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Question: A man has five resistors each of value \(\dfrac{1}{5}\Omega \). What is the maximum resistance he ca...

A man has five resistors each of value 15Ω\dfrac{1}{5}\Omega . What is the maximum resistance he can obtain by connecting them?
A. 1Ω1\Omega
B. 5Ω5\Omega
C. 12Ω\dfrac{1}{2}\Omega
D. 25Ω\dfrac{2}{5}\Omega

Explanation

Solution

A man is given five resistors with equal resistance, each has resistance 15Ω\dfrac{1}{5}\Omega . Connect these five resistors both in series and in parallel configurations and find the maximum value of these. When resistors are connected in series, then the equivalent resistance will be the sum of all the resistances. When the resistors are connected in parallel, then the reciprocal of the equivalent resistance is the sum of reciprocals of all the resistances. Use this info to solve the question.

Complete step by step answer: We are given that a man has five resistors each of value 15Ω\dfrac{1}{5}\Omega .
We have to find the maximum resistance he can obtain by connecting them.
Let the five resistors be R1,R2,R3,R4,R5{R_1},{R_2},{R_3},{R_4},{R_5}
And the resistance of these five resistors is the same, R1=R2=R3=R4=R5=15{R_1} = {R_2} = {R_3} = {R_4} = {R_5} = \dfrac{1}{5}
When these resistors are connected in series, then the equivalent resistance will be
Req=R1+R2+R3+R4+R5 R1=R2=R3=R4=R5=15 Req=15+15+15+15+15=1Ω  {R_{eq}} = {R_1} + {R_2} + {R_3} + {R_4} + {R_5} \\\ {R_1} = {R_2} = {R_3} = {R_4} = {R_5} = \dfrac{1}{5} \\\ \to {R_{eq}} = \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} = 1\Omega \\\
When the resistors are connected in series configuration, the equivalent resistance will be 1Ω1\Omega
When these resistors are connected in parallel, then the equivalent resistance will be
1Req=1R1+1R2+1R3+1R4+1R5 R1=R2=R3=R4=R5=15 1Req=1(15)+1(15)+1(15)+1(15)+1(15) 1Req=51+51+51+51+51 1Req=25 Req=125Ω  \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + \dfrac{1}{{{R_4}}} + \dfrac{1}{{{R_5}}} \\\ {R_1} = {R_2} = {R_3} = {R_4} = {R_5} = \dfrac{1}{5} \\\ \to \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} \\\ \to \dfrac{1}{{{R_{eq}}}} = \dfrac{5}{1} + \dfrac{5}{1} + \dfrac{5}{1} + \dfrac{5}{1} + \dfrac{5}{1} \\\ \to \dfrac{1}{{{R_{eq}}}} = 25 \\\ \therefore {R_{eq}} = \dfrac{1}{{25}}\Omega \\\
When these resistors are connected in parallel, the equivalent resistance will be 125Ω\dfrac{1}{{25}}\Omega
As we can see 1Ω1\Omega is greater than 125Ω\dfrac{1}{{25}}\Omega
So when the resistors are connected in a series configuration, the man can obtain maximum resistance which is 1Ω1\Omega
Therefore, the correct option is Option A.

Note: Be careful while calculating the equivalent resistance when the resistors are connected together by a configuration. You may get confused by the equivalent resistance and equivalent capacitance because the formula of series equivalent resistance will be the formula of parallel equivalent capacitance and the formula of parallel equivalent resistance will be the formula of series equivalent capacitance.