Question

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

Answer 1

Step 1. Given data

Four resistance are $40\Omega ,60\Omega ,90\Omega ,110\Omega$

Step 2 . Finding the currents, $I_{1}and{I}_2$

By using the formula,

$Current=\frac{Voltage}{Resis\tan ce}$

$$\begin{gathered} I_1=\frac{40}{40+60}A \\ {I}_1=\frac{40}{100}A \end{gathered}$$ [Current across the $40\Omega$ and $60\Omega$ resistance]

$$\begin{gathered} I_2=\frac{40}{90+110}A \\ {I}_2=\frac{40}{200}A \end{gathered}$$ [Current across the $90\Omega$ and $110\Omega$ resistance]

Step 3. Finding the potential difference across BD, $∆V$

$Voltage=Current\times Resis\tan ce$

Potential across B, $V_B=I_1\times 60$

$$\begin{gathered} V_B=\frac{40\times 60}{100} \\ {V}_B=24V \end{gathered}$$ [At $B$, Resistance$=60\Omega$]

Potential across D, $V_D=I_2\times 110$

$V_D=\frac{40\times 110}{200}$ [At $D$, Resistance$=110\Omega$]

$V_D=22V$

$$\begin{gathered} ∆V=V_B-V_D \\ ∆V=24-22 \end{gathered}$$

$∆V=2volt$

Therefore, the potential difference across $BD$ is $2volt$.