question_answer

Look at the circuit given below. Six resistors, each of value \[4\,\Omega \] are joined together. Calculate equivalent resistance across the points A and B. If a cell of emf 2 V is connected across AB, compute the current through the arms AB and DF of the circuit.

Circuit diagram

              

Answer:

\[{{R}_{AB}}=2\,\Omega ,\] \[{{I}_{FD}}=0,\] \[{{I}_{AB}}=0.5\,A\] In the given circuit, \[\frac{{{R}_{GF}}}{{{R}_{FH}}}=\frac{{{R}_{CD}}}{{{R}_{DE}}}\Rightarrow \frac{4}{4}=\frac{4}{4}=1\] Thus, it is balanced Wheatstone bridge. Therefore, \[{{V}_{F}}={{V}_{D}},\] this makes \[{{R}_{FD}}\] ineffective. The equivalent circuit will be given as The resistors on the arm GH and CE are in series \[\Rightarrow \] \[{{R}_{GH}}=4+4=8\Omega \] and \[{{R}_{CE}}=4+4=8\Omega \] As, the arm GH, CE and AB are in parallel combination. Therefore, the equivalent resistance across the points A and B is,                         \[\frac{1}{{{R}_{AB}}}=\frac{1}{8}+\frac{1}{8}+\frac{1}{4}=\frac{1+1+2}{8}=\frac{4}{8}\] \[\Rightarrow \]   \[\frac{1}{{{R}_{AB}}}=2\,\Omega \] So, the current (I) through the arm AB is, if a cell of 2 V emf connected across,                         \[{{I}_{AB}}=\frac{V}{R}\]      \[[\text{given,}\,V=2\,V]\] \[\Rightarrow \]   \[{{I}_{AB}}=\frac{2\,V}{4\,\Omega }=0.5A\] Since, \[{{V}_{D}}={{V}_{F}}.\] Therefore, the current \[{{I}_{FD}}=0\]