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 \[=\frac{R'\times R''}{R'+R''}=\frac{20\times 20}{20+20}=10\,\Omega \]You need to login to perform this action.
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