4.5 Volts
There is insufficient information in the question to answer it. You need to provide either the voltage across the resistor, or the power dissipated by the resistor. please restate the question.
In order to determine what size of resistor is required to operate an LED from a 9V battery, first start by knowing the current and voltage required for the LED. That information is available in the LED's specifications. For discussion purposes, lets assume a typical LED at 2.5V and 50mW. The translates to a forward current of 20mA. Build a simple series circuit containing a 9V battery, a resistor of an as yet unknown value, and the LED. By Kirchoff's current law, the current in the LED is the same as the current in the resistor, which is also the same as the current in the battery. This is 20ma. By Kirchoff's voltage law, the voltage across the LED plus the voltage across the resistor equals the voltage across the battery. This is 6.5V. (9 - 2.5) By Ohm's law, resistance is voltage divided by current, so the resistor is 6.5 / 0.02, or 325 Ohms. The nearest standard value to that is 330 Ohms. Cross check the power through the resistor. Power is voltage times current, or 6.5V times 0.02A, or 0.13W. A half watt resistor is more than adequate for this job.
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.
Voltage x current. In a resistor for example it is the voltage drop across it that is relevant, it may be part of a circuit.
I = V / R = (14.4 / 1.6) = 9 AmperesNote: Stand back! The resistor dissipates 129.6 watts.
The ammeter is reading zero because there is no current flowing. This is because one of the resistors is faulty; the faulty resistor has an "open circuit" (open circuit means there is a broken connection). We know that: Ohms law is: V = I x R (voltage = current x resistance) Therefore because there is zero current in each resistor there will be zero voltage across each resistor. However we also know that: Kirchhoff's voltage law is: V1 +V2 +V3 + … = Vs (the sum of the voltage drops accross each component in a circuit MUST equal the supply (or battery) voltage). But if all the resistors are zero volts, then what component equals the supply (or battery) voltage? The battery voltage is developed across the open circuit… therefore the resistor which is faulty will have a voltage across it equal to the battery voltage. That easy to measure with a volt meter! hope this helps
There is insufficient information in the question to answer it. You need to provide either the voltage across the resistor, or the power dissipated by the resistor. please restate the question.
In order to determine what size of resistor is required to operate an LED from a 9V battery, first start by knowing the current and voltage required for the LED. That information is available in the LED's specifications. For discussion purposes, lets assume a typical LED at 2.5V and 50mW. The translates to a forward current of 20mA. Build a simple series circuit containing a 9V battery, a resistor of an as yet unknown value, and the LED. By Kirchoff's current law, the current in the LED is the same as the current in the resistor, which is also the same as the current in the battery. This is 20ma. By Kirchoff's voltage law, the voltage across the LED plus the voltage across the resistor equals the voltage across the battery. This is 6.5V. (9 - 2.5) By Ohm's law, resistance is voltage divided by current, so the resistor is 6.5 / 0.02, or 325 Ohms. The nearest standard value to that is 330 Ohms. Cross check the power through the resistor. Power is voltage times current, or 6.5V times 0.02A, or 0.13W. A half watt resistor is more than adequate for this job.
It depends on the current going through it. Ohm's law: Voltage equals current times resistance.
Two milliamperes. Ohm's law: Current equals voltage divided by resistance.
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.
12 volts...! The voltage drop across a 2 ohm resistor depends on the current flowing through it. As voltage (E) equals current (I) times resistance (R), if 1/2 amp is flowing through your 2 ohm resistor, 1/2 times 2 = 1 volt. If 1 amp is flowing through your 2 ohm resistor, 1 times 2 = 2 amps. Piece of cake. If the two ohm resistor is the only component in the circuit, it will drop whatever the applied voltage is. Put a 2 ohm resistor across a 6 volt battery, it drops 6 volts. If you put your 2 ohm resistor across a 9 volt battery, it drops 9 volts. Another way to say voltage drop may help. The voltage drop across a resistor is the voltage it "feels" when in a circuit. And that last couple of examples says that very well. In a circuit where a given resistor is the only component, it drops all the voltage in the circuit. It "feels" all the voltage in the circuit. In a circuit where there are 2 resistors of equal value in series, each one drops or "feels" half of the applied voltage. (The sum of the voltage drops equals the applied voltage.) As you work more with simple circuits using resistors in different arrangements with a given voltage source, try thinking of the voltage drop of a resistor as the voltage it "feels" when the circuit is energized.
Voltage x current. In a resistor for example it is the voltage drop across it that is relevant, it may be part of a circuit.
The question is a bit ambiguous, but I will try to address it. If the 6 ohm resistance is in series with another resistance then some of the 5 volts would be dropped across the 6 ohm resistance and the remainder of the voltage would be dropped across the other resistance. To calculate the voltage, use the 'resistor voltage divider equation' (Google it). If the 5 volts is applied across only a 6 ohm resistance, then the top of the resistor is at 5 volts and the bottom of the resistor would be at 0 volts. The resistor would drop all of the voltage.
I = V / R = (14.4 / 1.6) = 9 AmperesNote: Stand back! The resistor dissipates 129.6 watts.
The power dissipated across a resistor, or any device for that matter, is watts, or voltage times current. If you don't know one of voltage or current, you can calculate it from Ohm's law: voltage equals resistance times current. So; if you know voltage and current, power is voltage times current; if you know voltage and resistance, watts is voltage squared divided by resistance; and if you know current and resistance, watts is current squared times resistance.
Half that, or 2 amps. The basic rule in circuits is that voltage (E) equals current (I) times resistance (R). Here's how that expression of Ohm's law looks: E= I x R That means that current equals voltage divided by resistance, as is shown here: I = E / R This expression says that resistance is inversely proportional to current (with voltage staying the same). Further, if resistance goes up, current goes down. If resistance doubles (goes up by a factor of 2), which it does in the case specified in the question, then current is cut in half (goes down by a factor of 2). Half of 4 amps is 2 amps, and that's where the answer came from.