Question

if two resistances are joined in series then equivalent resistance is S .if they are connected un parallel then equivalent resistance is P. and S=nP.FIND n

Answer 1

$let \: resistances \: be \: r _{1} \: and \: r _{2} \\ when \: in \: series \: \\ s = r_{1} + r_{2} - eq1\\ when \: in \: parallel \\ p= \frac{r_{1}r _{2} }{r _{1} + r _{2} } - eq2 \\ eq1 \div eq2\\ = \frac{s}{p} = \frac{(r _{1} + r _{2} )^{2} }{r _{1}r _{2}} \\ so \: n \: = \: \frac{ {(r _{1} + r _{2}) }^{2} }{r _{1}r _{2} }$
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Answer 2

HEY BUDDY..!!!

HERE'S THE ANSWER...

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▶️ First we'll solve them individually

# For series , let 2 resistors be R 1 , R 2

=> We know net resistance for resistors in series is equal to sum of all resistors combined in series.

⏺️ So ,

=> S = R 1 + R 2

# For parallel ,

=> We know net resistance for resistors in series is equal to sum of reciprocal of all resistors combined in series.

⏺️ So ,

=> P = 1 / R 1 + 1 / R 2

=> P = R 1 . R 2 / R 1 + R 2..

▶️ Now putting values in S = n P

=> R 1 + R 2 = n ( R 1 . R 2 / R 1 + R 2 )

=> ( R 1 + R 2 )^2 = n ( R 1 . R 2 )

⏺️ For minimum , R 1 = R 2 = R

=> ( R + R )^2 = n ( R . R )

=> ( 2 R )^2 = n R^2

=> 4 R^2 = n R^2

=> [ n = 4 ]

HOPE HELPED..

GOOD NIGHT...

:-)