Question

Find the value of x in the following attachment. Please solve this. Class 10th
Chapter 12 Electricity
​

Answer 1

Answer:

The value of x is 11.29 Ω.

Explanation

First of all, we will calculate the resistance of the circuit placed in parallel.

We know for parallel,

 $\frac{1}{R}$ = $\frac{1}{R_{1} }$ + $\frac{1}{R_{2} }$ + $\frac{1}{R_{3} }$

Here, R₁ = 12 Ω

          R₂ = 6 Ω

          R₃ = 3 Ω

$\frac{1}{R}$ = $\frac{1}{12}$ + $\frac{1}{6}$ + $\frac{1}{3}$

   = $\frac{1+2+ 4}{12}$

   = $\frac{7}{12}$

=> R = $\frac{12}{7}$ Ω

Calculating the resistance in series,

We know R = R₁ + R₂

Here  R₁ = 2 Ω

          R₂ = $\frac{12}{7}$ Ω

=> R = 2 + $\frac{12}{7}$

       = $\frac{26}{7}$ Ω

As x Ω resistor is in series with the combined resistance of 3.714 Ω, the total resistance of the circuit is (x + 3.714) Ω.

Total current = 0.4 A                     (Given)

Total voltage = 6 V                        (Given)

According to Ohm's Law,

V = I R

=> 6 = 0.4 (x + $\frac{26}{7}$ )

    $\frac{6}{0.4}$ = (x + $\frac{26}{7}$ )

    15 = (x + $\frac{26}{7}$ )

    15 - $\frac{26}{7}$  = x

    x = $\frac{79}{7}$ Ω = 11.29 Ω.