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(C) The equivalent resistance of three $1\, \Omega$ resistors connected in parallel is given by the formula $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{

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(C) The equivalent resistance of three $1\, \Omega$ resistors connected in parallel is given by the formula $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.Since $R_1 = R_2 = R_3 = 1\, \Omega$,we have $\frac{1}{R_p} = 1 + 1 + 1 = 3$,which implies $R_p = \frac{1}{3}\, \Omega$.This parallel combination is connected in series with a $\frac{2}{3}\, \Omega$ resistor.The total resistance $R_{eq}$ in a series circuit is the sum of individual resistances: $R_{eq} = R_p + R_{series}$.Therefore,$R_{eq} = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1\, \Omega$.

Three resistances of $1\, \Omega$ each are connected in parallel. Such a connection is then connected in series with a $\frac{2}{3}\, \Omega$ resistor. The resultant resistance will be ........... $\Omega$.

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