Step 1: Understanding the Concept:
The ratio $\frac{L}{R}$ represents the time constant of a series Inductor-Resistor (LR) circuit.
The time constant governs how quickly the current builds up or decays in the circuit.
Since it characterizes a duration of time, its physical dimension must be time.
Step 2: Key Formula or Approach:
We can derive the units directly from standard formulas:
Voltage across an inductor: $V = L \frac{di}{dt} \implies L = \frac{V \cdot t}{I}$.
Ohm's Law for resistance: $V = IR \implies R = \frac{V}{I}$.
Substitute these into the expression $\frac{L}{R}$.
Step 3: Detailed Explanation:
Let's analyze the dimensions of Inductance $L$ and Resistance $R$.
From $L = \frac{V \cdot dt}{dI}$, the unit of $L$ is $\frac{\text{Volt} \cdot \text{second}}{\text{Ampere}}$ ($\text{V}\cdot\text{s/A}$ or Henry).
From $R = \frac{V}{I}$, the unit of $R$ is $\frac{\text{Volt}}{\text{Ampere}}$ ($\text{V/A}$ or Ohm).
Now, construct the unit for the ratio $\frac{L}{R}$:
\[ \text{Unit of } \left(\frac{L}{R}\right) = \frac{\text{Unit of } L}{\text{Unit of } R} \]
\[ \text{Unit of } \left(\frac{L}{R}\right) = \frac{\frac{\text{V} \cdot \text{s}}{\text{A}}}{\frac{\text{V}}{\text{A}}} \]
When we divide these fractions, the Volts and Amperes cancel out:
\[ \text{Unit of } \left(\frac{L}{R}\right) = \left(\frac{\text{V} \cdot \text{s}}{\text{A}}\right) \times \left(\frac{\text{A}}{\text{V}}\right) \]
\[ \text{Unit of } \left(\frac{L}{R}\right) = \text{second (s)} \]
Therefore, the SI unit is the second.
Step 4: Final Answer:
The SI unit is second.