It increases the time constant ($\tau$), causing the current to change more slowly.

The current starts to increase, and the inductor generates a back emf that opposes the change in current.

It leads to a decrease in the inductor's stored energy over time, as the energy is converted to heat.

The inductor acts like a short circuit, and the current reaches its maximum value, determined by the battery's emf and the resistance.

It decreases the time constant ($\tau$), causing the current to reach its steady-state value more quickly and reduces the maximum current.

The inductor induces an emf that opposes the change in current, according to Lenz's Law.

1. Identify a closed loop in the circuit. 2. Sum the voltage drops across each element in the loop, considering the sign. 3. Set the sum equal to zero: $\mathcal{E} - IR - L \frac{dI}{dt} = 0$.

1. Identify the inductance (L) and the equivalent resistance ($R_{eq}$) in the circuit. 2. Use the formula: $\tau = \frac{L}{R_{eq}}$.

1. Apply Kirchhoff's loop rule to obtain a differential equation. 2. Solve the differential equation for the current as a function of time, I(t). 3. Analyze how the current, voltage, and energy change over time using exponential functions.

1. After a long time, treat the inductor as a short circuit (a wire with zero resistance). 2. Calculate the current through the circuit using Ohm's law, considering only the battery's emf and the total resistance.

1. Recognize that the current through the inductor cannot change instantaneously. 2. If the initial current is zero, the current remains zero immediately after closing the switch.

1: Battery ($\mathcal{E}$), 2: Resistor (R), 3: Inductor (L), 4: Switch