Explanation: The time constant ($\tau$) of a Resistor-Capacitor (RC) circuit is calculated using the formula $\tau = RC$, where R is the total resistance and C is the total capacitance. First, calculate the total resistance ($R_T$) for the two 470-kilohm resistors in series: $R_T = R_1 + R_2 = 470 \text{ k}\Omega + 470 \text{ k}\Omega = 940 \text{ k}\Omega$ Next, calculate the total capacitance ($C_T$) for the two 100-microfarad capacitors in series. Capacitors in series combine like resistors in parallel: $C_T = \frac{C_1 \times C_2}{C_1 + C_2} = \frac{100 \mu F \times 100 \mu F}{100 \mu F + 100 \mu F} = \frac{10000 \mu F^2}{200 \mu F} = 50 \mu F$ Finally, calculate the time constant: $\tau = R_T \times C_T = (940 \times 10^3 \Omega) \times (50 \times 10^{-6} F)$ $\tau = 47000 \times 10^{-3} \text{ seconds} = 47 \text{ seconds}$ Option B is the correct answer. The other options would result from incorrect calculations for total resistance, total capacitance, or unit conversions. For instance, using 470 kΩ directly with 100 μF would yield 47 seconds, but that ignores the other resistor and capacitor. Incorrectly adding capacitors as if they were parallel would result in 200 μF, leading to an incorrect time constant.