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:                                              [AIPMT 2001]

A)  \[10\,\,\Omega \]

B)      \[5\,\,\Omega \]

C)  \[20\,\,\Omega \]           

D)       \[40\,\,\Omega \]

Correct Answer: A

Solution :

[a] Key Idea: In a Wheatstone'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

\[=\frac{R'\times R''}{R'+R''}=\frac{20\times 20}{20+20}=10\,\Omega \]