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.You need to login to perform this action.
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