Answer
Verified
407.1k+ views
Hint:To find the equivalent resistance between two points the first step we need to do is to find the shortest past by which current can travel and then we need to find out what resistance is out of circuit in that path. We know that current always take the shortest path so it is very important to first know what path our current is taking.
Complete answer:
In the above question, the circuit seems very complex at first sight. So the first thing we need to do is reduce the circuit to a simpler form. When the current starts from point A it will take one of the two paths either it will travel from point A to D via APQSD or from path APRSD, because no other path is possible to complete the circuit. In both the paths the resistance connected near the point C and point B will be useless because no current will be supplied to them and hence, we will not take them into consideration.
The below figure shows all the points with the path specified above.
Now we need to make another diagram that will be the equivalent diagram for the above one.
Now in the above figure we can see that the resistance between point P and S via R and Q are in series.
We know that to find equivalent resistance in series we have to simply add both the resistance
So \[{{R}_{eq(PQS)}}=20\Omega \]
And \[{{R}_{eq(PRS)}}=20\Omega \]
Now the figure becomes,
Now both the $20\Omega $ resistance are in parallel connection with each other, so we know that the equivalent resistance for parallel connection is by
$\dfrac{1}{{{R}_{eq}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}$
So ${{R}_{eq}}=\dfrac{{{R}_{1}}\times {{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}$
Now putting the values, we get,
${{R}_{eq}}=\dfrac{20\times 20}{20+20}$
${{R}_{eq}}=10\Omega $
Now all the resistance between the points A to D becomes in series,
Hence, we can add all of them and get the final equivalent resistance.
$R=10+10+10$
$R=30\Omega $
Hence, we can conclude that option (C) $R=30\Omega $ is the correct answer.
Note:
When a question about finding the equivalent resistance is given, always redraw the diagram in a simpler form and try to reduce as much resistance as possible. Keep everything to minimum so that the circuit will be small and not a complex one. After making an equivalent circuit find the circuits with series and parallel connection and then solve the question by finding the equivalent resistance between them.
Complete answer:
In the above question, the circuit seems very complex at first sight. So the first thing we need to do is reduce the circuit to a simpler form. When the current starts from point A it will take one of the two paths either it will travel from point A to D via APQSD or from path APRSD, because no other path is possible to complete the circuit. In both the paths the resistance connected near the point C and point B will be useless because no current will be supplied to them and hence, we will not take them into consideration.
The below figure shows all the points with the path specified above.
Now we need to make another diagram that will be the equivalent diagram for the above one.
Now in the above figure we can see that the resistance between point P and S via R and Q are in series.
We know that to find equivalent resistance in series we have to simply add both the resistance
So \[{{R}_{eq(PQS)}}=20\Omega \]
And \[{{R}_{eq(PRS)}}=20\Omega \]
Now the figure becomes,
Now both the $20\Omega $ resistance are in parallel connection with each other, so we know that the equivalent resistance for parallel connection is by
$\dfrac{1}{{{R}_{eq}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}$
So ${{R}_{eq}}=\dfrac{{{R}_{1}}\times {{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}$
Now putting the values, we get,
${{R}_{eq}}=\dfrac{20\times 20}{20+20}$
${{R}_{eq}}=10\Omega $
Now all the resistance between the points A to D becomes in series,
Hence, we can add all of them and get the final equivalent resistance.
$R=10+10+10$
$R=30\Omega $
Hence, we can conclude that option (C) $R=30\Omega $ is the correct answer.
Note:
When a question about finding the equivalent resistance is given, always redraw the diagram in a simpler form and try to reduce as much resistance as possible. Keep everything to minimum so that the circuit will be small and not a complex one. After making an equivalent circuit find the circuits with series and parallel connection and then solve the question by finding the equivalent resistance between them.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE