In order to determine this, it will be necessary to find which resistor 'maxes out' at the lowest voltage. This can be found using the equation Vi=sqrt (Pi*Ri) for each resistor, where Pi is the power rating of resistor i and Ri is the value of resistor i. Once this is found, the power dissipation of each other resistor can be found using the equation Pi=(Vl^2)/Ri, where Vl is the voltage that maxes out the resistor which maxes out at the lowest voltage, and Ri is the resistance of each resistor. The equivalent power rating would then be the sum of the power dissipated across each resistor.
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.
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
If two 1-ohm resistors are connected in parallel, their resistance is 0.5 ohms. If they are connected in series, their resistance is 2 ohms. It is not possible to connect only two resistors in series parallel.
the voltage across that resistor will increase if it is in series with the other resistors. the current through that resistor will increase if it is in parallel with the other resistors.
Yes. When resistors are connected in "parallel" (all the left ends connected together and all the right ends connected together) the effective resistance is always less then the smallest resistor in the group. For example If you connected a 2 ohm in parallel with a 4 ohm the effective resistance is 1.33 ohm. To your question; if you connect N equal resistors R in parallel the effective resistance would be R/N . The formula for calculating effective resistance R of a group R1, R2, R3, ... in parallel is: 1/R = 1/R1 + 1/R2 + 1/R3 + .... Note; write the right side as a single fraction by getting a common denomenator then invert to get R.
It depends upon the connection of the resistors, if the resistors are connected in parallel then the voltage is same where as in case of resistors connected in series the voltage is different across different resistors.
The equivalent resistance of resistors connected in series is simply the sum of their individual resistances. Therefore, the equivalent resistance of three 8.0-W resistors connected in series is 24.0 W.
It represents that two resistors are connected in parallel.
If three equal resistors are connected in parallel, the equivalent resistance will be one-third of the resistance in series. This lower resistance will result in a higher current flowing through the resistors when connected in parallel compared to when they are in series. Therefore, the power dissipated by the resistors in parallel will be greater than 10W.
Resistors are in series if they are connected end-to-end, creating one path for current to flow. Resistors are in parallel if they are connected side by side, providing multiple paths for current to flow. You can determine if resistors are in series or parallel by examining how they are connected in a circuit.
Resistors connected in parallel have the same voltage across them, while resistors connected in series have the same current passing through them. In a parallel configuration, the total resistance decreases as more resistors are added, while in a series configuration, the total resistance increases.
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.
When resistors are connected in series, the total resistance is the sum of the individual resistances. When resistors are connected in parallel, the total resistance is less than the smallest individual resistance.
Both resistors will have the voltage of the battery.
The resistors should be connected in parallel .
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
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.