This is a problem related to a series RL circuit where a voltage of 90V is applied across a resistor $R = 60 \, \Omega$ and an inductor $L = 30 \, H$. We are asked to find the expression for the current through the inductor, $i_L(t)$.

Step-by-step analysis:

1. Time Constant ( $\tau$ ):
   For an RL circuit, the time constant $\tau$ is given by: $\tau = \frac{L}{R}$ Plugging in the values: $\tau = \frac{30}{60} = 0.5 \, \text{seconds}$

2. Steady-State Current:
   The steady-state current $I_{ss}$ in the circuit can be calculated using Ohm's law: $I_{ss} = \frac{V}{R} = \frac{90}{60} = 1.5 \, \text{A}$

3. Current Expression:
   In a series RL circuit, the current $i_L(t)$ as a function of time when a step voltage is applied is given by: $i_L(t) = I_{ss} \left( 1 - e^{-\frac{t}{\tau}} \right)$ Substituting the known values of $I_{ss} = 1.5 \, A$ and $\tau = 0.5 \, s$: $i_L(t) = 1.5 \left( 1 - e^{-2t} \right)$

Thus, the correct expression for the current is: $\boxed{1.5 \left( 1 - e^{-2t} \right) \, A}$

This matches option a in the given choices.

Would you like more details or have any questions?

Here are 5 related questions to expand this topic:

1. How does the time constant $\tau$ affect the response of an RL circuit?
2. What is the energy stored in the inductor at steady state?
3. How would the current change if the resistance $R$ is increased?
4. How is the voltage across the inductor related to the current?
5. What happens to the circuit if the inductor is replaced by a capacitor?

Tip: In RL circuits, the current grows exponentially with time, reaching its maximum value as the inductor opposes changes in current initially but allows a steady flow after a long time.