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Which has more resistance three resistors in series orf three resistors in parallel?
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I have three identical resistors all at 100 ohms in what combination can I put them to obtain a resistance of 66.7?
If you have three 100 ohm resistors, and you want an equivalent resistor of 66.7 ohms, put two resistors in series, and then parallel the third resistor across the first two. Resistors in series: R1 + R2 Resistors in parallel: R1 * R2 / (R1 + R2) This example: Two 100 ohm resistors in series: 100 + 100 = 200 A 100 ohm resistor in parallel with a 200 ohm resistor: 100 * 200 / (100 + 200) = 66.7
Twelve identical wires of resistance 6 ohm each are arranged to form the edges of cube the effective resistance bw the opp corners of cube is?
The effective resistance between opposite corners of a cube comprised of twelve 6 ohm resistors, one at each edge, is 5 ohms. There are several ways to solve this. One approach is to build a system of 12 equations in 12 unknowns, and solve them. Another approach is this... Consider that there are three resistors leaving the input node, and there are three resistors entering the output node. In between those three resistors, there are six resistors in a criss-cross matrix. (Draw it out, flattened, to see this.)Inspecting the six resistors in the center, you note that they are completely symmetrical. Since they are symmetrical, you can conclude that the voltage at the junction between the three input resistors and the six others is the same voltage. The same goes for the three output resistors. Said another way, the voltage across the three input resistors and the three output resistors is the same. Given two or more nodes in a circuit having the same voltage, you can draw a wire connecting them, i.e. a resistor of zero ohms. This does not change the characteristics of the circuit in any way, because zero voltage across any resistance is still zero amperes. Now that you have made these connections, look at the circuit. It has simplified to three parallel resistors, in series with six parallel resistors, in series with three parallel resistors. Three 6 ohm resistors in parallel is 2 ohms. Six 6 ohm resistors in parallel is 1 ohm. Three more 6 ohm resistors in parallel is 2 ohms. The total resistance is 2 + 1 + 2 ohms, or 5 ohms.
Which is the total resistance of a circuit that contains twenty 100 ohms resistors connected in parallel?
When resistors of the same value are wired in parallel, the total equivalent resistance (ie the value of one resistor that acts identically to the group of parallel resistors) is equal to the value of the resistors divided by the number of resistors. For example, two 10 ohm resistors in parallel give an equivalent resistance of 10/2=5Ohms. Three 60 ohm resistors in parallel give a total equivalent resistance of 60/3 = 20Ohms. In your case, four 200 Ohm resistors in parallel give 200/4 = 50 Ohms total.
How do you calculate the equivalent resistance of a circuit of the shape of a cube and a having a resistor of i kilo ohm in every branch?
Given twelve 1 KOhm resistors, connected in the shape of a cube, in order to determine the net resistance between opposite corners, first draw the cube in two dimensions. (Try this at each step before continuing, so you can understand the lesson as it unfolds.)There are three resistors leaving the initial vertex, and three resistors entering the final vertex. In between those six resistors, are six more resistors, each pair connected together on one end, and to two other resistors on the other end.If every resistor has the same value, then (by symmetry), the voltage on the ends of the first three resistors must be the same. Similarly, the voltage on the ends of the last three resistors must be the same.If two points in a circuit have the same voltage, then (for purposes of analysis) you can consider them to be shorted together. That short does not change the results, as there is no current flowing through that short.With the bottom ends of the first three resistors shorted, and with the top ends of the last three resistors shorted, the circuit degrades into three resistors in parallel, in series with six more resistors in parallel, in series with three more resistors in parallel.Three 1 KOhm resistors in parallel have a net resistance of 333 ohms. Six have a net resistance of 167 ohms. Two 333 ohm resistors and one 167 ohm resistor in series have a net resistance of 833 ohms, or 5/6 of 1 KOhms.Note: This technique does not work if the resistors are not all the same value. In that case, you would need to solve 12 equations in 12 unknowns, looking at the partial currents in each branch.
Each of the resistors in a circuit is 3 ohms. If the current in the middle resistor is zero what is the equivalent resistance?
First, the question doesn't say if the resistors are in series or parallel, or series-parallel. Second, the current given is zero, which can only be true if the circuit has no applied voltage. (It's turned off.) This will be true regardless of the circuit configuration. We were told the "middle resistor" in the question, but that's still a bit "iffy" for us. We need to know how it's wired. Since we don't, we'll look at the three possibilities. If all three resistors are in series, the total resistance is the sum of all the resistors. It's this: Rt = R1 + R2 + R3 ... or Rt = 3 + 3 + 3 = 9 ohms A shortcut can be applied when identical resistors are in series. The total resistance will be the value of one multiplied by the number of them in series. In this case, 3 x 3 = 9 ohms. If the resistors are all in series, the total resistance is this: Rt = 1 / ( {1 / R1} + {1 / R2} + {1 / R3} ...) or Rt = 1 / ( {1/3} + {1/3} + {1/3}) = 1 / (3/3) = 1 / 1 = 1 ohm We can shortcut that when we have identical resistors in parallel. The total resistance will be the value of one of them divided by the number that are in parallel. So we'd have: Rt = (3 / 3) = (1 / 1) = 1 ohm If two are in series with one across them in parallel, the total resistance is found for each individual parallel branch and then the parallel branches (which have been reduced to a single equivalent resistance) can be taken into the parallel resistors equation and the total equivalent resistance calculated. In this case, one branch has two series resistors of three ohms. The total for that branch is 6 ohms, which we find by just adding them up. Now we have a 6 ohm (equivalent) resistor in parallel with a 3 ohm resistor. Take them into the equation and calculate. It's like this: Rt = 1 / ({1/6 } + { 1/3 }) = 1 / ({ 1/6 } + { 2/6 }) = 1 / ( { 3/6 }) = 1/ (1/2) = 2 ohms
Three 8.0-W resistors are connected in series What is their equivalent resistance?
Three 8.0-W resistors are connected in parallel. What is their equivalent resistance?
What is the resistance between any two of the corners if Three 3 ohm resistors are connected such that it make triangle.?
2 ohms. It is like connecting two 3 ohm resistors in series and then these two series resistors are connected in parallel with third 3 ohm resistor in parallel
How do you calculate the joint resistance in a series circuit with three resistances?
The resistance of a series circuit is simply the sum of the individual resistors.
I have three identical resistors all at 100 ohms in what combination can I put them to obtain a resistance of 66.7?
If you have three 100 ohm resistors, and you want an equivalent resistor of 66.7 ohms, put two resistors in series, and then parallel the third resistor across the first two. Resistors in series: R1 + R2 Resistors in parallel: R1 * R2 / (R1 + R2) This example: Two 100 ohm resistors in series: 100 + 100 = 200 A 100 ohm resistor in parallel with a 200 ohm resistor: 100 * 200 / (100 + 200) = 66.7
What is the equivalent resistance when resistors are in a cube?
The equivalent resistance, from corner to corner, of 12 resistors connected in a cube is 5/6 that of a single resistor.Proof:Start from one corner and flow current through to the opposite corner. You have three resistors. Each of those three resistors is connected to two resistors, in a crisscross pattern. Those six resistors are then connected to three resistors which are connected to the other corner. By symmetry, the voltages at the upper junctions are the same, and then same can be said for the lower junction. You can then simplify the circuit by shorting out the upper junctions and (separately) the lower junctions. This means the circuit is equivalent to three resistors in parallel, in series with six resistors in parallel, in series with three resistors in parallel. This is 1/3 R plus 1/6 R plus 1/3 R, or 5/6 R.
What is the total resistance in a circuit that contains three 60 resistors connected in series?
The total effective resistance of resistors in series is the sum of the individual resistances.Three 60-ohm resistors in series have a total effective resistance of (60 + 60 + 60) = 180 ohms.
Twelve identical wires of resistance 6 ohm each are arranged to form the edges of cube the effective resistance bw the opp corners of cube is?
The effective resistance between opposite corners of a cube comprised of twelve 6 ohm resistors, one at each edge, is 5 ohms. There are several ways to solve this. One approach is to build a system of 12 equations in 12 unknowns, and solve them. Another approach is this... Consider that there are three resistors leaving the input node, and there are three resistors entering the output node. In between those three resistors, there are six resistors in a criss-cross matrix. (Draw it out, flattened, to see this.)Inspecting the six resistors in the center, you note that they are completely symmetrical. Since they are symmetrical, you can conclude that the voltage at the junction between the three input resistors and the six others is the same voltage. The same goes for the three output resistors. Said another way, the voltage across the three input resistors and the three output resistors is the same. Given two or more nodes in a circuit having the same voltage, you can draw a wire connecting them, i.e. a resistor of zero ohms. This does not change the characteristics of the circuit in any way, because zero voltage across any resistance is still zero amperes. Now that you have made these connections, look at the circuit. It has simplified to three parallel resistors, in series with six parallel resistors, in series with three parallel resistors. Three 6 ohm resistors in parallel is 2 ohms. Six 6 ohm resistors in parallel is 1 ohm. Three more 6 ohm resistors in parallel is 2 ohms. The total resistance is 2 + 1 + 2 ohms, or 5 ohms.
A 24 volt battery supplies three devices wired in parallel that each have a resistance of 9 ohms. What is the voltage drop across each device?
A simple circuit has three resistors connected in series. The resistors are 14 ohms 12 ohms and 9 ohms. What is the total resistance of the circuit?
How can three resistors of 2 ohm 3 ohm and 4 ohm be connected to give a total resistance of 4 ohm?
2 in series with 3&4 in parallel
The total resistance of a circuit with three 30 ohm resistors in parallel is ohms?
If the parallel resistors are equal, then the total resistance (in this case, with three resistors) will decrease by a factor of 3. I suggest you verify this with the standard formula for parallel resistance: 1/R = 1/R1 + 1/R2 + 1/R3, replacing the value 30 for R1, R2, and R3, and calculating R, the combined resistance.
Which is the total resistance of a circuit that contains twenty 100 ohms resistors connected in parallel?
When resistors of the same value are wired in parallel, the total equivalent resistance (ie the value of one resistor that acts identically to the group of parallel resistors) is equal to the value of the resistors divided by the number of resistors. For example, two 10 ohm resistors in parallel give an equivalent resistance of 10/2=5Ohms. Three 60 ohm resistors in parallel give a total equivalent resistance of 60/3 = 20Ohms. In your case, four 200 Ohm resistors in parallel give 200/4 = 50 Ohms total.
Three equal resistors connected in series across a voltage source dissipate a power of 10W what sould be the power desipated if same resistors are connected in parallel?
If the voltage and resistance values remain the same the power dissipated will be 90 W.
