it explodes and burns everyone in the room.
SMOKE!!! Yiii-haaaa! An ammeter is always as low a resistance as possible. That way, the current that you measure in a circuit using the meter will be the same as if the meter were not there. So putting an ammeter in parallel with a circuit means that you just short circuited the circuit you were intending to measure. Poof! There goes your ammeter!An ammeter has a very low internal resistance. So, if it is connected in parallel with a load, it will short-out that load. The resulting high current flowing through the ammeter may severely damage the meter (and possible harm the user), although most are fitted with fuses to protect them.
"Ohms Law" defines resistance (R) as the the ratio of voltage (V) to current (I).R = V/IIf you move those variables around, you can get the formula:I = V/RSo you can see that when resistance increases, current flow will decrease.CommentResistance is most definitely not defined as 'the ratio of voltage to current', although that ratio may tell you what it happens to be.Resistance isn't a variable in the Ohm's Law equation. It is a constant because it is unaffected by either current or resistance.
Ohm's law states that "The current is directly proportional to the applied EMF (voltage) and inversely proportional to the resistance in the circuit." <<>> if resistor exists, resistance decreases according to ohm's law, current is directly proportional to voltage and current is inversely proportional to resistance it means as current increases, voltage increases. resistance increases, current decreases so as voltage if there is no resistor, there should be no resistance except internal resistance of voltmeter and ammeter
depends on the simple circuit. please describe it.
The remaining bulb will be brighter than it was when both bulbs were working - due to the increased voltage.
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.
It depends on the configuration of the ammeter. If you have an ammeter with red on one side, green on the other, and a balance point in the center, the needle will go into the red when when load is increased in relation to the charge. If the load is decreased in relation to the charge, the needle will go into the green. If you have an ammeter with a numerical value (i.e. 30, 40, 50, etc.) then the needle will go up to the value of the load present in the system not in relation to the charge.
If the resistance of the variable resistor increases, the current flowing through the circuit decreases. As a result, the bulb will emit less light or may not light up at all, depending on the magnitude of the resistance increase.
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.
the bulb will glow and ammeter will show the reading
The dependent variable.
When a common ohmic resistor is heated, its resistance typically increases.
Ohms law is V=I X R. If resistance (R) is reduced and current (I) is constant, then voltage (V) must decrease. You can see from the equation that they are proportional to one another. If, however, R is reduced and V is held constant, then I must increase (I and R are inversely proportional). The only way V can increase is if either or both I and R increase.
When the voltage is increased across a metal film resistor, the current flow will also increase. Ohm's law states that the current flowing through a resistor is directly proportional to the voltage across that resistor. I = V/R Let us assume an initial voltage drop across a 4.99K ohm metal film resistor is 5V. The current flow through the resistor is calculated to be: I = 5/4990 = 0.001 Amps or 1 mA If that voltage were to say double to 10V: I = 10/4990 = 0.002 Amps or 2 mA Using these values it is also possible to calculate the power dissipated by the resistor. P = I*V = 0.002 * 10 = 0.02 Watts This power calculation determines the minimum physical case size needed for the resistor to function within these conditions. Anything smaller, the resistor will fail.
An independent variable is when you do not control what happens In an experiment,however; a dependent variable is when you actually control the experiment,
An independent variable is when you do not control what happens In an experiment,however; a dependent variable is when you actually control the experiment,
what happens if we remove Rc in RC coupled amplifier