Short answer: yes.
Most modern multimeters will not be damaged by external power when measuring resistance. But they will give erroneous readings. It is best to remove the power and disconnect the measured resistance from the larger circuit.
A multimeter determines resistance by applying a small voltage, and measuring the resulting current. If the resistor has an external voltage source, then it will interfere with the measurement. Furthermore, if the resistance is connected to a larger circuit, then the resistance of this larger circuit will also be involved.
Power is inversely proportional to resistance. Ohm's law: Current is voltage divided by resistance Power law: Power is voltage times current, therefore power is voltage squared divided by resistance.
All resistances will emit heat energy when a current flows. The heat production rate (or power) can be found by any of these formulas: Power = Current * Voltage Power = Current2 * Resistance Power = Voltage2 / Resistance. Power is given in Watts when Current is in Amps, Voltage in Volts, and Resistance in Ohms.
when source resistance and load resistance are equal maximum power is transfered
First, this statement stands as long as voltage is constant. If you held the current constant then power would increase as resistance increases.V=IR. For a fixed voltage if you increase the resistance (R) then the current (I) will decrease - following the formula.Power = VI so as the resistance increases the value of VI (power) decreases as V is constant and I gets smaller.Therefore the power is decreasing as the resistance increases (when voltage is held constant).Hope this helps.
A lamp has two resistances: a 'hot' resistance (its operating resistance) and its 'cold' resistance (its resistance when switched off), and the hot resistance is significantly higher than its cold resistance.You can calculate its 'hot' resistance from its rated power and its rated voltage (assuming that it is being supplied at its rated voltage), by manipulating the following equation, to make Rthe subject: P= V2/RYou will, though, have to measure its cold resistance.
A: Checking continuity is the same as checking resistance, therefore yes someone can determine the resistance at that particular current level
an ohmmeter contains its own power source that it uses to test the circuit. if power is still on in the circuit, at the least the meter will give incorrect readings, at the worst the meter may be damaged rendering it useless.
Continuity is checking for a completed circuit including a short circuit. Checking resistance would be checking in ohms resistance of a circuit, motor windings or an open circuit.
power=i square*resistance or power=v suare/resistance
On Circuit Resistance = Close to the total load Resistance. Off Circuit Resistance = Near Infinitive High Resistance.
Since power = voltage2/resistance, reducing the resistance will increase the power of the circuit. Incidentally, power is not 'consumed'; it's energy that's consumed.
Power=current squared times resistance
Power is inversely proportional to resistance. Ohm's law: Current is voltage divided by resistance Power law: Power is voltage times current, therefore power is voltage squared divided by resistance.
All resistances will emit heat energy when a current flows. The heat production rate (or power) can be found by any of these formulas: Power = Current * Voltage Power = Current2 * Resistance Power = Voltage2 / Resistance. Power is given in Watts when Current is in Amps, Voltage in Volts, and Resistance in Ohms.
yes, batteries have high internal resistance. The higher the resistance the lower power you get out of the batter. Therefore if you no power you have very high resistance.
No. The power (in watts) would decrease, due to the greater effective resistance.
if the resistance is decreased and the current stays the same, then the power decreases.