
Two identical resistance are first connected in series and then in parallel. Find the ratio of equivalent resistance in both cases.
Answer
566.4k+ views
Hint In series combination, resistances are connected end by end equally and the current through each resistance is the same. In parallel combination one end of each resistance connected at the same terminal and other to the other terminal, then voltage across each resistance is the same as voltage of source.
Complete step by step solution
Let $R$ be the value of each resistance.
For series combination, resistances are connected end by end and equivalent resistance is given by formula
${R_S} = {R_1} + {R_2} + ... + {R_N}$
Here we have two resistance of value $R$ each.
Then, ${R_S} = R + R = 2R$ (1)
For parallel combination, we know that equivalent resistance is given by
$\dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ... + \dfrac{1}{{{R_N}}}$
Here we have two resistance of value $R$ each.
Then, $\dfrac{1}{{{R_P}}} = \dfrac{1}{R} + \dfrac{1}{R}$ or ${R_P} = \dfrac{{R \times R}}{{R + R}} = \dfrac{R}{2}$ (2)
Here we need to find the ratio of ${R_P}$ and ${R_S}$.
From equation (1) and (2), we get
\[\dfrac{{{R_P}}}{{{R_S}}} = \dfrac{R}{2} \times \dfrac{1}{{2R}} = \dfrac{1}{4}\]
Hence the answer is $4:1$.
Note If given resistance are not equal our may vary with very high difference value. In parallel combination equivalent resistance is always less than minimum resistance of all resistances connected in parallel combination.
Complete step by step solution
Let $R$ be the value of each resistance.
For series combination, resistances are connected end by end and equivalent resistance is given by formula
${R_S} = {R_1} + {R_2} + ... + {R_N}$
Here we have two resistance of value $R$ each.
Then, ${R_S} = R + R = 2R$ (1)
For parallel combination, we know that equivalent resistance is given by
$\dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ... + \dfrac{1}{{{R_N}}}$
Here we have two resistance of value $R$ each.
Then, $\dfrac{1}{{{R_P}}} = \dfrac{1}{R} + \dfrac{1}{R}$ or ${R_P} = \dfrac{{R \times R}}{{R + R}} = \dfrac{R}{2}$ (2)
Here we need to find the ratio of ${R_P}$ and ${R_S}$.
From equation (1) and (2), we get
\[\dfrac{{{R_P}}}{{{R_S}}} = \dfrac{R}{2} \times \dfrac{1}{{2R}} = \dfrac{1}{4}\]
Hence the answer is $4:1$.
Note If given resistance are not equal our may vary with very high difference value. In parallel combination equivalent resistance is always less than minimum resistance of all resistances connected in parallel combination.
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