(N/A) When three resistors $R_1, R_2,$ and $R_3$ are connected in series,the same current $I$ flows through each resistor. The potential difference across each resistor is $V_1 = IR_1, V_2 = IR_2,$ and $V_3 = IR_3$. The total potential difference $V$ is $V = V_1 + V_2 + V_3 = I(R_1 + R_2 + R_3)$. If $R_s$ is the equivalent resistance,then $V = IR_s$. Thus,$R_s = R_1 + R_2 + R_3$.
$(b)$ The given circuit shows three resistors connected in parallel with values $R_1 = 6 \Omega, R_2 = 10 \Omega,$ and $R_3 = 15 \Omega$.
For parallel connection,the equivalent resistance $R_p$ is given by:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{10} + \frac{1}{15}$
$\frac{1}{R_p} = \frac{5 + 3 + 2}{30} = \frac{10}{30} = \frac{1}{3}$
Therefore,$R_p = 3 \Omega$.