Question

2. Equivalent resistance of circuit diagram
8.33 is 5 ohm. Calculate the value of x. And explain how
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Answer 1

Given ,

The parrallel combination of 6 ohm , x and 2 ohm is in series with 4 ohm

The equivalent resistance of circuit is 5 ohm

We know that , the equivalent resistance in parrallel combination is given by

$\boxed{ \tt{\frac{1}{R} = \frac{1}{ R_{1}} + \frac{1}{ R_{2}} + ... + \frac{1}{ R_{n}}}}$

And

The equivalent resistance in series combination is given by

$\boxed{ \tt{R = R_{1} + R_{2} +. .. + R_{n} }}$

Thus ,

1/R = 1/6 + 1/x + 1/2

1/R = (2x + 12 + 6x)/12x

1/R = (8x + 12)/12x

R = 12x/(8x + 12)

Now , the combination of R is in series with 4 ohm

Thus ,

R' = 12x/(8x + 12) + 4

R' = (12x + 32x + 48)/(8x + 12)

R' = (44x + 48)/(8x + 12)

Since , the equivalent resistance of the circuit is 5 ohm

Thus ,

R' = Resistance of the circuit

(44x + 48)/(8x + 12) = 5

44x + 48 = 40x + 60

4x = 12

x = 12/4

x = 3

Therefore , the value of x is 3 ohm

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Answer 2

Answer:

The value of x is 3Ω.

Explanation:

Let the 4Ω resistance be denoted R₁, 6Ω resistance be denoted R₂, unknown resistance x be R₃, 2Ω resistance be R₄, and R be the equivalent resistance.

Given:

R₁ = 4Ω, R₂ = 6Ω, R₄ = 2Ω, R = 5Ω

To find:

R₃ (x) =?

Solution:

As we can see from the given circuit diagram, resistors R₂, R₃, and R₄ are connected in parallel.

The equivalent resistance for resistors connected in parallel is given as the sum of reciprocals of individual resistances.

Let R₅ be the equivalent resistance of the parallel combination. Therefore,

$\frac{1}{R_5} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$

$\frac{1}{R_5} = \frac{R_3R_4+R_2R_3+R_2R_4}{R_2R_3R_4}$

$R_5 = \frac{R_2R_3R_4}{R_3R_4+R_2R_3+R_2R_4}$

Substituting the given values, we get

$R_5 = \frac{6*x*2}{(x*2)+(6*x)+(6*2)}$

$R_5 = \frac{12x}{2x+6x+12}$

$R_5 = \frac{12x}{8x+12}$

Now, resistors R₁ and the equivalent resistance R₅ are connected in series.

The equivalent resistance for resistors connected in series is given as the sum of the individual resistances. Therefore,

R = R₁ + R₅

Substituting the respective values, we get

$5 = 4 + \frac{12x}{8x+12}$

$\frac{12x}{8x+12} = 1$

12x = 8x + 12

4x = 12

x = 12/4

x = 3Ω

Therefore, the value of the unknown resistance is 3Ω.

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