If `L` is the inductance, `'I'` is current in the circuit, `(1)/(2) Li^(2)` has the dimesions of

Options

A. Work
B. Power
C. Pressure
D. Force

Answer

To find the dimensions of the expression \(\frac{1}{2} L I^2\), where \(L\) is inductance and \(I\) is current, we can follow these steps:

### Step 1: Identify the dimensions of inductance \(L\)
The dimension formula for inductance \(L\) is given by:
\[ [L] = M L^2 T^{-2} I^{-2} \]
where \(M\) is mass, \(L\) is length, \(T\) is time, and \(I\) is electric current (in amperes).

### Step 2: Identify the dimensions of current \(I\)
The dimension of current \(I\) is:
\[ [I] = I \]

### Step 3: Calculate the dimensions of \(I^2\)
Since we need \(I^2\), we square the dimension of current:
\[ [I^2] = I^2 \]

### Step 4: Combine the dimensions of \(L\) and \(I^2\)
Now, we can combine the dimensions of \(L\) and \(I^2\):
\[ \text{Dimensions of } \frac{1}{2} L I^2 = [L] \cdot [I^2] = (M L^2 T^{-2} I^{-2}) \cdot (I^2) \]

### Step 5: Simplify the expression
When we multiply the dimensions, we get:
\[ M L^2 T^{-2} I^{-2} \cdot I^2 = M L^2 T^{-2} \]

### Step 6: Final result
Thus, the dimensions of \(\frac{1}{2} L I^2\) are:
\[ M L^2 T^{-2} \]
This is the dimension of work or energy.