question_answer

                A bridge circuit is shown in figure the equivalent resistance between A and B will be:     

A)                 \[21\,\Omega \]              

B)                 \[7\Omega \]                    

C)                 \[\frac{252}{85}\Omega \]          

D)                 \[\frac{14}{3}\Omega \]

Correct Answer: D

Solution :

                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                 \[\frac{P}{Q}=\frac{R}{S}\]