Question

What is the equivalent resistance between $\text{A}$ and $\text{B}$ in the given network?

• A. $\frac{4}{3}\Omega$
• B. $\frac{3}{4}\Omega$
• C. $2\Omega$
• D. $\frac{1}{2}\Omega$

Answer 1

The correct option is A $\frac{4}{3}\Omega$

Let us reduce the circuit as the two $2\Omega$ resistors are in parallel.
So, $R_3=\frac{R}{n}=\frac{2}{2}=1\Omega$

Here,
$R_1\times R_3=R_2\times R_4=1\times 2=2\Omega$

Since, the product of resistances in the opposite arms are equal, so it is a balanced Wheatstone bridge.

Hence, no current flows through arm $\text{PQ}$.

Redrawing the circuit:

Equivalent resistance of the circuit,

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

$\therefore R_{\text{eq}}=\frac{4}{3}\Omega$

Hence, the correct answer is option (a).