A) 0.36 A
B) zero
C) 0.ISA
D) 0.72 A
Correct Answer: A
Solution :
The balanced condition for Wheatstone's bridge is \[\frac{P}{Q}=\frac{R}{S}\] as is obvious from the given values. No, current flows through galvanometer is zero. Now, P and R are in series, so resistance \[{{R}_{1}}=P+R\] \[=10+15=25\,\Omega \] Similarly, Q and S are in series, so resistance \[{{R}_{2}}=Q+S\] \[=20+30=50\,\Omega \] Net resistance of the network as\[{{R}_{1}}\]and\[{{R}_{2}}\]are in parallel \[\frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\] \[\therefore \] \[R=\frac{25\times 50}{25+50}=\frac{50}{3}\Omega \] Hence, \[I=\frac{V}{R}=\frac{6}{\frac{50}{3}}=0.36A\]
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