Question

Find the total resistance of the combination of resistors shown in the figure below.
(R1 = 0.800 µΩ, R2 = 15.0 µΩ, and R3 = 2.50 µΩ.)

Answer 1

Answer:

$2.158 \mu \Omega$

Explanation:

First of all, we start by calculating the total resistance of the combination R1 and R2.

These two resistors are connected in series, so their total resistance is given by:

$R_{12}=R_1+R_2$

where

$R_1=0.800 \mu \Omega$ is the resistance of resistor 1

$R_2=15.0\mu \Omega$ is the resistance of resistor 2

Substituting,

$R_{12}=0.800+15.0=15.8 \mu \Omega$

Then, we observe that the combination 1-2 and resistor 3 are connected in parallel. Therefore the total resistance of these 2 combinations is given by

$\frac{1}{R}=\frac{1}{R_{12}}+\frac{1}{R_3}$

where

$R_{12}=15.8\mu \Omega$

$R_3=2.50\mu \Omega$ is the resistance of resistor 3

Substituting,

$\frac{1}{R}=\frac{1}{15.8}+\frac{1}{2.50}=0.463\\R=\frac{1}{0.463}=2.158 \mu \Omega$

Answer 2

Total resistance of the combination of resistors is 2.15 µΩ

Equivalent resistance :

From given figure,

It is observed that resistance $R_{1}$ and $R_{2}$ are connected in series and parallel with resistance $R_{3}$.

Given that, $R_{1}= 0.800 ,R_{2} = 15.0 ,R_{3} = 2.50$

       $R_{s}=R_{1}+R_{2}=0.800+15=15.800$

              $R_{T}=\frac{R_{s}*R_{3}}{R_{s}+R_{3}}$

             $R_{T}=\frac{15.8*2.5}{15.8+2.5} \\\\R_{T}=\frac{39.5}{18.3}= 2.15$

Total resistance of the combination of resistors is 2.15 µΩ

Learn more about the equivalent resistance here:

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