question_answer

In die circuit shown, if a conducting wire is connected between points A and B, the current in this wire will:                                                                                                                                                                                                                                                                                                     [AIPMT (S) 2006]

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 :

[d] 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.

\[\therefore \]      \[8{{i}_{1}}=4{{i}_{2}}\]

\[\Rightarrow \]   \[{{i}_{2}}=2{{i}_{1}}\]

Also      \[i={{i}_{1}}+{{i}_{2}}\]

\[\Rightarrow \]   \[\frac{3V}{8}={{i}_{1}}+2{{i}_{1}}\]

\[\Rightarrow \]   \[{{i}_{1}}=\frac{V}{8}\,A\]

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.