question_answer

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 Wheatstone’s bridge in middle arm galvanometer must be connected. In Wheatstone’s bridge, cell and galvanometer arms are interchangeable.

In both the cases, condition of balanced bridge is \[\frac{P}{Q}=\frac{R}{S}\]