No, the total resistance increases.
it won't flow
YOU DO NOT "measure resistance on house current".You never measure resistance of anything that has any path to any source ofpower ... not to a wall outlet, a battery, a windmill, a solar panel, etc. You onlymeasure resistance when all power is REMOVED from the circuit or componentyou're measuring. Then, the range you choose for the ohmmeter depends onthe component or circuit you're measuring, NOT on how it's powered when it'sturned on.
I think it is due to the resistance to the electrical appliance, the resistance offered by the electrical appliance works as a load in the circuit and when load is removed it causes sparking.
explain why all the load must be removed when to conduct the insulation and resistance testing
A non-auto circuit breaker is a circuit breaker with the trip element removed. Basically, it is a modified circuit breaker that is now a disconnect switch (glorified disconnect switch).
if we remove a resistor from the parallel connection the effective resistance value will be increased.
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.
If you are referring to an electrical circuit, a series circuit is wired in such a way that if one object is removed from the circuit, the circuit is broken and everything within the circuit loses power. In a parallel circuit different components of the circuit can be removed without disabling power to the rest of the devices within the circuit.
In a series circuit, if one component is removed or defective, the circuit will be broken and no current will flow. In a parallel circuit, if one component is removed or defective, the current will simply bypass that component and continue to flow through the other branches.
In a parallel circuit, each load added subtracts from total resistance. When one or more loads is removed from a parallel circuit, the total resistance is increased, reducing the total amperage draw. The less resistance a load has, the more current can pass through. This is part of Ohm's law. The mathematical equation that describes Ohm's law is: I=V/R , where I is the current in amperes, V is the potential difference in volts,and R is a circuit parameter called the resistance For example : The humble light-bulb is rated by the watts it uses. The amount of watts used by a light-bulb is calculated using Ohm's law. With the resistance of the bulb's filament and the voltage the bulb is designed to operate with, one can derive the amperage the bulb will draw. The amperage is then multiplied by the voltage to show wattage. Using Ohm's law : With the resistance of a 40watt 120volt light-bulb, only 0.33amps is able to pass through the bulb's 363ohm filament at 120volts. A lamp that has two 40watt bulbs inplace, and the two bulbs are in parallel, the circuit will have a resistance of 179ohms and draw 0.67amps which is 80watts at 120volts.
when all the resistance removed from the circuit
assume the following configuration: battery connected to 2 parallel resistors with an ammeter in series with the battery... observe the current measurement ... remove one of the resistors .... observe the current again, it will have decreased: if the resistors were of equal value, the current will decrease to half of its original value when one of the resistors is removed. Mathematics: V=IR (V- voltage, I - current, R - resistance in a parallel circuit, R=(R1*R2)/(R1+R2) where R1 and R2 are the values of resistance of the resistors. Before removal- Ib=V*(R1+R2)/(R1*R2) After removal (assume R2 is removed)- Ia=V/R1 so Ia/Ib=(R1*R2)/(R1*(R1+R2)) or Ia=Ib*(R2/(R1+R2) if R1=R2 then Ia=Ib*R2/(2*R2) or Ia=Ib/2 so the current after is 1/2 of that before.
To solve any D.C. circuit by using Thevenin Theorem,First of all load resistance RL is disconnected from the circuit and open circuit voltage across the circuit is calculated (known as Thevenin equivalent voltage)Secondly, the battery is removed by leaving behind its internal resistance. Now we calculate equivqlent resistance of the circuit ( called Thevenin equivalent resistance).Now we connect Thevenin Voltage in series with Equivalent resistance of the circuit and now connect load resistance across this circuit to calculate current flowing through the load resistance.Whereas in the case of using Norton theorem, we again remove the load resistance if any, and then short circuit these open terminals and calculate short circuit current Isc.Second step is same as in Thevenin theorem i.e. remove all sources of emf by replacing their internal resistances and calculate equivqalent resistance of the circuit.Lastly, join short circuit current source in parallel with equivalent resistance of the circuit. Now, we can calculate votage across the resistance which was connected in parallel with Isc.So, by knowing the open circuit voltage, we can calculate current flowing the resistance and on the other hand , by knowing the short curcuit current , we can calculate voltage across the resistance.
it can exchange the current is much larger and has Avery low resistance when it is turned on.the mosfet is not removed from the circuit when the supply is on because it flow the high current.
The bulb you remove will go out :) Overall current will also be reduced proportional to the resistance of the bulb being removed. Lets say you have two 60 W incandescent bulbs in parallel and they each are drawing 1/2 Amp (60W = 120 Volts x 1/2 Amp). The resistance of each bulb is 240 Ohms (120 Volts / .5 Amps). The parallel resistance is 120 Ohms so 1 Amp is being drawn. When one of the two bulbs is removed the resistance changes from 120 Ohms to 240 Ohms, reducing the current from 1 Amp to 1/2 Amp.
It is a series circuit. In a series circuit, the components are connected in a single path so if one lamp is removed, the circuit is broken and the other lamp will not receive electricity.
By connecting it to a circuit with the power still applied. Ohm meters work best when the component under test is removed from the circuit to eliminate parallel paths of current through other components.