A multimeter set to measure current (in amperes or milliamperes) would be connected in series to the resistor to measure the current flowing through it. The multimeter would be placed in series with the resistor to accurately measure the current passing through it.
Yes, definately yes. The Magnet will pull the spinning disk on the electricity meter resulting for that disk to slow down and THE RESULT OF MONTHLY ELECTRICITY BILL will decrease 25%. Just place the magnet on both sides of the Electricity Meter then It will slow down. WARNING: Do not get caught by the Readers or you will busted for jail! WORKS ON ME! :D
1 cubic meter is big enough to hold 264.17 gallons of liquid.
Balancing the meter rule on the wedge edge-wise ensures that the weight of the meter rule is evenly distributed on both sides, leading to a stable and accurate reading. Placing it edgewise helps minimize the effect of friction and other external forces that may affect the balance of the rule.
That is 1 metre x 1 metre x 1 metre, and it is commonly called a "stere"
The volume of 1 meter times 1 meter times 1 meter = 1 cubic meter.
It depends on the purpose for installing the resistor. If the intent is to decrease current flow, the resistor must be connected in series with the load. If the purpose is to increase current flow, the resistor must be connected in parallel with the load. To connect a resistor in series, connect the resistor to one side of the power source, in line with the load. This will decrease circuit current flow. To connect a resistor in parallel, connect the resistor between the positive and negative sides of the power source, which will effectively connect the resistor across the load . This will increase current flow through the circuit. However, before connecting a component in parallel, make sure the increase in current flow will not exceed the current rating of the circuit or fuses/breakers will blow.
ammeter in series at any side as required since it is bilateral and voltmeter is connected in parallel to measure voltage drop across it
Resistors can be installed in any direction in a circuit. It doesn't matter in which direction you are installing a resistor. Resistor will resist the current in both sides of the connection. You can also apply capacitor in any direction for alternating current. Where as if you are concerned with direct current, a direction is needed to apply the connection
Yes, a resistor has two sides, and furthermore they are interchangeable.
Resistor is behind glove box (Push inward on glove box hinges to release) Pull out both sets of wires from resistor plug-in. (push release clips on both sets to pull out wires) Remove 2 small screws holding the resistor. Remove the resistor ( test the wire that has the small diode ) with a multimeter to ensure it has power on both sides of the diode. If there is current on your multmeter the resistor is okay...... if no current is present on your multimeter the diode is burnt and the resistor unit needs replacement.
One trace of the oscilloscope traces have to be inverted to display the voltage and current with the correct phase relationship because you are probably connecting to both sides of a small resistor in series with the load. The supply side shows voltage, while the load side shows voltage drop for the load. You are showing differential voltage across the resistor in the second trace and, of course, its negative.If, however, you are using a current probe, perhaps a clamp-on ammeter, I would question if you have it connected correctly.
If you add another resistor or just increase the resistance the current will decrease. I think the statement you are talking about means that whatever the current is in the series circuit it will be the same everywhere in that circuit, on both sides of the resistance. The resistance lowers the current in the entire circuit, not just after the resistance.
You use a combination of Ohm's law ( V = I * R ) and the power formula ( P = V * I ).Substituting you get: P = ( I * R ) * I = I**2 * RDividing both sides by R you get: P / R = I**2 * R / R = I**2Taking the square root of both sides you get: SQRT( P / R) = ISwapping sides of the equals sign you get: I = SQRT( P / R )Thus the maximum current can be found using the equation I = SQRT( P / R ).
Capacitors do not get "full" like a glass of water. The current into a capacitor is the rate of change of charge, so it's equal to C * dV/dt or something. If the voltage is constant, there's no current. If the voltage on both sides of the resistor is the same, there's no current through the resistor and hence into the capacitor, so that's the steady-state - what you call "full" - the capacitor charged to the supply voltage.
All sides are connected and there are no curved sides.
If anything is "cubic" it must have 6 sides
The area of square is : 144.0