A) 0.75 A
B) 1 A
C) 2 A
D) 1.5 A
Correct Answer: B
Solution :
Required arrangement is shown in figure. The equivalent circuit will look like (since the two resistances of \[1\Omega \]and \[2\Omega \] are in series, which form \[3\Omega \]which is in parallel with \[3\,\Omega \]resistance). Therefore, the effective resistance is \[\frac{(1+2)\times 3}{(1+2)+3}=\frac{3}{2}\,\Omega \] \[\therefore \]Current in the circuit, \[I=\frac{3}{(3/2)}=2A\] \[\therefore \]Current in \[3\Omega \] resistor\[=\frac{I}{2}=1A\]You need to login to perform this action.
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