# 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

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.

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.

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.

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

Shiva said:
1 decade ago

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.

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.

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

=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

Manisha said:
1 decade ago

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.

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.

Ajay said:
1 decade ago

R for two lowest value resistor = 1.2/2 kohm

= 600 ohm

other four resistor vallu is higher then this two.

if it 601 ohm then

R = 601*4+1200

= 600.66 ohm

Answer D is wrong

= 600 ohm

other four resistor vallu is higher then this two.

if it 601 ohm then

R = 601*4+1200

= 600.66 ohm

Answer D is wrong

Post your comments here:

Your comments will be displayed after verification.

Quick links

Quantitative Aptitude

Verbal (English)

Reasoning

Programming

Interview

Placement Papers

© IndiaBIX™ Technologies