Find the equivalent resistance, R_EQ between A-B for the following resistor combination circuit.
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To find the equivalent resistance, R_EQ for the following ladder network resistor (it is just a combination of series and parallel resistors connected together) combination circuit, we will start from the right-hand side. 

We know that equation for computing series circuits is given as- 

R_t = R₁ + R₂ + R₃ + R₄ +...... where R_t is the total resistance, and R₁, R₂, R₃, and R₄ are the resistance values of the individual resistors that are connected in series.

Using the simplified equation for series circuits, for three series resistors, we can find the equivalent resistance of the R₇, R₈ and R₉ combination and call it R_A.

Therefore, the total resistance can be calculated as R_A as shown above-

R_A = R₇ + R₈ + R₉ 

R_A = 2 + 2 + 2 

R_A = 6Ω

This resistive value of R_A = 6Ω is now in parallel with R₆.

We know that equation for computing parallel circuits is given as- 

R_t = 1 / R₁ + 1 / R₂ + 1 / R₃ +  1 / R₄....... where R_t is the total resistance, and R₁, R₂, R₃, and R₄ are the resistance values of the individual resistors that are connected in series.

Using this equation for parallel circuits, for two parallel resistors, we can find the total resistance of the R_A, and R₆ combination and call it R_B.

Therefore the total resistance can be calculated as R_B as shown above-

$\frac{1}{{R}_{B}}=\frac{1}{{R}_{A}}+\frac{1}{{R}_{6}}$

$\frac{1}{{R}_{B}}=\frac{1}{6}+\frac{1}{2}\Leftrightarrow \frac{2+6}{6\times 2}\Leftrightarrow \frac{8}{12}\Rightarrow \frac{2}{3}$

${R}_{B}=\frac{3}{2}$

R_B = 1.5Ω

This resistive value of R_B = 1.5Ω is now in series with R₃ & R₅.

Using the simplified equation for series circuits, for three series resistors, we can find the equivalent resistance of the R_B, R₃ and R₅ combination and call it R_C.

Therefore, the total resistance can be calculated as R_C as shown above-

R_C = R_B + R₃ + R₅ 

R_C = 1.5 + 2 + 2

R_C = 5.5Ω

This resistive value of R_C = 5.5Ω is now in parallel with R₂.

Using the equation for parallel circuits, for two parallel resistors, we can find the total resistance of the R_C, and R₂ combination and call it R_D.

Therefore the total resistance can be calculated as R_D as shown above-

$\frac{1}{{R}_{D}}=\frac{1}{{R}_{C}}+\frac{1}{{R}_{2}}$

$\frac{1}{{R}_{D}}=\frac{1}{5.5}+\frac{1}{2}\Leftrightarrow \frac{2+5.5}{5.5\times 2}\Leftrightarrow \frac{7.5}{11}\Rightarrow \frac{75}{110}$

${R}_{D}=\frac{110}{75}$

R_D = 1.46Ω

This resistive value of R_D = 1.46Ω is now in series with R₁ & R₄.

Using the simplified equation for series circuits, for three series resistors, we can find the total resistance of the R_D, R₁ and R₄ combination and call it R_t.

Therefore, the total resistance can be calculated as R_t as shown above-

R_t = R_D + R₁+ R₄

R_t = 1.46 + 2 + 2

R_t = 5.46Ω

So, the complex combinational resistive network above comprising of nine individual resistors connected together in series and parallel combinations can be replaced with just one single equivalent resistance (R_EQ) of value 5.46Ω.

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Updated on: 10-Oct-2022

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