Question

The figure shows a network in which the cell is ideal and it has an emf $E$. The potential difference across the resistance $2R$ is

• A. $2E$
• B. $\frac{4E}{7}$
• C. $\frac{E}{7}$
• D. $\frac{3E}{7}$

Answer 1

Answer: (B)

In the given circuit, resistors $4R$ and $2R$ are connected in parallel while resistance $R$ is connected in series to it.
Hence, equivalent resistance is, of parallel combination
$\frac{1}{R^{\prime }}=\frac{1}{4R}+\frac{1}{2R}$
${\stackrel{}{\Rightarrow }}^{}$ $R\text{'}=\frac{8}{6}R$$=\frac{4}{3}R$
$⟹{R"}^{}=R+\frac{4}{3}R=\frac{7R}{3}$(series combination)
Given, emf is E volts, thereforeusing ohm's law,
$i=\frac{E}{R^{\mathrm{\text{'}\text{'}}}}=\frac{3E}{7R}$
Potential difference across R is
$V=ir=\frac{3E}{7R}\times R=\frac{3E}{7}$
Potential difference across 2R is
$V^{\prime }=E-\frac{3E}{7}=\frac{4E}{7}$