if not disconnected you will measure the resistance of the circuit in parallel with the resistor.
The total resistance of a set of resistors in parallel is found by adding up the reciprocals of the resistance values, and then taking the reciprocal of the total. By removing a resistor the total current will lower. If you short out the parallel circuit as suggested it will take out the fuse that should be protecting the circuit.AnswerShorting-out a resistor in a parallel circuit, will act to short out the entire circuit, therefore, significantly increasing, not lowering, the current! And, as the previous answer indicates, this short-circuit current will operate any protective devices, such as a fuse.In a parallel circuit current does not lower but it will be increase if shorting-out one resistor in the two resistor parallel circuit, the circuit will become very low resistive and the larger current will flow through the short path.
It does not matter. Kirchoff's Current Law states that the signed sum of the currents entering a node is zero. A consequence of that law is that the current in every part of a series circuit is the same. The only thing that resistor location affects is the potential voltage of the LED terminals with respect to the rest of the circuit. Certainly, if you are driving the LED with high voltage, such as 120VAC, you should consider the resistor location so as to reduce electrocution hazard but, the LED's performance is not affected by resistor location in the circuit.
The centrifugal switch on a motor's start winding should open at 75 to 80 percent of the motors run speed RPM.
Add an ammeter in parralel with the circuit. An ammeter in parallel with a circuit to be measured will not measure any current. In fact, placing an ammeter in parallel will cause a short. The ammeter must be placed in series with the circuit to be measured. It should be noted that this technique only allows for small magnitudes of current to be measured. One should not try to measure more than 10 amps using this method. However, this is not the goal of the question asked. I am not 100% sure what "intercepting the supply" means but I think the person asking the question means without breaking the circuit? In this case one could use a clamp on ammeter. There are several varieties that can measure low currents and there are those that can measure 100s of amps.
Yes it is. The filament in a standard incandescent bulb is a type of resistor. An incandescent light bulb contains tungsten which reduces electricity and converts electricity to heat and light. All incandescent bulbs are resistors, but only a fraction of resistors are bulbs. If you want to see if a bulb is a resistor, try adding another bulb in series without changing the voltage. Both bulbs will be very dim. Another way to test this is to get a multimeter and set the meter to the resistance setting. If you get any value other than zero, then it is functioning as a resistor.
hen the wire inside the resistor will melt and the circuit will be broken
We know definition of Resistance, that resistor always opposes to flow of current. resistor should have input signals from source , so it generates passivity in circuit
That depends on the quality (price) of the meter. Ideally it should have no effect.
Resistors are tiny devices on circuit boards. Prices should range but should not be expensive but dependent on the materials used and the nature of conductivity.
It will not be possible to measure current unless the circuit is powered and operating.
The total resistance of a set of resistors in parallel is found by adding up the reciprocals of the resistance values, and then taking the reciprocal of the total. By removing a resistor the total current will lower. If you short out the parallel circuit as suggested it will take out the fuse that should be protecting the circuit.AnswerShorting-out a resistor in a parallel circuit, will act to short out the entire circuit, therefore, significantly increasing, not lowering, the current! And, as the previous answer indicates, this short-circuit current will operate any protective devices, such as a fuse.In a parallel circuit current does not lower but it will be increase if shorting-out one resistor in the two resistor parallel circuit, the circuit will become very low resistive and the larger current will flow through the short path.
It does not matter. Kirchoff's Current Law states that the signed sum of the currents entering a node is zero. A consequence of that law is that the current in every part of a series circuit is the same. The only thing that resistor location affects is the potential voltage of the LED terminals with respect to the rest of the circuit. Certainly, if you are driving the LED with high voltage, such as 120VAC, you should consider the resistor location so as to reduce electrocution hazard but, the LED's performance is not affected by resistor location in the circuit.
answ2. There should be a complete circuit for electrons to travel round it. Otherwise you'd build up charge in the one spot.If the circuit contains resistors, then these will slow the electrons down; just temporarily whilst they bounce round in the resistor, for in there are less paths for free electrons.The energy of an electron is measured in volts, and some of the energy is absorbed in the resistor and will show up as a voltage lost in the passage through the resistor.A side consequence of electrons moving in a circuit is that a small magnetic field will be generated, but in normal electronic circuits, this is so minute as to be ignored.
The circuit won't do anything until both terminals of the battery are connected to the correct [two different] points in the circuit. As long as either terminal of the battery remains disconnected, the circuit is "OFF".
You almost NEVER do. 1) The circuit should be off and/or disconnected when using an ohmmeter. 2) It should be in parallel with the component as far as the rest of the circuit is concerned, but alone in series with the device its measuring.
a power supply (eg: battery) and a load (eg: resistor) even a piece wire shorting two terminals of the power supply is a closed circuit
Calculate the impedence of your 3V circuit in ohms. Figure the voltage you want to drop. In this case, you need to lose 9V. This is three times the voltage you are supplying, so the resistor should have three times the resistence of your 3V circuit. If your circuit has 100 ohms of resistence (impedence), then you would connect a 300 ohm resistor in series with your circuit.