How to solve this question **Problem Description:**The four resistors in the figure have an equivalent resistance of 12 Ω. The resistances are as follows:- $ R_1 = 8.0 \, \Omega $- $ R_2 = 4.0 \, \Omega $- $ R_3 = 4.0 \, \Omega $Calculate the value of $ R_x $.**Diagram Explanation:**The diagram on the right shows a circuit with four resistors:1. $ R_1 $ and $ R_x $ are connected in parallel.2. The combination of $ R_1 $ and $ R_x $ is in series with $ R_2 $ and $ R_3 $.**Solution Steps:**1. **Parallel Combination:** The equivalent resistance $ R_p $ of $ R_1 $ and $ R_x $ in parallel is given by: \[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_x} \]2. **Series Combination:** The total equivalent resistance $ R_{eq} $ is: \[ R_{eq} = R_p + R_2 + R_3 \] Given $ R_{eq} = 12 \, \Omega $, substitute the known values: \[ 12 = R_p + 4 + 4 \] \[ 12 = R_p + 8 \] \[ R_p = 4 \, \Omega \]3. **Calculate $ R_x $:** Using the parallel resistance formula for $ R_p = 4 \, \Omega $: \[ \frac{1}{4} = \frac{1}{8} + \frac{1}{R_x} \] \[ \frac{1}{R_x} = \frac{1}{4} - \frac{1}{8} \] \[ \frac{1}{R_x} = \frac{2}{8} - \frac{1}{8} \] \[ \frac{1}{R_x} = \frac{1}{8} \] \[ R_x =

Answered: The four resistors in the figure have…

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Publisher:Raymond A. Serway, Chris Vuille

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How to solve this question

**Problem Description:**

The four resistors in the figure have an equivalent resistance of 12 Ω. The resistances are as follows:
- $ R_1 = 8.0 \, \Omega $
- $ R_2 = 4.0 \, \Omega $
- $ R_3 = 4.0 \, \Omega $

Calculate the value of $ R_x $.

**Diagram Explanation:**

The diagram on the right shows a circuit with four resistors:
1. $ R_1 $ and $ R_x $ are connected in parallel.
2. The combination of $ R_1 $ and $ R_x $ is in series with $ R_2 $ and $ R_3 $.

**Solution Steps:**

1. **Parallel Combination:**
The equivalent resistance $ R_p $ of $ R_1 $ and $ R_x $ in parallel is given by:
\[
\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_x}
\]

2. **Series Combination:**
The total equivalent resistance $ R_{eq} $ is:
\[
R_{eq} = R_p + R_2 + R_3
\]

Given $ R_{eq} = 12 \, \Omega $, substitute the known values:
\[
12 = R_p + 4 + 4
\]
\[
12 = R_p + 8
\]
\[
R_p = 4 \, \Omega
\]

3. **Calculate $ R_x $:**
Using the parallel resistance formula for $ R_p = 4 \, \Omega $:
\[
\frac{1}{4} = \frac{1}{8} + \frac{1}{R_x}
\]
\[
\frac{1}{R_x} = \frac{1}{4} - \frac{1}{8}
\]
\[
\frac{1}{R_x} = \frac{2}{8} - \frac{1}{8}
\]
\[
\frac{1}{R_x} = \frac{1}{8}
\]
\[
R_x =

Expert Solution

Step 1

Given Data

Resistance of the resistors are $R_{1} = 8.0 Ω, R_{2} = 4.0 Ω, R_{3} = 4.0 Ω$

Equivalent resistance of the circuit is $R_{s} = 12 Ω$

Concept

In the given circuit $R_{1}$ and $R_{x}$ are connected in parallel combination and equivalent resistance of these two resistors is connected in series with $R_{2}$ and $R_{3}$ .

Formula to be used

$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{x}}$

$R_{s} = R_{p} + R_{2} + R_{3}$

where $R_{p}$ is the equivalent resistance of the parallel combination of $R_{1}$ and $R_{x}$

$R_{s}$ is the equivalent resistance of the circuit.

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The four resistors in the figure have an equivalent resistance of 122. The resistances are as follows: R¡ = 8.02, R, = 4.02, and R3 - 4.02. Calculate the value of R, . min R, - Ω