A battery of $6\,V$ is connected to the circuit as shown below. The current I drawn from the battery is:
JEE MAIN 2022, Medium
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Balanced wheat stone bridge in circuit so there is no current in $5\,\Omega$ resistor so it can be removed from the circuit.
$R _{eq }=\frac{6 \times 12}{6+12}+2$
$=\frac{6 \times 12}{18}+2$
$R _{e q }=6\,\Omega$
$I =\frac{ V }{ R _{a q }}=\frac{6}{6}=1\,Amp$
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