A bridge circuit is shown in figure the equivalent resistance between A and B will be: [AIPMT 2000] |
A) \[21\,\,\Omega \]
B) \[7\,\,\Omega \]
C) \[\frac{252}{85}\,\Omega \]
D) \[\frac{14}{3}\Omega \]
Correct Answer: D
Solution :
[d] Key Idea: In the balanced condition of bridge circuit, no current will flow through \[7\,\,\Omega \] resistance. |
The bridge circuit can be shown as: |
The balanced condition of bridge circuit is given by |
\[\frac{P}{Q}=\frac{3}{4},\frac{R}{S}=\frac{6}{8}=\frac{3}{4}\] |
\[\therefore \] \[\frac{P}{Q}=\frac{R}{S}\] |
Thus, it is balanced Wheatstone's bridge, so potential at F is equal to potential at H. Therefore, no current will flow through \[7\,\Omega \] resistance. So, circuit can be redrawn as shown above. |
P and Q are in series, so their equivalent resistance \[=3+4=7\,\Omega \] |
R and S are also in series, so their equivalent resistance \[=6+8=14\,\,\Omega \] |
Now \[7\,\,\Omega \] and \[14\,\,\Omega \] resistances are in parallel, |
So |
\[{{R}_{AB}}=\frac{7\times 14}{7+14}=\frac{7\times 14}{21}=\frac{14}{3}\Omega \] |
Note: Normally, in Wheatstones bridge in middle arm galvanometer must be connected. In Wheatstones bridge, cell and galvanometer arms are interchangeable. |
In both the cases, condition of balanced bridge is \[\frac{P}{Q}=\frac{R}{S}\] |
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