If an ammeter read 2A, another ammeter would probably also read 2A.
Keep in mind that every measurement involves error, and the equipment has a stated accuracy. If, for instance, the ammeter used had a 1% tolerance, than 2A actually means anywhere between 1.98A and 2.02A.
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.
Put a known small resistance (maybe 1 ohm) in parallel to it. Then use the equation I=V/R (R being the known resistance, and V being the voltage readout) So when using a 1 ohm resistor, if the voltmeter reads 2.35V, it means I=2.35/1 = 2.35 A
2000ma is equal to 2 amps. Set you meter to a current range ABOVE 2 amps.
i=v/r can be used to it
Inductance = Magnetic Flux/Current = [ML2T-2A-1]/[A] = [ML2T-2A-2] So, Dimensional Formula of Inductance = [ML2T-2A-2]
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.
The multiplicative inverse of -2a is 1/(-2a) = -1/2a
-2a + 5 = 1 -2a = -5 + 1 -2a = -4 a = -4/-2 a = 2
1/(2a)
If I've read your question correctly, you need to subtract: a2 +2a -7 a2 -4a2 +5a2 -6 = 2a2 -6 Note, if x - y = z, then y = x - z; so: 2a2 -6 - (a2 -2a +1) = 2a2 -6 - a2 +2a -1 = a2 +2a -7
3a-2a = 1
(2a + 1)(2a - 5)
2a + 1 = 3subtract 1 from both sides2a = 2divide both sides by 2a = 1
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.
2a-3ab = -1
It is 2a - 16.
(2a+1)(a+1)