Question

The correct combination of three resistances $1\Omega$, $2\Omega$ and $3\Omega$ to get equivalent resistance $\frac{11}{5}\Omega$ is?

• A. All three are combines in parallel
• B. All three are combines in series
• C. $1\Omega$ and $2\Omega$ in parallel and $3\Omega$ is in series to both
• D. $2\Omega$ and $3\Omega$ are combined in parallel and $1\Omega$ is in series to both

Answer 1

Answer: (A)

Step 1: Given that:

First resistance($R_1$) = 1Ω

Second resistance($R_2$) = 2Ω

Third resistance($R_3$) = 3Ω

Equivalent resistance($R_{eq}$) = $\frac{11}{5}\text{Ω}$

Explanation of correct option:

Option d)

When 2$\text{Ω}$ and 3$\text{Ω}$ are connected in parallel, the equivalent resistance will be;

$\frac{1}{R^{\prime }}=\frac{1}{2\text{Ω}}+\frac{1}{3\text{Ω}}$

$\frac{1}{R^{\prime }}=\frac{3+2}{6}$

$\frac{1}{R^{\prime }}=\frac{5}{6}\text{Ω}$

$R^{\prime }=\frac{6}{5}\text{Ω}$

Now, R' and 1Ω are connected in series, thus,

Their equivalent resistance will be

$R_{eq}=\frac{6}{5}\text{Ω}+1\text{Ω}$

$=\frac{6+5}{5}$

$=\frac{11}{5}\text{Ω}$

Explanation of incorrect option:

Option a)

When the resistances are connected in parallel, the equivalent resistance is given by:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Then,

$\frac{1}{R_{eq}}=\frac{1}{1\text{Ω}}+\frac{1}{2\text{Ω}}+\frac{1}{3\text{Ω}}$

$\frac{1}{R_{eq}}=\frac{6+3+2}{6}$

$\frac{1}{R_{eq}}=\frac{11}{6}\text{Ω}$

$R_{eq}=\frac{6}{11}\text{Ω}$

Option b)

When the resistances are connected in series:

For the series combination of resistances, the equivalent resistance is given by;

$R=R_1+R_2+R_3$

Thus,

The equivalent resistance of the given resistances when they are connected in series will be;

$R_{eq}=1\text{Ω}+2\text{Ω}+3\text{Ω}$

$=6\text{Ω}$

Option c)

When 1$\text{Ω}$ and 2$\text{Ω}$ are connected in parallel,

Their equivalent resistance will be,

$\frac{1}{R^{\prime }}=\frac{1}{1\text{Ω}}+\frac{}{2\text{Ω}}$

$\frac{1}{R^{\prime }}=\frac{2+1}{2}$

$\frac{1}{R^{\prime }}=\frac{3}{2}$

$R^{\prime }=\frac{2}{3}\text{Ω}$

Now R' is connected in series with 3$\text{Ω}$ then the equivalent resistance will be

$R_{eq}=\frac{2}{3}\text{Ω}+3\text{Ω}$

$R_{eq}=\frac{11}{3}\text{Ω}$

Thus,

Option d) 2Ω and 3Ω are combined in parallel and 1Ω is in series to both is the correct option.