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 circuit with an alternating voltage applied across a coil of inductor \( L \), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The alternating voltage is given by \( E = E_0 \sin(\omega t) \).
- The inductance of the coil is \( L \).

2. **Apply Kirchhoff's Law**:
- According to Kirchhoff's law, the sum of the electromotive forces (emf) in a closed loop is zero. The instantaneous induced emf across the inductor is given by:
\[ E_i = -L \frac{dI}{dt} \]
- The net emf in the circuit can be expressed as:
\[ E - E_i = 0 \quad \Rightarrow \quad E = L \frac{dI}{dt} \]

3. **Substitute the Expression for Voltage**:
- Substitute the expression for \( E \):
\[ E_0 \sin(\omega t) = L \frac{dI}{dt} \]

4. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ \frac{dI}{dt} = \frac{E_0}{L} \sin(\omega t) \]

5. **Integrate Both Sides**:
- Integrate both sides with respect to time \( t \):
\[ \int dI = \int \frac{E_0}{L} \sin(\omega t) dt \]
- The left side integrates to \( I \), and the right side can be integrated using the integral of sine:
\[ I = \frac{E_0}{L} \left(-\frac{1}{\omega} \cos(\omega t)\right) + C \]
- Here, \( C \) is the constant of integration.

6. **Simplifying the Expression**:
- Rearranging gives:
\[ I = -\frac{E_0}{\omega L} \cos(\omega t) + C \]

7. **Determine the Constant of Integration**:
- If we assume that at \( t = 0 \), the current \( I = 0 \), we can find \( C \):
\[ 0 = -\frac{E_0}{\omega L} \cos(0) + C \quad \Rightarrow \quad C = \frac{E_0}{\omega L} \]

8. **Final Expression for Current**:
- Substituting \( C \) back into the equation gives:
\[ I = -\frac{E_0}{\omega L} \cos(\omega t) + \frac{E_0}{\omega L} \]
- This can be rewritten as:
\[ I = \frac{E_0}{\omega L} (1 - \cos(\omega t)) \]

9. **Expressing in Terms of Sine**:
- Using the identity \( \cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right) \):
\[ I = \frac{E_0}{\omega L} \sin\left(\omega t - \frac{\pi}{2}\right) \]
- Letting \( I_0 = \frac{E_0}{\omega L} \), we get:
\[ I = I_0 \sin\left(\omega t - \frac{\pi}{2}\right) \]

### Final Result:
The current flowing through the circuit at any instant is given by:
\[ I = I_0 \sin\left(\omega t - \frac{\pi}{2}\right) \]