Two identical resistance are first connected in series and then in parallel. Find the ratio of equivalent resistance in both cases.
Answer
650.1k+ 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.
Recently Updated Pages
What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

A Paragraph on Pollution in about 100-150 Words

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

