The current through the
A) \[3\Omega \]resistor is 0.50 A
B) \[3\Omega \]resistor is 0.25 A
C) \[4\Omega \]resistor is 0.50 A
D) \[4\Omega \]resistor is 0.25 A.
Correct Answer: D
Solution :
[d]: \[{{R}_{BC}}\](right hand side) \[=\frac{8\times 8}{8+8}\Omega \] \[=4\Omega \,(as2\Omega +4\Omega +2\Omega =8\Omega )\] \[{{R}_{AD}}\](right hand side) is again \[4\Omega \]. Equivalent resistance of the circuit, \[R=3\Omega +4\Omega +2\Omega =9\Omega \] Current drawn from battery\[I=\frac{V}{R}=\frac{9}{9}=1A\] At A, I is equally divided (I/2) between \[8\Omega \]. resistance and the remaining circuit of \[8\Omega \]. At B, (I/2) is equally divided (I/4) between the \[8\Omega \]. resistor and the remaining circuit of resistance \[8\Omega \]. Thus, current through \[4\Omega \]resistor is I/4, i.e., 0.25 A.
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