Question

Four resistances $40\Omega ,60\Omega ,90\Omega \text{and}110\Omega$ make the arms of a quadrilateral ABCD. Across $AC$ is a battery having emf $40\text{V}$ and negligible internal resistance. The potential difference across BD, in Volts, is

Answer 1

Current through AB is, $i_1=\frac{40}{40+60}=0.4\text{A}$

Current through AD is,$i_2=\frac{40}{90+110}=\frac{1}{5}\text{A}$

Now,

$V_A-V_B=40i_1=40\times 0.4$

$\therefore V_A-V_B=16-\left(i\right)$

Similarly,
$V_A-V_D=90i_2=90\times \frac{1}{5}$

$\therefore V_A-V_D=18-\left(ii\right)$

Subtracting $\left(i\right)$ from $\left(ii\right)$, we get,

$\therefore V_B-V_D=2\text{V}$