i=v/r can be used to it
decrease to half of its original value
The Thevenin equivalent circuit of this battery is 1.5V and 0.6 ohms in series. A more exact answer cannot be given without knowing the actual resistance of the 2 meters (I assumed infinite for the voltmeter and zero for the ammeter, as would be for ideal meters).However I would NEVER attempt this test as you describe it, many types of batteries will explode like bombs when shorted (as they would be when an ammeter was placed across them)! The correct way to do this test safely is with just a voltmeter and an adjustable high wattage resistor.
If you're connecting it properly, then I would have to guess that the multimeter is defective.
Assuming the new lamp is in series, the ammeter reading falls because the total resistance has increased. By how much depends on how the lamp resistance depends on voltage. If the lamp is added in parallel to the first, then the ammeter reading doubles.
To measure the current in a DC circuit an ammeter may be used. This ammeter may consist of a sensitive meter with a shunt in parallel with it to divert part of the current. In case even more current is expected than the full scale reading of the meter an additional shunt may be connected in parallel with that arrangement.
If the resistor is removed from the circuit, the total resistance in the circuit decreases. This causes the total current in the circuit to increase, which would result in an increase in the ammeter reading.
The ammeter is reading zero because there is no current flowing. This is because one of the resistors is faulty; the faulty resistor has an "open circuit" (open circuit means there is a broken connection). We know that: Ohms law is: V = I x R (voltage = current x resistance) Therefore because there is zero current in each resistor there will be zero voltage across each resistor. However we also know that: Kirchhoff's voltage law is: V1 +V2 +V3 + … = Vs (the sum of the voltage drops accross each component in a circuit MUST equal the supply (or battery) voltage). But if all the resistors are zero volts, then what component equals the supply (or battery) voltage? The battery voltage is developed across the open circuit… therefore the resistor which is faulty will have a voltage across it equal to the battery voltage. That easy to measure with a volt meter! hope this helps
it explodes and burns everyone in the room.
the bulb will glow and ammeter will show the reading
An ammeter reads the current that is flowing through a branch of a circuit. If there is a break within that same branch of the circuit, current will not be able to flow through that branch of the circuit as it forms an incomplete loop, so the ammeter will read 0 A of current. If there is a break in a circuit in a branch that is not connected to the ammeter however, the ammeter will give a higher reading of the current. This is assuming that the break in the other branch does not short out the branch with the ammeter attached, and that the circuit can still form a complete loop without that branch.
An open switch in a circuit will stop all current flow so the ammeter should read zero amps.
The voltage remains the same across the circuit as it is a parallel connection. So, the current across the upper half of the circuit where the ammeter is connected is calculated as I = V/R = 12.04 (total voltage)/12 (Resistance R1) = 1 A. Hence, the ammeter will read 1 A.
it shows some reading even though battery is low because of there will some flux remains in the magnets
An ammeter is the instrument used to measure current in an electric circuit. It is connected in series in the circuit and provides a reading of the amount of electric current flowing through it.
Connect ammeter in series and voltmeter in parallel to the circuit
decrease to half of its original value
The strength of an electric current is measured in amperes (A) using an ammeter. An ammeter is a device that is connected in series in a circuit to measure the flow of current. The higher the current flowing through a circuit, the higher the amperage reading on the ammeter.