Since the current in a resistor is the same as the current in the leads/wires on either side of the resister, I might use a clamp meter such as an Amprobe to measure current, if the current was alternating (AC). Otherwise, I would have to break one of the leads and insert an ammeter or a multimeter with an amp setting into the circuit. Afterwards the broken connection would have to be repaired.
The current would be about 20 volts.
No, because the power dissipated in a resistor is proportional to the square of the current through the resistor but only directly proportional to the resistance of the resistor (I^2 * R) and the current through the lower value resistor will be higher than the current through the higher value resistor, the lower value resistor will usually dissipate more power.
Power dissipated by the resistor = I^2 * R or V^2 / R, where R = its resistance value, I = the current in the resistor, and V = the voltage drop across the two terminals of the resistor. You need to measure or find the information of either I (using an ammeter) or V (a voltmeter).
Multimetres are generaly used to measure the following quantities. Voltage: multimeter is connected in parallel with the Load; Current: multimeter is connected in series with Load; Resistance: the resistor must be taken out of the circuit first, then, the probes from the multimeter are connected across the resistor,
If you are looking for the resistance of each resistor in either a series circuit or a parallel circuit you must measure the current I and the voltage V for each resistor. Then calculate its resistance using Ohms Law R = V / I where I = current (Amps), V = voltage (Volts) and R= resistance (Ohms).
A: NO the only way is to lift one lead and measure.
Without load there is no current so it is impossible to measure it.
Depends on the current. Put a resistor in-line with the current, then measure the voltage across the resistor. V=RI. So, divide the measured voltage by resistor value. Be careful with the size of the resistor, as Power dissipated in a resistor is R*I^2 or V^2/2. So, a 1-Amp current into a 1 Ohm resistor will result in a 1Watt power dissipated in the resistor. If it's too small, it'll burn. Also, notice that if you do that, you haven't measured the current in the original circuit. You've measured the current when an extra resistor is installed in the original circuit, and that's different.
you can if you know the current measure the voltage across the resistor E=I*R there is actually no way to measure the resistance without passing current thru it P=I*E E=I*R any 2 will give you the other 2
To measure the value of a resistor, apply a voltage and measure the voltage across the resistor and the current through the resistor. Use Ohm's law: Resistance equals Voltage divided by Current. Start with a small voltage and increase gradually until a reading is obtained, but be careful that the power dissipation (watts = volts times amperes) of the resistor is not exceeded. Simpler solution: Use an ohmeter.
An ammeter is a low voltage voltmeter in parallel with a small resistance resistor. Current flow through the resistor creates a voltage drop across it which is then measured by the voltmeter.
The resistor is 1/3 of an ohm. A 9 volt drop across the resistor would cause a draw of 27 amps through the resistor. The wattage you would need for that resistor is at least a 243 watts.
ohm meter puts current thru a resistor to measure voltage drop E / I = R if the circuit already has current flowing the numbers are meaningless
The current would be about 20 volts.
A current is passed through a resistor and the voltage drop is measured on the other side. It follows Ohm's law by a given current and voltage you can measure the resistance of a resistor by V=IR.
Current flow would be the same through the resistor, since it's in series, but the voltage would be slightly reduced based on the resistance. If you have 2 resistors in parallel, the current will divide through each resistor, and the voltage stays the same. PLL Ohm's law and water flow - PLL
An ohmmeter is a calibrated circuit that puts current through the resistor to measure its value. If something else is also putting current through the resistor that the meter doesn't know about, the meter calibration is disturbed and a false reading results. It may even damage the meter.