The current decreases due to I=V/R. The ammeter reading will decrease as R is increased.
i=v/r can be used to it
Micro ammeter has higher resistance. use ohm's law.. Say the voltage applied is 1 V. then for the ammeter.. I=1A, V=1V R=1 ohm for the micro ammeter.. I=0.000001, V=1V R=1000000 ohms
You'll get a reading on the ammeter showing 15 mA
decrease to half of its original value
the bulb will glow and ammeter will show the reading
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.
An ammeter or a voltmeter gives the reading of the flow of current, which is due to the movements of ions. When a rusted nail is immersed in the solution it will add the no of ions into the solution. This will increase the ion concentration and therefore will change the reading of a voltmeter and ammeter.
In a series circuit, the current is the same throughout all components. Therefore, both ammeter 1 and ammeter 2 will read the same current, as they are measuring the same flow of electrons in the circuit. The readings on both ammeters will be equal, so neither will have a bigger reading. The voltage drop across the lamps may differ, but that does not affect the ammeter readings.
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.
An open switch in a circuit will stop all current flow so the ammeter should read zero amps.
The decrease of light intensity on an LDR will cause the resistance of the LDR to increase, which will result in a decrease in current flow through the circuit. As a result, the ammeter reading will decrease.