Question

Four resistances $10\Omega ,5\Omega ,7\Omega$ and $3\Omega$ are connected so that they form the sides of a rectangle $AB,BC,CD$, and $DA$ respectively. Another resistance of $10\Omega$ is connected across the diagonal $AC$. 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 $AC$.
$\therefore \frac{1}{R_1}=\frac{1}{10}+\frac{1}{10}$
$\Rightarrow R_1=5\Omega$
Now, $R_1$ is in series with resistor $(5\Omega )$ in arm $CB$.
$\therefore R_2=5+5=10\Omega$
Again $R_2$ is in parallel with resistance $(10\Omega )$ in arm $AB$
$\therefore \frac{1}{R}=\frac{1}{10}+\frac{1}{10}$
$\Rightarrow R=5\Omega$