Yes, a light bulb obeys Ohm's law since its resistance is constant as long as the voltage across it is constant. Ohm's law states that the current through a conductor is directly proportional to the voltage across it, and inversely proportional to the resistance of the conductor.
Unit measure for resistence is called as "OHM". Its law is called as "OHM'S LAW".
If a lightbulb has a resistance of 250 ohms, the voltage required for the bulb to draw a current of 0.5 A is 125 V. (Ohm's law: voltage equals current times resistance)Unfortunately, its more complicated than that...Is the resistance of 250 ohms the hot resistance or the cold resistance? It matters. It matters very much.Light bulbs have a dramatic positive resistance to temperature coefficient. It is not uncommon for the instantaneous on power to be 10 or 20 times the nominal value.So, if the 250 ohms is the measured resistance while operating at a current of 0.5 A, then 125 V is the correct answer. If the resistance is the cold resistance, you need to go back and find out the hot resistance at the desired operating point.
Hot, really hot. Typically in the range of about 2000C to 2500C (3600F - 4500F). It would be pretty difficult to directly measure the temperature of the filament so you have to use something other than a thermometer for the measurement. It should be possible to estimate the temperature of the filament from the light spectrum. Basically, treat the light bulb like a perfect black body radiator and use Plank's Law (http://en.wikipedia.org/wiki/Planck%27s_law) and the light spectrum of the bulb (for example http://www.graphics.cornell.edu/online/measurements/source-spectra/index.html) to calculate the temperature. Another way, which is a bit easier, is to use basic electronic theory to calculate the temperature of the filament that is required to produce the manufacturer's specifications for the bulb. For example, consider a typical 100 watt, 120 VAC light bulb with a tungsten filament. The bulb consumes (and radiates) 100 watts of power. A light bulb is a purely resistive load so Power=(Voltage)*(Current). Plugging in 100 watts as the power, 120 as the voltage (actually, that's the RMS voltage), and solving for current we get an RMS current of 0.83 amps. Since the bulb is just a resistor it obeys Ohm's Law; Voltage=(Current)*(Resistance). Our voltage is 120 and we determined the current to be 0.83 amps, so the resistance of the bulb when it is operating is 120/0.83= 144 ohms. If you take a 100W incandescent light bulb and measure its resistance at room temperature you get a value of about 15 Ohms. The difference between the room temperature resistance and the resistance when the bulb is operating is due to the affect of temperature on the filament's resistance. Metals (and conductors in general) increase their resistance as they are heated. The resistance at a particular temperature can be calculated with; R=Rref*(1 + alpha(T-Tref)] Where, R is the resistance at temperature T degrees Celsius. Rref is the resistance at a standard temperature Tref (often 0C or 20C). Alpha is the "temperature coefficient of resistance" for the material. For tungsten alpha=0.0044/C with a Tref of 20C (68F). If we assume that the 15 Ohm resistance at room temperature is close enough to the value at 20C (68F) then we can use Rref=15 ohms and R is the 144 ohms we calculated from the wattage and voltage of the bulb. Plugging these numbers into the equation; 144=15*(1 + 0.0044*(T - 20)) Solving for T (the temperature required to get the filament's resistance to 144 ohms) we get T=1975C (about 3600F). That's pretty hot! The filaments temperature will change if the applied voltage changes. The temperature will also be different from light bulb to light bulb (even if they have the same voltage and wattage ratings) since no two bulbs are exactly alike. An individual bulb will also change as it ages and as a function of the temperature outside the bulb.
You can use Ohm's Law to calculate the current of a light bulb by dividing the voltage across the light bulb by its resistance, which is typically provided on the bulb itself or its packaging. The formula is: Current (I) = Voltage (V) / Resistance (R).
Yes, a concave mirror obeys the law of reflection. The angle of incidence is equal to the angle of reflection when light rays hit the mirror surface and bounce off.
No.
If you plot a graph of current against a range of voltages applied to an incandescent lamp, the result will be a curvedline. This tells us that the current is not proportional to the voltage and, so, the lamp does not obey Ohm's Law.However, the ratio of voltage to current will indicate the resistance for that particular ratio.
The voltage of a circuit with a resistance of 250 ohms and a current of 0.95 amps is 237.5 volts. Ohms's law: Voltage = Current times Resistance
To find the resistance in ohms of a 194 12-volt bulb, you can use Ohm's Law (R = V/I). The 194 bulb typically has a current rating of about 0.25 amps. Using this information, the resistance would be approximately 48 ohms (R = 12V / 0.25A).
No, the wattage is determined by the resistance of the filament in the light bulb. The formula to determine the wattage is Watts = Voltage (squared)/Resistance in Ohms. To find the resistance of a 120 volt light bulb use the formula, Resistance in Ohms = Voltage (squared)/Watts. So for a 100 watt bulb at 120 volts the resistance is 120 volts x 120 volts = 14400/100 = 144 ohms. For a 60 watt bulb at 120 volts the resistance is 120 volts x 120 volts = 14400/60 = 240 ohms. As you can see this holds true to Ohm's law, current is inversely proportional to the resistance of the circuit. The higher the resistance of a load, the harder it is for the current to flow. In this case less current results in less light being emitted from the filament in the light bulb.
Unit measure for resistence is called as "OHM". Its law is called as "OHM'S LAW".
By Ohm's law, resistance is voltage divided by current, so the resistance of a light bulb can be measured by observing the voltage across it simultaneously with observing the current through it. Interestingly, the hot resistance is significantly different that the cold resistance, so measuring resistance with an ohmmeter will not give a meaningful resistance. This is because the resistance of a light bulb has a positive temperature coefficient. Take a typical 60 W 120V light bulb, for instance... Its cold resistance is about 16 Ohms. Calculate current and power at 120 V and you get 7.5 A and 900 W. The truth is that at 60 W, the bulb pulls 0.5 A and has a resistance of 240 Ohms.
If a lightbulb has a resistance of 250 ohms, the voltage required for the bulb to draw a current of 0.5 A is 125 V. (Ohm's law: voltage equals current times resistance)Unfortunately, its more complicated than that...Is the resistance of 250 ohms the hot resistance or the cold resistance? It matters. It matters very much.Light bulbs have a dramatic positive resistance to temperature coefficient. It is not uncommon for the instantaneous on power to be 10 or 20 times the nominal value.So, if the 250 ohms is the measured resistance while operating at a current of 0.5 A, then 125 V is the correct answer. If the resistance is the cold resistance, you need to go back and find out the hot resistance at the desired operating point.
ohms law.
Everything obeys Ohm's law - antennas, cables, transformers, integrated circuits, etc.AnswerIt is not true that 'everything' obeys Ohm's Law. For a device to obey Ohm's Law, the ratio of voltage to current MUST remain constant for variations in voltage. This is why Ohm's Law is a law of constant proportionality.
V = IR Voltage = Current * Resistance so 9 = 0.25 * R Hence R =36 Ohms
As an example imagine a 60 W light bulb running off 120 V. The current will be 1/2 A. Now assume that you put a second 60 W bulb in series with the first. Now the resistance of the single 60 W bulb is 240 Ohms. The 1/2 A flowing through the bulb heats up the filament wire and causes an amount of light to be emitted. Resistance in series adds, so the total resistance for two of the bulbs in series is 480 Ohms and by Ohms law V = I x R, the current through each bulb will be 1/4 amp and hence the bulbs will be less bright.