Question

Five equal resistances of $10\Omega$ are connected between $A$ and $B$ as shown in figure. The resultant resistance is

• A. $10\Omega$
• B. $5\Omega$
• C. $15\Omega$
• D. $6\Omega$

Answer 1

Answer: (B)

According to the given circuit $10\Omega$ and $10\Omega$ resistances are connected in series.
$\therefore R^{\prime }=10+10=20\Omega$
Again $10\Omega$ and $10\Omega$ resistances are connected in series
$\therefore R^{\prime \prime }=10+10=20\Omega$
$R^{\prime }$, $R^{\prime \prime }$ and $10\Omega$ all connected in parallel than
$\therefore \frac{1}{R_{eq}}=\frac{1}{R^{\prime }}+\frac{1}{R^{\prime \prime }}+\frac{1}{10}$
$=\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\frac{1+1+2}{20}$
$=\frac{4}{20}=\frac{1}{5}$ ;
$R_{eq}=5\Omega$.