A 221 V battery, an inductor, and a 33 ohms resistor are connected in series, as shown in the diagram below. A two way switch makes it possible to include or exclude the battery. The switch that had been in position 1 long enough for the circuit to reach

After the switch is moved to position 2, the circuit can be simplified as follows: The inductor acts as a short circuit for a short period of time as it opposes any change in the current flowing through it. Thus, the resistor is effectively in parallel with the battery. Using the voltage divider rule, we can calculate the voltage across the resistor as: V = V_batt * R / (R + L/R) where V_batt = 221 V, R = 33 ?, and L/R is the time constant of the circuit. The time constant is given by: ? = L/R where L is the inductance of the inductor. The value of L is not given, so we cannot calculate the time constant. Assuming that the time constant is much smaller than the time it takes for the switch to move to position 2, we can assume that the circuit has reached a steady state before the switch is moved. After the switch is moved to position 2, the circuit starts to discharge through the resistor. The voltage across the resistor decreases exponentially with a time constant of ? = L/R. After four time constants have passed, the voltage across the resistor has decreased to: V = V_batt * e^(-4) where e is the mathematical constant \approx imately equal to 2.71828. Therefore, the voltage across the resistor after the switch has been in position 2 for four time constants is: V = 221 * e^(-4) ? 28.9 V.