Question

Four resistances $10 \Omega, 5 \Omega, 7 \Omega$ and $3 \Omega$ are connected so that they form the sides of a rectangle $A B, B C, C D$, and $D A$ respectively. Another resistance of $10 \Omega$ is connected across the diagonal $A C$. The equivalent resistance between $A$ and $B$ is

• A. $2\,\Omega$
• B. $5\,\Omega$
• C. $7\,\Omega$
• D. $10\,\Omega$

Answer 1

Answer: (B)

$3 \Omega$ resistor and $7 \Omega$ resistor are in series.

Therefore resultant is $=10 \Omega(7+3)$.
This $10 \Omega$ equivalent resistance is in parallel with resistance $(10 \Omega)$ in arm $A C$.
$\therefore \frac{1}{R_{1}}=\frac{1}{10}+\frac{1}{10}$
$\Rightarrow R_{1}=5 \Omega$
Now, $R_{1}$ is in series with resistor $(5 \Omega)$ in arm $C B$.
$\therefore R_{2}=5+5=10 \Omega$
Again $R_{2}$ is in parallel with resistance $(10 \Omega)$ in arm $A B$
$\therefore \frac{1}{R}=\frac{1}{10}+\frac{1}{10} $
$\Rightarrow R=5\, \Omega$