question_answer

                In a Wheatstone's bridge resistance of each of the four sides is \[10\,\,\Omega \]. If the resistance of the galvanometer is also \[10\,\,\Omega \] then effective resistance of the bridge will be:                                                  

A)                 \[10\,\,\Omega \]           

B)                 \[5\,\,\Omega \]                             

C)                 \[20\,\,\Omega \]           

D)                 \[40\,\,\Omega \]

Correct Answer: A

Solution :

                Key Idea: In a Wheatstdne's bridge, if \[\frac{P}{Q}=\frac{R}{S},\] then resistance of galvanometer will  be ineffective.                 The given circuit can be shown as.                                 From figure, \[\frac{P}{Q}=\frac{10}{10}=1\]                 \[\frac{R}{S}=\frac{10}{10}=1\]                         \[\therefore \frac{P}{Q}=\frac{R}{S}\]                 Therefore, the galvanometer will be ineffective.                 The above Wheatstone?s bridge can be redrawn as                                 Resistances P and Q are in series, so                 R' = 10 + 10 = 20 \[\Omega \]                 Resistances R and S are in series, so                 R?? = 10 + 10 = 20 \[\Omega \]                 Now, R' and R" are in parallel hence, net resistance of the circuit                 \[=\frac{R'\times R''}{R'+R''}=\frac{20\times 20}{20+20}=10\,\Omega \]