Question

Find equivalent resistance of the given circuit.

• A. $\frac{90\Omega }{43}$
• B. $\frac{80\Omega }{43}$
• C. $\frac{60\Omega }{43}$
• D. $\frac{50\Omega }{43}$

Answer 1

The correct option is A

$\frac{90\Omega }{43}$

Given: $R_1=2\Omega$, $R_2=3\Omega$, $R_3=5\Omega$, $R_4=4\Omega$ and $R_5=6\Omega$.

Now $R_1$ and $R_2$ are in series and $R_3$ and $R_4$ are in series.

Series1 $R_{eq1}=R_1+R_2=2\Omega +3\Omega =5\Omega$

Series 2 $R_{eq2}=R_3+R_4=5\Omega +4\Omega =9\Omega$

Therefore, the given circuit can be deduced to:

Now all these three are in parallel to each other.

So, for parallel connection:
$\frac{1}{R_p}=\frac{1}{R_{eq1}}+\frac{1}{R_{eq2}}+\frac{1}{R_5}$

$=\frac{1}{5\Omega }+\frac{1}{9\Omega }+\frac{1}{6\Omega }=\frac{18+10+15}{90\Omega }$

$\therefore R_p=\frac{90\Omega }{43}$.

So, equivalent resistance of the given circuit is $\frac{90\Omega }{43}$.