Math, asked by killerwastaken, 3 months ago

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Answers

Answered by BrainlyYuVa
2

Solution

Given :-

  • Two circuit are A. & B. , where some resister connected in parallel & series .

Find :-

  • Effective Resistance of both circuit .

Explanation

In First Circuit

, & connected in series ,

then effective Resistance will be 1/R

Using Formula

\boxed{\underline{\tt{\red{\:\dfrac{1}{R}\:=\dfrac{1}{R^1}\:+\dfrac{1}{R^2}\:+\dfrac{1}{R^3}}}}}

==> 1/R = 1/R¹ + 1/R² + 1/R³

Keep all above Values,

==> 1/R = 1/90 + 1/45 + 1/180

==> 1/R = (2 + 4 + 1)/180

==> 1/R = 5/180

==> 1/R = 1/36

Or,

==> R = 36 Ohm

___________________________

In Second circuit

R1 & R⁴ connected in series & & connected in Parallel.

We Know,

\boxed{\underline{\tt{\green{\:\dfrac{1}{R}\:=\:\dfrac{1}{R^1}\:+\:\dfrac{1}{R^2}\:+\dfrac{1}{R^3}}}}}

==> 1/R' = 1/30 +1/50

==> 1/R' = (50 +30)/1500

==> 1/R' = 80/1500

==> 1/R' = 8/150

Or,

==> R' = 150/8

==> R' = 18.75

Now Calculate series , and R' now also perform as a series resistance,

==> R = 18.75 + 20 + 20

==> R = 58.75 ohm

Hence

  • Effective Resistance of first circuit = 36 ohm

  • Effective Resistance of second Circuit = 58.75 ohm

___________________

Answered by vinodrenusanju
0

 1) Rs = R1 + R2 + R3

         = 90 + 45 + 180

    Rs = 315//

V = IR

9 = I(315)

9/315 = I

0.028 = I

 2) Rs = R1 + R2 + R3 + R4

          = 20 + 30 + 50 + 20

      Rs = 120//

V = IR

10 = I (120)

10/120 = I

0.083 = I

TOTAL RESISTANS IS

R = Rs + Rs

R = 315 + 120

R = 435//

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