If an external Voltage is applied to a multimeter while the multimeter`s function switch is in the Ohm meter position then the Ohmmeter is apt to be destroyed or disabled. Therefore the simplest answer is that the power must be turned off in order to avoid damage to the Ohm meter and possibly to yourself caused by Voltages present in the circuit while the power is own. Another reason of course is because a valid Ohm reading can not be obtained while Voltages are present across the resistance that is being measured because the Ohmmeter itself applies a Voltage across the resistance and then a sampling of the current through the unknown resistance is used by the Ohmmeter`s circuitry to cause a calibrated deflection of the Ohmmeter and thus display on the meter`s dial the Ohmic value of the unknown resistance.
For the same reasons as above it is also a good practice to discharge all the capacitors in the circuit after the power is turned off and before any in circuit resistance measurements are made.
The power will be the product of the square of the current and the resistance of the load. The fact that the circuit is a parallel circuit is irrelevant to this question.P = I2R = 0.032 x 1000 =0.9 W
Consider t resistors with same Ohmic values. If they are in series total resistance Rt = R1 + R2. if they are in parallel then total resistance Rt = 1/R1 + 1/R2. Series connection will have higher resistance.
It depends somewhat on context. A dc circuit consists of of a flow of current to the load and back to the source. This is often called the loop. The loop resistance would be the total resistance measured in ohms. In most power circuits the above would hold true in an ac circuit as well.
"Volts" is electrical pressure applied to a circuit; whereas, "ohms" is electrical resistance to that pressure. One cannot determine ohms from voltage without knowing either the current (in "amps") or power (in "watts"). A normal 120V household circuit can handle a maximum of 20 amps, so using ohm's law of resistance = voltage / current, the minimum resistance required in a 120V household circuit would be 6 ohms. Any less than 6 ohms will cause the circuit breaker to trip.
If it's a pure inductor, no power consumption. However it must be wound using unobtanium wire which has zero resistance, and the core must be vacuum. Air is nearly lossless.
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.
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.
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 = (energy used)/(time to use it)Power dissipated by an electrical circuit =(voltage across the circuit) x (current through the circuit)or(resistance of the circuit) x (square of the current through the circuit)or(square of the voltage across the circuit)/(resistance of the circuit)
The unit of power measured is watt, irrespective of resistance, capacitance or inductance of the circuit.
Did you mean "maximum RESISTANCE" or "maximum VALUE"? If the former, then, you have a ZERO reading, meaning there is high resistance, and no electrical connectivity. If the latter, you have a ONE (or 100%) reading, meaning there is NO resistance, or absolute electrical connectivity. As an analogy, if you turn on a plugged-in, working, lamp, then it has NO resistance, such that power flows easily through the cord; if it didn't turn on, then there IS resistance, such that no power flows, possibly due to broken wire, bad switch, burned-out bulb, or blown fuse.
if the resistance is decreased and the current stays the same, then the power decreases.
An Ohmmeter.
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if the switch in the circuit is switched off, the power is turned off on the object.
The power will be the product of the square of the current and the resistance of the load. The fact that the circuit is a parallel circuit is irrelevant to this question.P = I2R = 0.032 x 1000 =0.9 W
The power factor never depends on the resistance of a circuit. It depends on the equivalent inductance and capacitance in the circuit, and on the frequency of the power supply, even if the resistance is zero.