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