: ammeter: .1 / (120 + .1) = .08%; Current will be .999Amps, power to the light bulb will be 119.9 watts Volt meter: 1,000,000/ (120 + 1,000,000) = 99.9%; current will be 120micro Amps, power to the light bulb will be 14.4 milliwatts (the light bulb will not appear to be on).
