Question

Help Me With This Question Please.

Thank you UwU​

Answer 1

8Ω, 9Ω, xΩ are in series

$\therefore \: R _{1} = 8 + 9 + x = (17 + x)Ω$

3Ω and 5Ω are in series

$\therefore \: R _{2} = 3+ 5 = 8Ω$

Now equivalent resistance of R1 and R2 which are parallel in R3 = 6Ω

[See The Attachment]

$\therefore \frac{1}{R _{3}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} \\ \\ \therefore \: \frac{1}{6} = \frac{1}{(17 + x)} + \frac{1}{8} \\ \\ = \frac{1}{17 + x} = \frac{1}{6} - \frac{1}{8} = \frac{1}{24} \\ \\ \therefore17 + x = 24 \\ x = 24 - 17 = 7 \\ x = 7Ω$

Hope This Helps!!

Answer 2

The value of the resistance X is 7 ohms.

Equivalent resistance in series

The equivalent resistance in series is calculated as follows;

R₁ = 8 + 9 + x = 17 + x

R₂ = 3 + 5 = 8

Equivalent resistance in the circuit

The equivalent resistance in the circuit is calculated as follows;

$\frac{1}{R} = \frac{1}{R_1}+ \frac{1}{R_2} \\\\R = \frac{R_1 R_2}{R_1 + R_2}$

$R(R_1 + R_2 ) = R_1 R_2$

6(17 + x + 8) = 8(17 + x)

150 + 6x = 136 + 8x

14 = 2x

x = 7

Thus, the value of the resistance X is 7 ohms.

Learn more about series circuit here: https://brainly.com/question/19865219