Question

The unit of L/R is (where L = inductance and R = resistance)

• A. sec
• B. sec\(^{-1}\)
• C. volt
• D. ampere

Hint:

Time constant \(\tau = L/R\) gives time for current to reach 63.2% of final value.

Answer 1

The Correct Option is A

To determine the unit of \(\frac{L}{R}\), where L is the inductance and R is the resistance, we can analyze the units of each component:

1. Inductance (L) is measured in henrys (H). The fundamental units of inductance are: 
   \(\text{H} = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^2 \cdot \text{A}^2}\)
2. Resistance (R) is measured in ohms (\(\Omega\)). The fundamental units of resistance are: 
   \(\Omega = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^3 \cdot \text{A}^2}\)

Now, let's find the unit of the expression \(\frac{L}{R}\):

\(\frac{L}{R} = \frac{\text{H}}{\Omega} = \frac{\frac{\text{kg} \cdot \text{m}^2}{\text{s}^2 \cdot \text{A}^2}}{\frac{\text{kg} \cdot \text{m}^2}{\text{s}^3 \cdot \text{A}^2}} = \text{s}\)

When we simplify the above expression, we find that the units of \(\frac{L}{R}\) reduce to seconds (s).

Therefore, the correct answer is: sec.

Let's rule out the other options:

• sec^-1: This would imply a unit of frequency, which is not applicable here.
• volt: This is a unit of electromotive force and does not apply to L/R.
• ampere: This is a unit of electric current, also not applicable in this context.

Thus, the unit of \(\frac{L}{R}\) is indeed sec.