# What is the resistance when three resistors are connected in parallel, if the same connected in series has a resultant resistance of R?

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We know that when two resistors a and b are connected in series the equivalent resistance of the system is a + b.

Now we are given that three resistors connected in series give an equivalent resistance of R. Now this can be achieved using different combinations, let’s take the simplest where the resistances of the three are equal, so each resistor is of R/3 ohm.

When two resistors a and b are connected in parallel the equivalent resistance is given as (1/a + 1/b) ^ (-1). So for the three resistors connected in parallel the resistance is (1/(R/3) + 1/(R/3) + 1/(R/3)) ^-1 = (3 / (R/3)) ^-1 = R/9.

This value for equivalent resistance is not constant but will change depending on the original resistors we choose. For example, if we had taken R/2, R/4 and R/4 or any other similar combinations, the result would have been different. The result given above is only one example.

When the resistances are connected in seies the same cirrent is paasinf through the 3 resistors R1, R2 anand R3. Therefore

v = v1+v2+v3. Or

Therefore R = R1+R2+R3

When we connect the resistos in parallel, the same voltage is accross the 3 circutes and the sum of all current in all 3 resistors is equal to the current of the system.

I = I1+I2+I3

V/R' = V/R1+V/R2+V/R3

Or R' = 1/{1/R1+1/R2+1R3)

R' = 1/ (1/R++1/R+1/R) = **R/3**, if R1=R2 = R2.

Thus the resistance is reduced in parallel connection.