An alternating voltage `E=E_0 sin omega t` , is applied across a coil of inductor L. The current flowing through the circuit at any instant is

Options

A. `(E_0)/(omegaL)sin (omegat+pi//2)`
B. `(E_0)/(omegaL)sin (omegat-pi//2)`
C. `E_0 omegaL sin (omegat-pi//2)`
D. `E_0 omegaL sin (omegat+pi//2)`

Answer

To solve the problem of finding the current flowing through a coil of inductor \( L \) when an alternating voltage \( E = E_0 \sin(\omega t) \) is applied, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between voltage and current in an inductor**:
The voltage across an inductor is given by the formula:
\[ V_L = L \frac{dI}{dt} \]
where \( V_L \) is the voltage across the inductor, \( L \) is the inductance, and \( \frac{dI}{dt} \) is the rate of change of current with respect to time.

2. **Substitute the given voltage into the equation**:
We know that the applied voltage is:
\[ E = E_0 \sin(\omega t) \]
Therefore, we can set up the equation:
\[ E_0 \sin(\omega t) = L \frac{dI}{dt} \]

3. **Rearrange the equation to isolate \( dI \)**:
Rearranging the equation gives:
\[ \frac{dI}{dt} = \frac{E_0 \sin(\omega t)}{L} \]
Now, we can separate the variables for integration:
\[ dI = \frac{E_0}{L} \sin(\omega t) dt \]

4. **Integrate both sides**:
We will integrate both sides. The left side integrates to \( I \), and the right side requires integrating \( \sin(\omega t) \):
\[ I = \int \frac{E_0}{L} \sin(\omega t) dt \]
The integral of \( \sin(\omega t) \) is:
\[ -\frac{1}{\omega} \cos(\omega t) \]
Therefore, we have:
\[ I = -\frac{E_0}{L \omega} \cos(\omega t) + C \]
where \( C \) is the constant of integration.

5. **Express the current in terms of sine function**:
We can express the cosine function in terms of sine:
\[ \cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right) \]
Thus, we can rewrite the current as:
\[ I = \frac{E_0}{L \omega} \sin\left(\omega t - \frac{\pi}{2}\right) + C \]

6. **Determine the constant of integration**:
If we assume that at \( t = 0 \), the current \( I \) is zero (initial condition), we can find \( C \):
\[ 0 = \frac{E_0}{L \omega} \sin\left(0 - \frac{\pi}{2}\right) + C \]
Since \( \sin\left(-\frac{\pi}{2}\right) = -1 \):
\[ 0 = -\frac{E_0}{L \omega} + C \implies C = \frac{E_0}{L \omega} \]

7. **Final expression for current**:
Substituting \( C \) back into the equation for \( I \):
\[ I = \frac{E_0}{L \omega} \sin\left(\omega t - \frac{\pi}{2}\right) + \frac{E_0}{L \omega} \]

### Final Result:
The current flowing through the circuit at any instant is:
\[ I(t) = \frac{E_0}{L \omega} \sin\left(\omega t - \frac{\pi}{2}\right) \]