Question

Find potential difference across AB in figure.

• A. $6V$
• B. $4.5V$
• C. $\frac{10}{3}V$
• D. $4V$

Answer 1

Answer: (C)

$\frac{10}{3}V$

Let $R$ be the equivalent resistance of the circuit. The circuit is infinite hence, by adding one more section will not be effective on the equivalent resistance of the circuit. Now, consider the part of the circuit as shown in figure (a),
From figure, the equivalent resistance of the circuit is
$R=R_p+4$
where, $R_p=\frac{3R}{3+R}$

$R=\frac{3R}{3+R}+4$

$R=\frac{3R+12+4R}{3+R}$

$3R+R^2=3R+12+4R$

$R^2-4R-12=0$

$(R-6)(R+2)=0$

$R=6\Omega$

Now, this equivalent resistance is parallel to each $3\Omega$ resistor.
Hence,
$R_p=\frac{\left(3\right)\left(6\right)}{3+6}$

$R_p=\frac{18}{9}=2\Omega$

By potential division rule, voltage across $R_p$ is
$V_{AB}=\frac{V{R}_p}{R}$

$V_{AB}=\frac{\left(10\right)\left(2\right)}{6}$

$V_{AB}=\frac{10}{3}V$