Question

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Answer 1

`\star\:{\underline{\underline{\sf{\purple{ \: Question \: 19\: }}}}}`

`{\large{\textsf{\textbf{\underline{\underline{Given \: :}}}}}}`

☆ Number of resistors = 5

☆ Resistance of each resistors =
`\tt (1)/(5) \text{\O}mega`

`{\large{\textsf{\textbf{\underline{\underline{To \: Find \: :}}}}}}`

☆ Maximum resistance which can be made.

`{\large{\textsf{\textbf{\underline{\underline{Solution \: :}}}}}}`

• To get maximum resistance, connect the all the given resistances in series.

For,

Series combination :-

`\longrightarrow \sf R_(s) = R_(1) + R_(2) + R_(3) + R_(4) + R_(5)`

`\longrightarrow \sf R_(s)= (1)/(5) + (1)/(5) + (1)/(5) + (1)/(5) + (1)/(5)`

`\longrightarrow \sf R_(s) = (1 + 1 + 1 + 1 + 1)/(5)`

`\longrightarrow \sf R_(s) = \cancel{ (5)/(5) }`

`\longrightarrow \sf R_(s) = \purple{1 \: \text{\O}mega}`

`\therefore` Maximum resistance is 1 Ω.

`\star\:{\underline{\underline{\sf{\red{ \: Question \: 20\: }}}}}`

`{\large{\textsf{\textbf{\underline{\underline{Given \: :}}}}}}`

☆ Number of resistors = 5

☆ Resistance of each resistors =
`\tt (1)/(5) \text{\O}mega`

`{\large{\textsf{\textbf{\underline{\underline{To \: Find \: :}}}}}}`

☆ Minimum resistance which can be made.

`{\large{\textsf{\textbf{\underline{\underline{Solution \: :}}}}}}`

• To get minimum resistance, connect the all the given resistances in parallels.

For,

Parallel combination :-

`\longrightarrow \sf (1)/(R_(p))=(1)/(R_1)+(1)/(R_2)+(1)/(R_3)+(1)/(R_4)+(1)/(R_5)`

`\longrightarrow \sf(1)/(R_(p))= (1)/( (1)/(5) ) + (1)/( (1)/(5) ) + (1)/( (1)/(5) ) + (1)/( (1)/(5) ) + (1)/( (1)/(5) )`

`\longrightarrow \sf(1)/(R_(p))= 1 * (5)/(1) +1 * (5)/(1) +1 * (5)/(1) + 1 * (5)/(1) + 1 * (5)/(1)`

`\longrightarrow \sf(1)/(R_(p))= (5)/(1) + (5)/(1) + (5)/(1) + (5)/(1) + (5)/(1)`

`\longrightarrow \sf(1)/(R_(p))= 25`

`\longrightarrow \sf R_(p)= \red{ (1)/(25) \: \text{\O}mega}`

`\therefore` Minimum resistance is
`(1)/(25)` Ω.

`{\large{\textsf{\textbf{\underline{\underline{Note \: :}}}}}}`

☆ Figure in attachment.

`{\underline{\rule{290pt}{2pt}}}`