Equation of Equivalent Resistance in Parallel is 1/Equivalent Resistance = 1/Resistance 1+ 1/Resistance 2 +1/Resistance 3
If the resistance of the resistor is 3 Ohms then
1/Req= 1/3 + 1/3 = 1/3
1/Req= 3/3
Then cross multiply so
3Req = 3
Then isolate variable for Req (divide both sides by 3)
answer
1 Req= 3/3
=1 Ohm
Equivalent Resistance is 1 Ohm
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.
connect 2 2ohm resistors in parallel and connect it to a series 2ohm resistor
5CommentThe plural of ohm is ohms, not ohm's!
The total resistance is 5 ohms. Scroll down to related links and look at "Parallel Resistance Calculator".
Total resistance is 120 ohms. The 120VAC will be split evenly over this 120 ohm load, so every ohm of resistance gets a volt. So there will be a 40 volt drop across the 40 ohm resistor.
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.
Join two resistors in parallel (effective resistance = 1 ohm). Join the third in series with the parallel combination (2 + 1 = 3 ohm).
The combined resistance is 7.6049 ohms.
Let the equivalent resistance be R and let there be 3 resistors namely R1,R2 and R3, connected in a parallel way. Now, the relation is: 1/R = 1/R1 + 1/R2 + 1/R3
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.
There is no 'equivalent resistance' for three resistors connected in star.
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.
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
Equation for Equivalent Resistance in Series isReq= R1+R2+R3+...........If each resistor is equal to 3OhmsthenReq= R1+R2+R3Req=3+3+3Req=9 OhmsThe Equivalent resistance is 9 Ohms.
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
connect 2 2ohm resistors in parallel and connect it to a series 2ohm resistor
6 separate 18kOhm resistors wired in series make a total resistance of:18k+18k+18k+18k+18k+18k which is 108 kOhm.6 separate 18kOhm resistors wired in parallel make a total resistance of:1/[(1/18k) + (1/18k) + (1/18k) + (1/18k) + (1/18k) + (1/18k)] which is 3 kOhm.