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 Ohms 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. |
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