Electrical Engineering - Series-Parallel Circuits - Discussion

Discussion :: Series-Parallel Circuits - General Questions (Q.No.9)


A certain Wheatstone bridge has the following resistor values: R1 = 10 k, R2 = 720 , and R4 = 2.4 k. The unknown resistance is

[A]. 24
[B]. 2.4
[C]. 300
[D]. 3,000

Answer: Option D


No answer description available for this question.

Lavanya said: (Mar 4, 2011)  
Explanation please?

Ramu said: (Jun 10, 2011)  
This will ans is correct but procedure is wrong.

Wheat stone bridge Rx= (R2/R1) *R3.

Anand said: (Jul 27, 2011)  
(R1/R2)=(R3/R4),So R4=(R2/R1)*R3

Sai said: (Aug 5, 2011)  
I agree with you anand.

Tamilarasan said: (Aug 11, 2012)  
In a wheatstone bridge the ratio is equal (i.e. R1*R2=R3*R4).

So R2=R3*R4/R1. Sub the value. We will get the answer.

Keerthi said: (Oct 28, 2013)  
Here Rv means R1 or R3?

Meenakshi said: (Feb 15, 2014)  
Here Rv means variable resistance i.e., generally R3.

Formula R4 = R2R3/R1, but here R4 is given, we have to fnd R1, so eqn becomes R1 = R2R3/R4.

Therefore R1 = 720*10k/2.4K = 3000 ohms.

Ankit Tiwari said: (May 27, 2014)  
Normal formula apply here R1/R2 = R3/R4.

Mahe said: (Jun 29, 2014)  
In wheatstone bridge product of two opposite resistors is equal to the another product. Here R1 R2 and R3 wherever we can mention. So these kind of problems we need a circuit or figure.

Nouman said: (Dec 3, 2014)  
@Meenakshi : How do we find R1 ? when its already been given in the question.

Ravi said: (May 9, 2015)  
Formula: R1*R2 = R3*R4.

Therefore R3 = (R1*R2)/R4.

= (10k*720)/2.4 k.

= 3000 ohms.

Roopesh D M said: (Jun 17, 2015)  
Data is insufficient. Because we get different values of resistance depends on the bridge circuit.

We can take R1*R2 = R3*R4.

Or R1*R3 = R2*R4.

Or R1*R4 = R2*R4.

In each case we get different values, so without circuit we can't decide the unknown resistance value.

Chetan said: (Jun 25, 2015)  
In wheatstone bridge product of two opposite resistors is equal to the another product.

So R1*R2=R3*R4,

R1, R2, R4 are given, we need to find R3.

R3 = R1*R2/R4 = 10K*720/2.4K = 3000.

Piya said: (Jun 27, 2015)  
But in wheatstone bridge: R2/R1 = R4/R3.

Parneet Kaur said: (Jul 9, 2015)  
@Ankit Tiwari.

Your formula is wrong. The correct formula is r1*r2=r3*r4.

Megha said: (Jul 16, 2015)  
Nice answer.

Vijay said: (Oct 16, 2015)  
R1*R2 = R3*R4.

There for R3 = (R1*R2)/R4 720 ohm = 0.72 k ohm.

10k ohm*0.72k ohm/2.4k ohm.

3k ohm = 3k*1000.

Answer: 3000 ohm.

Puja said: (Jan 11, 2017)  
But, If we apply the formula R1/R2 = R3/R4 then answer is different from the answer of R1* R2= R3 * R4 formula. Please explain me again which method is right.

Aditya said: (Oct 22, 2017)  
No, the correct answer is 33.3kohm.

Jagan said: (Apr 12, 2018)  
I am not getting it. Please explain it.

Mari Raj said: (Nov 11, 2018)  
Unknown resister R3=(R1/R2)*R4.
So, R3 = 33.3kohm.

Hardiksinh said: (Mar 11, 2019)  
Yes, R3 = 33.33, because the formula is R1R4=R2R3. Then R3= R1R4/R2.

Darshil said: (Jul 29, 2019)  
The correct formula is R1/R2 = R3/R4.
So the answer comes 33.3kOhm.

W.B said: (Dec 31, 2019)  
R1/R2 = R3/R4.


Jojo D. said: (Mar 29, 2020)  
The answer should be 33.33 k Ohms.

To picture out the circuit, let Nodes A and B are connected across the positive and negative sides of the source respectively. The standard circuit configuration is drawn such that R1 and R2 are connected in series, with the other end of R1 connected to Node A and the other end of R2 connected to Node B. R3 and R4 are also connected in series with the other end of R3 connected Node A and the other end of R4 connected to Node B.

Now, we have a clear picture showing us how the four resistors are inter-connected in the circuit. Based on this configuration, we can derive that R1/R2=R3/R4.

Therefore, R3=(R1R4)/R2. Hence, the answer should be 33.33 kOhms.

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