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In order for a 30 volt 90 watt lamp to work properly in a 120 volt supply the required series resister in ohm is?
A 30 volt 90 watt lamp has 3 amps going through it. The series resistor also has 3 amps going through it, by Kirchoff's current law. The voltage across the resistor is 90 volts. With 3 amps, that is 30 ohms. (By the way... The resistor must be rated to carry 270 watts. That is a lot of power for a resistor.)
When resistors are connected in series then what happens to the watt?
Resistors are rated in wattage so the lowest wattage rating will be the wattage of the series circuit. It will be able to handle that power any more and the lowest wattage resistor could be damaged and fail.Another AnswerWhen two or more resistors are connected in series, the resistor with the lowest resistance will operate at the highest power. If the power developed by a resistor exceeds its rated power, then the resistor may burn out.
A 100 ohm 1 watt resistor and 800 ohm 2 watt resistor are connected in seriesmaximum dc output voltage that can be applie continuously to series circuit without exceeding power limit of resistor?
If a 100 ohm 1 watt resistor and an 800 ohm 2 watt resistor are connected in series, the maximum DC voltage that can be applied without exceeding the power limit of both resistors is 45 volts. Start with power rule and ohm's law... P = IE E = IR so P = I2R solve for current I2 = P/R I = (P/R)0.5 determine current at max power for each resistor R1 = 100, R2 = 800 I1 = (1/100)0.5 = 0.1 amperes I2 = (2/800)0.5 = 0.05 amperes The limiting resistor is R2, so calculate voltage of both at current flow of I2 V1+2 = (0.05)(900) = 45 volts
If a 100 ohm resistor is needed in an electric circuit carrying a current of 0.3 amperewhat resistor would you select?
A 100 ohm resistor carrying a current of 0.3 amperes would, by Ohm's Law, have a potential difference of 30 volts. A current of 0.3 amperes through a voltage of 30 volts would, by the Power Law, dissipate a power of 9 watts. You need a 10 watt resistor, alhough it is better to use a 20 watt resistor. E = IR 30 = (0.3)(100) P = IE 9 = (30)(0.3)
How do you calculate the amount of current that will go through the resistor a 12 volt battery a resistor of 470 ohms?
Ohm's Law: Current is voltage divided by resistance.12 volts divided by 470 ohms is 25.5 milliamperes.Power is voltage times current, so power in this case would be 12 volts times 25.5 milliamperes, or 0.306 watts. As a result, you need at least a one half watt resistor, and I would prefer a one watt resistor, because it is going to get a bit warm, and a margin of safety is always a good thing. This is particularly true when you consider tolerances, such as the battery actually putting out 13 or 14 volts and the resistor being on the low side, at 470 - 10%, or 423 ohms. (0.463 watts - too close to one half watt for comfort)
In order for a 30 volt 90 watt lamp to work properly in a 120 volt supply the required series resister in ohm is?
A 30 volt 90 watt lamp has 3 amps going through it. The series resistor also has 3 amps going through it, by Kirchoff's current law. The voltage across the resistor is 90 volts. With 3 amps, that is 30 ohms. (By the way... The resistor must be rated to carry 270 watts. That is a lot of power for a resistor.)
When resistors are connected in series then what happens to the watt?
Resistors are rated in wattage so the lowest wattage rating will be the wattage of the series circuit. It will be able to handle that power any more and the lowest wattage resistor could be damaged and fail.Another AnswerWhen two or more resistors are connected in series, the resistor with the lowest resistance will operate at the highest power. If the power developed by a resistor exceeds its rated power, then the resistor may burn out.
What happens of watt when two resistor is connected in series or parallel connection?
In case of parallel combination watt increases with respect to series combination. Let, V=12v,Ra=5 Ohm, Rb=3 Ohm then Rse=Ra+Rb=8 Ohm Rpa=Ra.Rb/(Ra+Rb)=1.875 Ohm Now, P=IV or V^2/R then Pse=12^2/8=18W Ppa=12^2/1.875=76.8W So from here we can easily understand that why watt increases in parallel combination w.r.t series combination.
How does a resistor TBA820 work?
The TBA820 is a 2 watt general purpose amplifier integrated circuit, not a resistor.
A 1.2 k ohms resistor is placed across an 18-V source The minimum acceptable power rating for the resistor is?
1/2 watt (In theory you could use a "0.27 Watt" resistor, however there would be no safety factory and there is no standard value resistor that size.)
A 100 ohm 1 watt resistor and 800 ohm 2 watt resistor are connected in seriesmaximum dc output voltage that can be applie continuously to series circuit without exceeding power limit of resistor?
If a 100 ohm 1 watt resistor and an 800 ohm 2 watt resistor are connected in series, the maximum DC voltage that can be applied without exceeding the power limit of both resistors is 45 volts. Start with power rule and ohm's law... P = IE E = IR so P = I2R solve for current I2 = P/R I = (P/R)0.5 determine current at max power for each resistor R1 = 100, R2 = 800 I1 = (1/100)0.5 = 0.1 amperes I2 = (2/800)0.5 = 0.05 amperes The limiting resistor is R2, so calculate voltage of both at current flow of I2 V1+2 = (0.05)(900) = 45 volts
What is 5W15RJ?
it may be a ceramic resistor. If it is a resistor, it'll be about 1 inch in length with 2 contact points. It sounds like a 5-Watt 15-Ohm resistor.
How do you reduce 48 volts dc to 12 volts dc with resistors?
Yes. You can use a voltage divider. Say, for instance, one 1KOhm resistor in series with a 3KOhm resistor. Connect the 3k resistor to the 48 volts and connect the 1k resistor to ground. The 1k resistor will have 12 volts acress it. These resistors need to be at least 1 watt each as they are going to dissipate 0.576 watts and get warm. Now, if you attempt to pull power from the 1k resistor, note that regulation will be poor because the impedance of the load will go in parallel with the 1k resistor and change its value.
You need 18 v and 1.5 watt zener diode for your power supply.but it is not available.you have only 0.25 watt zener diode.how will you increase your watt keeping voltage constant?
You could design a series regulator using the 0.25 W zener as a reference. There might even be an off the shelf 3 pin series regulator meeting your needs, then you wouldn't even need the zener.
How much is a watt?
A watt is one joule of energy used every second. In electronics, applying 1 volt across a 1 ohm resistor will produce 1 W of heat.
What is different between the resistor and pullup resistor?
A "pull-up" resistor is a resistor used to to perform a specific electronic function - it is not a different type of resistor. A very small current flows through a pull-up resistor so it does not need to be high wattage (1/8 watt is generally fine). The value of a pull-up resistor depends on the resistance of the sensor. If it is simply on or off (no resistance) then a typical pull-up resistor might be 10k ohms.
If a 100 ohm resistor is needed in an electric circuit carrying a current of 0.3 amperewhat resistor would you select?
A 100 ohm resistor carrying a current of 0.3 amperes would, by Ohm's Law, have a potential difference of 30 volts. A current of 0.3 amperes through a voltage of 30 volts would, by the Power Law, dissipate a power of 9 watts. You need a 10 watt resistor, alhough it is better to use a 20 watt resistor. E = IR 30 = (0.3)(100) P = IE 9 = (30)(0.3)
