# Physics Problem about Wheatstone Bridge.?

I completely forgot about this topic, and my professor didn't explain a single thing about this in his lecture.

Is it also called Wheatstone Bridge? It doesn't have any resistor in the middle.Also, how can you solve it?

Thanks a lot.

### 3 Answers

- Andrew SmithLv 75 months ago
This is not a Wheatstone bridge. All you have is a 35 ohm resistor in parallel with a 14 ohm resistor (10 ohm equivalent) and a 36 ohm resistor in parallel with a 12 ohm resistor (9 ohm equivalent.) These two are in series with each other. 19 ohms. Now I cannot imagine that your lecturer has failed to mention PARALLEL and SERIES circuits. Were you taking adequate notes?

- FiremanLv 75 months ago
A Wheatstone bridge is an electrical circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. The primary benefit of the circuit is its ability to provide extremely accurate measurements.

The unknown resistance Rx is to be measured; resistances R1, R2 and R3 are known and R2 is adjustable. When the measured voltage VG is 0, both legs have equal voltage ratios: R2/R1 = Rx/R3 and Rx= R3R2/R1.

However in the given diagrame:

=>14/12 = 1.17

& 35/36 = 0.97

=>R2/R1 ≠ R4/R3

=>This is not a Wheatstone bridge.

Further it can be solved as under:

Let Left point is A and the middle two points are B(since both are connected without any resister) and the right point is C.

=>35 & 14 Ω are in parallel between A & B, they can be replaced by their net resistance (R1) by

=>1/R1 = 1/35 + 1/14

=>R1 = 10 Ω

=>36 & 12 Ω are in parallel between B & C, they can be replaced by their net resistance (R2) by

=>1/R2 = 1/36 + 1/12

=>R1 = 9 Ω

Now R1 & R2 are in series between A & C, thus the net resistance (R) between A & C,

=>By R = R1 + R2

=>R = 10 + 9 = 19 Ω

- AshLv 75 months ago
It just simplifies to 35Ω //14Ω and 36Ω // 12Ω

Equivalent resistance will be 35Ω //14 + 36Ω // 12Ω

1/R₁ = 1/35Ω + 1/14Ω

1/R₁ = 2/70 + 5/70

1/R₁ = 7/70

R₁ = 10

1/R₂ = 1/36Ω + 1/12Ω

1/R₂ = 1/36 + 3/36

1/R₂ = 4/36

R₂ =9

R = R₁ + R₂ = 10 + 9 = 19 Ω