
The effective resistance between the points X and Y in the given circuit is
$\begin{align}
& \text{A}\text{. 8/3}\Omega \\
& \text{B}\text{. 8}\Omega \\
& \text{C}\text{. 4}\Omega \\
& \text{D}\text{. 2}\Omega \\
\end{align}$
Answer
584.4k+ views
Hint: Obtain the expression to obtain the total resistance of more than one resistance connected in series and parallel to each other. Simplify the circuit one by one by finding the resistance between two points first. This way we can find the resistance between any points.
Complete step-by-step answer:
If we have two resistance ${{R}_{1}}$ and ${{R}_{2}}$ connected in series, the total resistance of the circuit can be written as,
$R={{R}_{1}}+{{R}_{2}}$
Again, if we have two resistance ${{R}_{1}}$ and ${{R}_{2}}$ connected in parallel to each other, the total resistance of the circuit can be found out as,
$\dfrac{1}{R}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}$
To find the resistance between the points X and Y, let us first simplify the circuit.
In the circuit, the resistance in XP and PQ are in series. So, the total resistance will be, $4\Omega +4\Omega =8\Omega $
Now, this $8\Omega $resistance will be in parallel to the $8\Omega $ resistance in XQ.
Now, let the total resistance between XQ is ${{R}_{1}}$
So, we can write that,
$\begin{align}
& \dfrac{1}{{{R}_{1}}}=\dfrac{1}{8}+\dfrac{1}{8} \\
& \dfrac{1}{{{R}_{1}}}=\dfrac{1}{4} \\
& {{R}_{1}}=4\Omega \\
\end{align}$
Now, the $4\Omega $resistance in XQ and the $4\Omega $ resistance in QY are in series. So, the total resistance will be, $4\Omega +4\Omega =8\Omega $
Now, again this $8\Omega $ resistance is in parallel with the $4\Omega $ resistance between XY.
Let the total resistance between XY be R.
So, we can write that,
$\begin{align}
& \dfrac{1}{R}=\dfrac{1}{8}+\dfrac{1}{4} \\
& \dfrac{1}{R}=\dfrac{3}{8} \\
& R=\dfrac{8}{3}\Omega \\
\end{align}$
So, the total resistance between the points X and Y is $\dfrac{8}{3}\Omega $.
The correct option is (A).
Note: If we connect two or more resistance in series, the total resistance of the circuit will always become higher than any of the resistance connected in series. If we connect two or more resistance in parallel to each other, the total resistance of the circuit will always be less than any of the resistance connected in parallel.
Complete step-by-step answer:
If we have two resistance ${{R}_{1}}$ and ${{R}_{2}}$ connected in series, the total resistance of the circuit can be written as,
$R={{R}_{1}}+{{R}_{2}}$
Again, if we have two resistance ${{R}_{1}}$ and ${{R}_{2}}$ connected in parallel to each other, the total resistance of the circuit can be found out as,
$\dfrac{1}{R}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}$
To find the resistance between the points X and Y, let us first simplify the circuit.
In the circuit, the resistance in XP and PQ are in series. So, the total resistance will be, $4\Omega +4\Omega =8\Omega $
Now, this $8\Omega $resistance will be in parallel to the $8\Omega $ resistance in XQ.
Now, let the total resistance between XQ is ${{R}_{1}}$
So, we can write that,
$\begin{align}
& \dfrac{1}{{{R}_{1}}}=\dfrac{1}{8}+\dfrac{1}{8} \\
& \dfrac{1}{{{R}_{1}}}=\dfrac{1}{4} \\
& {{R}_{1}}=4\Omega \\
\end{align}$
Now, the $4\Omega $resistance in XQ and the $4\Omega $ resistance in QY are in series. So, the total resistance will be, $4\Omega +4\Omega =8\Omega $
Now, again this $8\Omega $ resistance is in parallel with the $4\Omega $ resistance between XY.
Let the total resistance between XY be R.
So, we can write that,
$\begin{align}
& \dfrac{1}{R}=\dfrac{1}{8}+\dfrac{1}{4} \\
& \dfrac{1}{R}=\dfrac{3}{8} \\
& R=\dfrac{8}{3}\Omega \\
\end{align}$
So, the total resistance between the points X and Y is $\dfrac{8}{3}\Omega $.
The correct option is (A).
Note: If we connect two or more resistance in series, the total resistance of the circuit will always become higher than any of the resistance connected in series. If we connect two or more resistance in parallel to each other, the total resistance of the circuit will always be less than any of the resistance connected in parallel.
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