
Five resistances of R$\Omega $ were taken. First three resistances are connected in parallel combination and rest two are connected in series combination, then the equivalent resistance is:
$
{\text{A}}{\text{. }}\dfrac{3}{7}R\Omega \\
{\text{B}}{\text{. }}\dfrac{7}{3}R\Omega \\
{\text{C}}{\text{. }}\dfrac{7}{8}R\Omega \\
{\text{D}}{\text{. }}\dfrac{8}{7}R\Omega \\
$
Answer
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Hint: For the given combination of resistances, first we need to find out the equivalent resistance for the three resistances connected in parallel with each other. Then we need to find equivalent resistances with the rest of two resistances connected in series with the parallel combination of first three resistances. This value will be the required answer.
Formula used:
Consider two resistances R and r connected in series with each other than their equivalent resistance is given as follows:
$R' = R + r$
If these two resistances are connected in parallel with each other than their equivalent resistance is given as follows:
$\dfrac{1}{{R'}} = \dfrac{1}{R} + \dfrac{1}{r}$
Complete answer:
We are given five resistance of R$\Omega $ each. We connect the first three resistances in parallel combination with each other while the rest two are connected in series combination with them. Let us first consider the resistances connected in parallel with each other. In parallel combination, let R’ be the equivalent resistance which can be calculated in the following way.
$
\dfrac{1}{{R'}} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{3}{R} \\
\Rightarrow R' = \dfrac{R}{3} \\
$
Now this equivalent of the three resistances connected in parallel with each other is connected in series with the other two resistances. Let R’’ be the equivalent resistance of this combination which can be obtained in the following way.
$R'' = R' + R + R = \dfrac{R}{3} + 2R = \dfrac{{7R}}{3}$
This is the required value of equivalent resistance for the given combination of five resistances.
Hence, the correct answer is option B.
Note:
It should be noted that the methods for calculating equivalent resistances for parallel and series combination of resistances is based on the fact that in series combination of resistances, the current remains the same while voltage gets distributed between resistances. In parallel combination of resistances, the voltages remain the same while the current gets distributed between the resistances.
Formula used:
Consider two resistances R and r connected in series with each other than their equivalent resistance is given as follows:
$R' = R + r$
If these two resistances are connected in parallel with each other than their equivalent resistance is given as follows:
$\dfrac{1}{{R'}} = \dfrac{1}{R} + \dfrac{1}{r}$
Complete answer:
We are given five resistance of R$\Omega $ each. We connect the first three resistances in parallel combination with each other while the rest two are connected in series combination with them. Let us first consider the resistances connected in parallel with each other. In parallel combination, let R’ be the equivalent resistance which can be calculated in the following way.
$
\dfrac{1}{{R'}} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{3}{R} \\
\Rightarrow R' = \dfrac{R}{3} \\
$
Now this equivalent of the three resistances connected in parallel with each other is connected in series with the other two resistances. Let R’’ be the equivalent resistance of this combination which can be obtained in the following way.
$R'' = R' + R + R = \dfrac{R}{3} + 2R = \dfrac{{7R}}{3}$
This is the required value of equivalent resistance for the given combination of five resistances.
Hence, the correct answer is option B.
Note:
It should be noted that the methods for calculating equivalent resistances for parallel and series combination of resistances is based on the fact that in series combination of resistances, the current remains the same while voltage gets distributed between resistances. In parallel combination of resistances, the voltages remain the same while the current gets distributed between the resistances.
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