Electrical Engineering - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 9)
9.
A certain Wheatstone bridge has the following resistor values: R1 = 10 k
, R2 = 720
, and R4 = 2.4 k
. The unknown resistance is



Discussion:
28 comments Page 1 of 3.
Jojo D. said:
5 years ago
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.
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.
ROOPESH D M said:
1 decade ago
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.
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.
Meenakshi said:
1 decade ago
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.
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.
Mahe said:
1 decade ago
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.
Chetan said:
1 decade ago
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.
So R1*R2=R3*R4,
R1, R2, R4 are given, we need to find R3.
R3 = R1*R2/R4 = 10K*720/2.4K = 3000.
Puja said:
9 years ago
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.
VIJAY said:
10 years ago
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.
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.
Haseeb said:
5 years ago
Wheatstone bridge.
R1R2=R3R4.
Here unknown resistance is R3.
So,
R3 = R1R2/R4.
= 10000 * 720/2400.
R3 = 3000 ohm.
R1R2=R3R4.
Here unknown resistance is R3.
So,
R3 = R1R2/R4.
= 10000 * 720/2400.
R3 = 3000 ohm.
(4)
Tamilarasan said:
1 decade ago
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.
So R2=R3*R4/R1. Sub the value. We will get the answer.
Ramu said:
1 decade ago
This will ans is correct but procedure is wrong.
Wheat stone bridge Rx= (R2/R1) *R3.
Wheat stone bridge Rx= (R2/R1) *R3.
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