A) \[\frac{1}{3}A\]
B) \[\frac{2}{3}A\]
C) \[1\,\,A\]
D) \[\frac{2}{9}\,\,A\]
Correct Answer: A
Solution :
The emf of the circuit is \[E={{E}_{1}}+{{E}_{2}}=4V+2V=6V\] In the given circuit, \[3\Omega \]. and \[6\Omega \] are connected in parallel, hence equivalent resistance is \[\frac{1}{R'}=\frac{1}{3}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\] \[\Rightarrow \] \[i=\frac{V}{R}=\frac{6}{6}=1A\] Total resistance of circuit is \[R=1\,\Omega +1\,\Omega +2\,\Omega +2\,\Omega =6\,\Omega \] From Ohm's law \[V=iR\] \[\Rightarrow \] \[i=\frac{V}{R}=\frac{6}{6}=1A\] The \[3\,\,\Omega \] and \[6\,\,\Omega \] resistors are in parallel, hence \[{{i}_{1}}{{R}_{1}}={{i}_{2}}{{R}_{2}}=V\] \[\therefore \] \[{{i}_{1}}\times 3={{i}_{2}}\times 6\] \[\Rightarrow \] \[{{i}_{1}}=2{{i}_{2}}\] and \[{{i}_{1}}+{{i}_{2}}=1\] \[2{{i}_{2}}+{{i}_{2}}=1\] \[3{{i}_{2}}=1\] \[\Rightarrow \] \[{{i}_{2}}=\frac{1}{3}A\]
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