# Electrical Engineering - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 24)
24.
Six resistors are in parallel. The two lowest-value resistors are both 1.2 k. The total resistance
is less than 6 k
is greater than 1.2 k
is less than 1.2 k
is less than 600
Explanation:
No answer description is available. Let's discuss.
Discussion:
8 comments Page 1 of 1.

Pessat said:   4 years ago
Let, assume 1.2 low, and 0ther 4 resistor value 2400.

1/1200 + 1/1200 + 1/2400 + 1/2400 + 1/2400 + 1/2400,
= 2400/8 = 300 ohm,
Option D.

Noor alsarraf said:   8 years ago
The two least value 1.2 k ohm.
We can choose the other four as 1.3 k ohm.
So 1/Rt = 1/1.2 + 1/1.2 + 1/1.3 + 1/1.3 + 1/1.3 + 1/1.3.
So, R t = 213.8 ohm.

Which is less than 600 ohm.

Chetan Krishna said:   9 years ago
Effective resistance of a parallel circuit is always less than the least resistance in the circuit so the combination of two 1.2k resistors can be replaced by a single resistance of 600 ohms.

So the new least resistance is 600 ohms. Therefore R < 600 ohms is the correct answer.

2 least resistor value is 1200 ohm.
Remaining 4 resistors, for each will consider 601 ohm,
Then,

1/R=(4/Ra)+(2/Rb) ie Ra=R1+R2+R3+R4, Rb=R5+R6.

1/R=(4*600)+(2*601)/(601*600).

1/R=100.11 ohm.

So the option D is correct.

Sumeet Pandey said:   1 decade ago
@Ajay.

You are doing a mistake don't consider resistance value for both of them as 1.2k ohm.

It is for both individually. Thanks.

Mujffar Ali said:   1 decade ago
Because the parallel combination of these two lowest resistors is

=1.2/2=600

now the parallel combination of these 5 resistors 4 remainig and 600 will be less than 600.
so ans d is correct

If it is 601 then are would be 601/4=150.

So total resistance is 150*600/150+600 i.e. 120 which is less than 600.

That is why ajay the answer is less than 600.