question_answer

In a network as shown in the figure, the potential difference across the resistance 2R is (the cell has an emf of E volts and has no internal resistance)

A)  2E                                         

B)  \[\frac{4E}{7}\]               

C)  \[\frac{E}{7}\]                  

D)         \[E\]

Correct Answer: B

Solution :

Key Idea: Potential difference across resistors connected in parallel is same. 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 \[\frac{1}{R}=\frac{1}{4R}+\frac{1}{2R}=\frac{6R}{8{{R}^{2}}}\]                                 \[R=\frac{8}{6}R=\frac{4}{3}R\]                 \[\Rightarrow \]               \[R\,=R+\frac{4}{3}R=\frac{7R}{3}\] Given, emf is E volts, therefore                                 \[i=\frac{E}{R}=\frac{3E}{7R}\] potential difference across R is                 \[V=ir=\frac{3E}{7R}\times R=\frac{3E}{7}\] Potential difference across 2R is                                 \[V=E-\frac{3E}{7}=\frac{4E}{7}\] Note: Potential drop across 4R resistor is same as that across 2R, since both are connected in parallel.