Just use Ohms Law: V=IR, that is, voltage (in Volt) = current (in Ampere) x resistance (in Ohms).
Voltage across a resistance = (resistance) x (current through the resistance) =4 x 1.4 = 5.6If the ' 1.4 ' is Amperes of current, then the required voltage is 5.6 volts.
In the circuit where the DC motor is added, it was not specified whether the motor was added in series or in parallel to circuit elements. If it was added in series, it will increase circuit resistance and it will cause circuit current to go down. In parallel, the motor will reduce total circuit resistance, and circuit current will increase.
There is insufficient information in the question to properly answer it. You need to specify, also, the voltage, in order to determine resistance from current. Please restate the question. Ohm's Law: Resistance = voltage divided by current.
If the current through a pure metallic conductor causes the temperature of that conductor to rise, then its resistance will increase. A practical example of this is an electric lamp. The cold resistance of a lamp is very much lower than the hot resistance.
ANY METER needs some kind of current flow to operate. Internal in the meter there are batteries that provide current that when passed trough a resistor will develop voltage as a function of the current. the meter will read this current and display the resistor size to cause this current to flow.
15 KOhms times 10 mA = 150 volts. 150 volts times 10 mA = 1.5 watts.
Voltage across a resistance = (resistance) x (current through the resistance) =4 x 1.4 = 5.6If the ' 1.4 ' is Amperes of current, then the required voltage is 5.6 volts.
how much resistance does a light bulb creat if iyt has a current of 25 mA around it in a 9 V circuit?
Current = voltage/resistance If those are the only components in the circuit, then Current = 9/12 = 0.75 Ampere = 750 mA
If you know the voltage and resistance, then current = voltage divided by resistance. Otherwise, you can attach an ammeter into the circuit (in series).
If the resistance is large enough, then there might not be enough voltage difference to allow much current. Since, Voltage = Current * Resistance, if resistance goes really large, and your voltage doesn't change, your current must decrease. An open circuit is where you do not have any current flowing, so whether no current verses very little current is the same is up to you.
Ohm's law applies: Current = Voltage / Resistance As such if you double the resistance of the light bulb you end up with half as much current.
In the circuit where the DC motor is added, it was not specified whether the motor was added in series or in parallel to circuit elements. If it was added in series, it will increase circuit resistance and it will cause circuit current to go down. In parallel, the motor will reduce total circuit resistance, and circuit current will increase.
It depends on the resistance of the circuit, as V=IR, so I=V/R (V=Voltage, I=Current, R=Resistance)
There is insufficient information in the question to properly answer it. You need to specify, also, the voltage, in order to determine resistance from current. Please restate the question. Ohm's Law: Resistance = voltage divided by current.
If the current through a pure metallic conductor causes the temperature of that conductor to rise, then its resistance will increase. A practical example of this is an electric lamp. The cold resistance of a lamp is very much lower than the hot resistance.
No it does not. A volt meter only reads the current that is passing through it.AnswerAll instruments draw some (albeit tiny) current from the circuit under test in order to operate. So, if this is what you mean by 'taking power from circuit', then the answer is yes, it does.Instruments also change the normal resistance of the circuit being tested -for example, ammeters increase the resistance of the circuit into which they are connected, while voltmeters decrease the circuit resistance across which they are connected. So adding a voltmeter (or an ammeter) to a circuit affects the operation of that circuit to some degree. To minimise this interference, it is important that an ammeter's internal resistance is very much lower than the circuit's resistance, and a voltmeter's resistance is very much higher than the circuit's resistance.