The dimension of `R/L` are

Options

A. `T^(2)`
B. `T`
C. `T^(-1)`
D. `T^(-2)`

Answer

To find the dimensions of \( \frac{R}{L} \), we will analyze the dimensions of resistance \( R \) and inductance \( L \) separately and then combine them.

### Step-by-Step Solution:

1. **Understanding Resistance (R)**:
- The formula for energy dissipated in a resistor is given by:
\[ E = I^2 R t \]
where \( E \) is energy, \( I \) is current, \( R \) is resistance, and \( t \) is time.
- Rearranging for \( R \):
\[ R = \frac{E}{I^2 t} \]

2. **Understanding Inductance (L)**:
- The energy stored in an inductor is given by:
\[ E = \frac{1}{2} L I^2 \]
- Rearranging for \( L \):
\[ L = \frac{2E}{I^2} \]

3. **Finding Dimensions**:
- The dimensions of energy \( E \) are:
\[ [E] = [M L^2 T^{-2}] \]
- The dimensions of current \( I \) are:
\[ [I] = [I] \]
- Therefore, substituting these into the expression for \( R \):
\[ [R] = \frac{[M L^2 T^{-2}]}{[I^2][T]} = [M L^2 T^{-3} I^{-2}] \]

4. **Substituting into Inductance**:
- For \( L \):
\[ [L] = \frac{[M L^2 T^{-2}]}{[I^2]} = [M L^2 T^{-2} I^{-2}] \]

5. **Finding Dimensions of \( \frac{R}{L} \)**:
- Now, we can find the dimensions of \( \frac{R}{L} \):
\[ \frac{R}{L} = \frac{[M L^2 T^{-3} I^{-2}]}{[M L^2 T^{-2} I^{-2}]} = \frac{[M L^2 T^{-3} I^{-2}]}{[M L^2 T^{-2} I^{-2}]} = [T^{-1}] \]

6. **Final Result**:
- Thus, the dimensions of \( \frac{R}{L} \) are:
\[ [\frac{R}{L}] = [T^{-1}] \]

### Final Answer:
The dimension of \( \frac{R}{L} \) is \( T^{-1} \).

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