The resistance of two or more resistors connected in series is the sum of the
individual resistances. (If any of the connections between them is sloppy and
involves some resistance at the connection, then that also has to be added in.)
When resistors are wired in series, their resistances are added to find the total resistance. If they are run in parallel, or series-parallel, the formula is different
Resistors are wired in series when they are connected in a line. The current flows through the resistors one after the other.
What would the measured ohms be for two 100 ohm resistors wired in series? Two 100 ohm resistors wired in series measure 200 ohms.
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.
30 ohms.
It depends on the value of the three resistors. If they are equal, then each resistor has 5 volts across it.
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?
What do you mean by a 'parallel delta' circuit -is there such a connection.
It represents that two resistors are 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.
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
Yes, fuses have a low resistance in order to not dominate the consumption of the circuit's power, such that the desired appliance wired in series with the fuse consumes the bulk of the power.