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(D) The circuit consists of three resistors connected in parallel, which are then connected in series with a fourth resistor.Let the value of each res

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$2\frac{2}{3}\, \Omega$

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(D) The circuit consists of three resistors connected in parallel, which are then connected in series with a fourth resistor.Let the value of each resistor be $R = 2\, \Omega$.First, calculate the equivalent resistance of the three parallel resistors $(R_p)$:$\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}$$R_p = \frac{R}{3} = \frac{2}{3}\, \Omega$.Now, this parallel combination is in series with the remaining resistor $R$.Therefore, the total equivalent resistance $R_{AB}$ is:$R_{AB} = R_p + R = \frac{2}{3} + 2 = \frac{2 + 6}{3} = \frac{8}{3}\, \Omega = 2\frac{2}{3}\, \Omega$.Thus, the correct option is $(d)$.

If all the resistors shown in the circuit have a value of $2\, \Omega$ each, find the equivalent resistance between points $A$ and $B$.

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