Answer
Verified
428.4k+ views
Hint: 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.
Distinguish the formed Wheatstone bridges in the given circuit which are balanced and hence eradicate the resistances through which no current flows. Then calculate the net resistance through A and C from the new resolved circuit.
Complete step by step solution:
Let us name all the resistors from 1 to 8 as:
Where $ {{R}_{1}}={{R}_{2}}={{R}_{3}}=.........={{R}_{8}}=R $
Here we can see that
$ \dfrac{{{R}_{8}}}{{{R}_{4}}}=\dfrac{{{R}_{6}}}{{{R}_{3}}}=1 $
Also,
$ \dfrac{{{R}_{1}}}{{{R}_{8}}}=\dfrac{{{R}_{2}}}{{{R}_{6}}}=1 $
So AOCD and AOCB form the balanced Wheatstone bridges.
This means that $ {{R}_{5}} $ and $ {{R}_{7}} $ have no flow of current through them and hence provide no resistance in the circuit.
The new circuit becomes:
In this new circuit we can clearly see that $ {{R}_{1}} $ and $ {{R}_{2}} $ , $ {{R}_{3}} $ and $ {{R}_{4}} $ , $ {{R}_{6}} $ and $ {{R}_{8}} $ are in series and their resultants are parallel to each other. This final resultant gives us the total resistance required between points A and C.
Now doing the calculations:
$ \begin{align}
& R'={{R}_{1}}+{{R}_{2}} \\
& =R+R \\
& =2R \\
& R''={{R}_{3}}+{{R}_{4}} \\
& =R+R \\
& =2R \\
& R'''={{R}_{6}}+{{R}_{8}} \\
& =R+R \\
& =2R \\
\end{align} $
Now,
$ \begin{align}
& \dfrac{1}{{{R}_{AC}}}=\dfrac{1}{R'}+\dfrac{1}{R''}+\dfrac{1}{R'''} \\
& =\dfrac{1}{2R}+\dfrac{1}{2R}+\dfrac{1}{2R} \\
& \dfrac{1}{{{R}_{AC}}}=\dfrac{3}{2R} \\
\end{align} $
Or
$ {{R}_{AC}}=\dfrac{2R}{3} $
Therefore, option (A) is the correct answer.
Note:
Balanced Wheatstone bridges are used to resolve a complicated circuit of resistors into a simpler form to calculate the net resistance, voltage or current between two points. The Wheatstone bridge is used for measuring the very low resistance values precisely. Wheatstone bridge along with an operational amplifier is used to measure the physical parameters like temperature, strain, light, etc.
Distinguish the formed Wheatstone bridges in the given circuit which are balanced and hence eradicate the resistances through which no current flows. Then calculate the net resistance through A and C from the new resolved circuit.
Complete step by step solution:
Let us name all the resistors from 1 to 8 as:
Where $ {{R}_{1}}={{R}_{2}}={{R}_{3}}=.........={{R}_{8}}=R $
Here we can see that
$ \dfrac{{{R}_{8}}}{{{R}_{4}}}=\dfrac{{{R}_{6}}}{{{R}_{3}}}=1 $
Also,
$ \dfrac{{{R}_{1}}}{{{R}_{8}}}=\dfrac{{{R}_{2}}}{{{R}_{6}}}=1 $
So AOCD and AOCB form the balanced Wheatstone bridges.
This means that $ {{R}_{5}} $ and $ {{R}_{7}} $ have no flow of current through them and hence provide no resistance in the circuit.
The new circuit becomes:
In this new circuit we can clearly see that $ {{R}_{1}} $ and $ {{R}_{2}} $ , $ {{R}_{3}} $ and $ {{R}_{4}} $ , $ {{R}_{6}} $ and $ {{R}_{8}} $ are in series and their resultants are parallel to each other. This final resultant gives us the total resistance required between points A and C.
Now doing the calculations:
$ \begin{align}
& R'={{R}_{1}}+{{R}_{2}} \\
& =R+R \\
& =2R \\
& R''={{R}_{3}}+{{R}_{4}} \\
& =R+R \\
& =2R \\
& R'''={{R}_{6}}+{{R}_{8}} \\
& =R+R \\
& =2R \\
\end{align} $
Now,
$ \begin{align}
& \dfrac{1}{{{R}_{AC}}}=\dfrac{1}{R'}+\dfrac{1}{R''}+\dfrac{1}{R'''} \\
& =\dfrac{1}{2R}+\dfrac{1}{2R}+\dfrac{1}{2R} \\
& \dfrac{1}{{{R}_{AC}}}=\dfrac{3}{2R} \\
\end{align} $
Or
$ {{R}_{AC}}=\dfrac{2R}{3} $
Therefore, option (A) is the correct answer.
Note:
Balanced Wheatstone bridges are used to resolve a complicated circuit of resistors into a simpler form to calculate the net resistance, voltage or current between two points. The Wheatstone bridge is used for measuring the very low resistance values precisely. Wheatstone bridge along with an operational amplifier is used to measure the physical parameters like temperature, strain, light, etc.
Recently Updated Pages
10 Examples of Evaporation in Daily Life with Explanations
10 Examples of Diffusion in Everyday Life
1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSE
What is the meaning of celestial class 10 social science CBSE
What causes groundwater depletion How can it be re class 10 chemistry CBSE
Under which different types can the following changes class 10 physics CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In the tincture of iodine which is solute and solv class 11 chemistry CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE