question_answer

                In die circuit shown, if a conducting wire is connected between points A and B, the current in this wire will:                                                                                                                                                                                                

A)                  flow from A to B

B)                             flow in the direction which will be decided by the value of V

C)                  be zero                               

D)                 flow from B to A

Correct Answer: D

Solution :

                Key Idea: Current will flow from higher to lower potential.                   Resistance \[4\,\,\Omega \] and \[4\,\,\Omega \] are connected in series, so their effective resistance is                 \[R'=4+3=8\Omega \]                 Similarly, \[1\,\,\Omega \]and \[3\,\,\Omega \] are in series                 So,          \[R''=1+3=4\Omega \]                 Now R? and R?? will be in parallel, hence effective resistance                 \[R=\frac{R'\times R''}{R'+R''}\]                 \[=\frac{8\times 4}{8+4}=\frac{32}{12}=\frac{8}{3}\,\Omega \]                 Current through the circuit, from Ohm?s law                 \[i=\frac{V}{R}=\frac{3V}{8}A\]                 Let currents \[{{i}_{1}}\] and \[{{i}_{2}}\] flow in the branches as shown.                 \[\therefore 8{{i}_{1}}=4{{i}_{2}}\] \[\Rightarrow {{i}_{2}}=2{{i}_{1}}\] \[Alsoi={{i}_{1}}+{{i}_{2}}\] \[\Rightarrow \frac{3V}{8}={{i}_{1}}+2{{i}_{1}}\] \[\Rightarrow {{i}_{1}}=\frac{V}{8}\,A\] \[and{{i}_{2}}=\frac{V}{4}A\]                 Potential drop at \[A,\,{{V}_{A}}=4\times {{i}_{1}}=\frac{4V}{8}=\frac{V}{2}\]                 Potential drop at B, \[{{V}_{B}}=1\times {{i}_{2}}=1\times \frac{V}{4}=\frac{V}{4}\]                 Since, drop of potential is greater in \[4\,\,\Omega \] resistance so. It will be at lower potential than B, hence, on connecting wire between points A and B, the current will flow from B to A.