A) \[205\Omega \]
B) \[10\Omega \]
C) \[3.5\Omega \]
D) \[5\Omega \]
Correct Answer: D
Solution :
In the given circuit, \[3\Omega \] and \[7\Omega \] resistors are in series, hence equivalent resistance is \[R'={{R}_{1}}+{{R}_{2}}=3\Omega +7\Omega =10\Omega \] This \[10\Omega \] resistor is in parallel with \[10\Omega \] resistance, hence equivalent resistance is \[\frac{1}{R''}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}\] \[\Rightarrow \] \[R''=5\Omega \] This \[5\Omega \] is in series with \[5\Omega \] resistor \[R''5\Omega +5\Omega =10\Omega \] Now, this \[10\Omega \] is in parallel with \[10\Omega \] between \[A\]and\[B\]. Hence, equivalent resistance is \[\frac{1}{{{R}_{4}}}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}\]\[\Rightarrow \]\[{{R}_{4}}=5\Omega \]
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