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 \[\frac{P}{Q}=\frac{R}{S}\]You need to login to perform this action.
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