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

Question

The figure shows a network in which the cell is ideal and it has an emf

E

. The potential difference across the resistance

2 R

is

(A)

2 E

(B)

\frac{4 E}{7}

(C)

\frac{E}{7}

(D)

\frac{3 E}{7}

Tardigrade Admin · Accepted Answer

In the given circuit, resistors

E

and

2 R

are connected in parallel while resistance

R

is connected in series to it. Hence, equivalent resistance is, of parallel combination

\frac{1}{R^{'}} = \frac{1}{4 R} + \frac{1}{2 R}

\overset{}{\Rightarrow}

R' = \frac{8}{6} R

= \frac{4}{3} R

⟹ R " = R + \frac{4}{3} R = \frac{7 R}{3}

(series combination) Given, emf is E volts, thereforeusing ohm's law,

i = \frac{E}{R^{''}} = \frac{3 E}{7 R}

Potential difference across R is

V = i r = \frac{3 E}{7 R} \times R = \frac{3 E}{7}

Potential difference across 2R is

V^{'} = E - \frac{3 E}{7} = \frac{4 E}{7}

Q. 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