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Any material can be fashioned into a sample with a resistance of 500 ohms, by

adjusting the length and thickness of the sample. Even an insulating material.

The familiar carbon composition resistors are available in perhaps 40 different

resistance values, all made of the same material.

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What is the total resistance is V is 10 volts R1 is 1000 ohms and R2 is 1000 Ohms?

If they're connected in series the total resistance is 2000 ohms. If they're connected in parallel the resistance is 500 ohms.


When two resister of 1000 ohms are connected in parallel their effective resistance will be?

500 ohms. RP = 1 / summation (1 / RI)


What does 500 ohms plus 500 ohms equal?

500 + 500 = 1000


In a parallel circuit the smallest resistor is 500 ohms. What is the largest value this parallel circuit can have?

500 ohms.


What is the amount of power loss on a transmission line that is transmitting 10 MW at 500 kV with 1000 ohms?

Line current = 10MW / 500kV = 20A Assuming the 1000 ohms is the resistance of the entire transmission line, end to end. Power loss = line current ^ 2 * line resistance = 20A ^ 2 * 1000 ohms = 400 KW


How many ohms is 500 mill ohms?

The correct term is 'milliohm', not 'mill ohm'. As a milliohm is one-thousandth of an ohm, 500 milliohms is 0.5 ohm.


What is the voltage drop for 500 meters 1.0mm copper cable drawing 230ma with 24volt supply?

The resistance of 1 mm (diameter) solid copper is about 21.8 ohms per kilometer.That's 10.6 ohms in 500 meters.When 230 mA flows through 10.6 ohms, the voltage drop isI · R = (.230)x(10.6) = 2.438 volts.The voltage of the supply is irrelevant.


What is the impedence of telephone line?

In the USA, should be 500 or 600 Ohms. It is 600 ohms as standard


How does the total resistance in a system change when additional resistance is added in series?

Sneighke answered: This discussion adds to the original question There are two types of resistance topologies: 1) Series and 2) parallel. To answer your question, resistance added in series always ADD together increasing the total resistance of the circuit. Conversely, adding parallel resistance reduces the total resistance of the circuit. So, for series circuits, R(total) = R1+R2+...Rx Parallel circuits are the exact mathematical inverse. The easiest way to determine parallel resistance is to add the inverse of resistance which is conductance, conductance being 1/r and is stated in Siemens (hold the jokes!...), then taking the inverse of the total conductance to convert back into Ohms. For example, if you have three resistors R1, R2, and R3, and they are parallel connected, the total resistance of the circuit is the inverse of the sum of conductance which would be written as 1 / (1/r1+1/r2+1/r3). By definition, conductance is the inverse of resistance. An example: Given three resistors of 5, 100, and 500 Ohms, In series, R(total) = 5+100+500 = 605 Ohms. In parallel, the total is always less than the lowest resistor: Converting to conductance (used to be called Mhos which is "Ohm" backwards, but has been replaced with the SI unit of Siemens): 5, 100, and 500 Ohms = 1/5+1/100+1/500 = 0.200+0.010+0.002 = 0.212 Siemens. Converting back into resistance, 1/conductance = 1/0.212 Siemens = 4.717 Ohms which as stated above, is less than the lowest resistance resistor. In fact, sometimes working with conductance is easier in series/parallel circuits and, in particular, calculating which values of resistors are required to yield a desired resistance; usually a non-standard resistance value needed for a specific purpose in a circuit. An example: Say you need a non-standard resistance of 698 Ohms. Since we know that parallel resistors create a value lower than the lowest parallel connected resistor, you would start with the next highest standard value and then add a parallel resistor to get you what you need. In this case, you would subtract the desired conductance from the starting resistor: 698 Ohms = 1/698 = 0.001427 Siemens or 1.4327 milliSiemens. If we had a standard value resistor of 750 Ohms (remember, you have to start higher): 750 Ohms = 1.3333 mS. To find the required parallel resistor to get us our 698 Ohms, subtracting the conductances 1.4327mS-1.333mS = 99.33uS (micro Siemens) [0.00009933 S]. Converting back into Ohms, 1/99.33uS = 10.07kOhms (10,070 Ohms) which is close to the standard value of 10kOhms. Doublechecking, Add the conductances: 10,000 Ohms = 100uS 750 Ohms = 1.3333mS Adding gives a total conductance of 1.4333mS. Thus the parallel equivalent = 1/Siemens = 1/0.0014333 = 697.7 Ohms which is within 0.04% of the 698 Ohms we need which is well within acceptable error and we have our 698 Ohm resistor by connecting 10,000 Ohms and 750 Ohms in parallel.


What size steel wire armoured cable for 40 amps 230 volts at 500 meters away?

On a 230 v supply assuming a volt-drop of 5% that is 11.5 v, so the maximum cable resistance is 11.5 / 40 in ohms, 0.2875 ohms so for 1000 metres of cable in total you need a cable that has 0.0002875 ohms of resistance per metre. Copper with a cross-section of 1 mm2 has resistance of 0.0168 ohms per metre so the size of cable needed is 0.0168 / 0.0002875 mm2, which is 58.43 mm2, so a cable of 60 mm2 should be selected.


A 6.50 mu F capacitor that is initially uncharged is connected in series with a 4500 Omega resistor and a 500 rm V emf source with negligible internal resistance Just after the circuit is completed?

Any capacitor that is initially uncharged, when presented with a step change in voltage, will have an instantaneous resistance of zero ohms. As a result, the instantaneous current with a circuit involving 4500 ohms and 500 volts will be about 0.111 amperes. It does not matter what the capacitance is.The stated voltage in the question is unclear, so 500 volts was assumed for the answer.


What is the total current is V is 10 Volts R1 is 500 Ohms and R2 is 1500 ohms?

We need to find R equivalent at first then divide V from it to find the total current.Case 1: If resistances are connected in series.Then, R(eq.) = R1 + R2 = 500 + 1500 = 2000 ohmSo, Total current, I = V/R(eq.) = 10/2000 = 1/200 or 0.002 A.Case 2: If resistances are connected in parallel.Then, R(eq.) = (R1 x R2)/(R1 + R2) = (500 x 1500)/(500 + 1500) = 400 ohmSo, Total current, I = V/R(eq.) = 10/400 = 1/40 or 0.025 A.