The current through an inductor of `1H` is given by `i =3tsin t`. Find the voltage across the inductor.

Options

A. `3sin t+3cos t `
B. `3cos t+3sin t `
C. `3sin t+3tcos t`
D. `3tcos t+sin t`

Answer

To find the voltage across the inductor when the current is given by \( i = 3t \sin t \), we will use the formula for the voltage across an inductor, which is given by:

\[ V_L = L \frac{dI}{dt} \]

where \( L \) is the inductance and \( \frac{dI}{dt} \) is the rate of change of current with respect to time.

### Step 1: Identify the given values
- Inductance \( L = 1 \, \text{H} \)
- Current \( i(t) = 3t \sin t \)

### Step 2: Differentiate the current with respect to time
We need to find \( \frac{dI}{dt} \). Using the product rule for differentiation, we differentiate \( i(t) = 3t \sin t \):

\[ \frac{dI}{dt} = \frac{d}{dt}(3t \sin t) = 3 \sin t + 3t \cos t \]

### Step 3: Substitute \( \frac{dI}{dt} \) into the voltage formula
Now we substitute \( \frac{dI}{dt} \) back into the voltage formula:

\[ V_L = L \frac{dI}{dt} = 1 \cdot (3 \sin t + 3t \cos t) \]

### Step 4: Simplify the expression
Since \( L = 1 \, \text{H} \), we get:

\[ V_L = 3 \sin t + 3t \cos t \]

### Final Answer
Thus, the voltage across the inductor is:

\[ V_L = 3 \sin t + 3t \cos t \]