An alternating e.m.f. of angular frequency `omega` is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency

Options

A. `(omega)/4`
B. `(omega)/2`
C. `omega`
D. `2 omega`

Explanation

To solve the problem, we need to find the angular frequency of the instantaneous power developed in a circuit with an alternating e.m.f. applied across an inductance. Here’s a step-by-step solution:

### Step 1: Define the Alternating e.m.f.
Let the alternating e.m.f. be given by:
\[ E(t) = E_0 \sin(\omega t) \]
where \(E_0\) is the maximum e.m.f. and \(\omega\) is the angular frequency.

**Hint:** Remember that the e.m.f. can be expressed as a sinusoidal function.

### Step 2: Determine the Inductive Current
For a pure inductor, the current lags the e.m.f. by \(90^\circ\) (or \(\frac{\pi}{2}\) radians). Therefore, the instantaneous current \(I(t)\) can be expressed as:
\[ I(t) = I_0 \sin\left(\omega t - \frac{\pi}{2}\right) \]
where \(I_0\) is the maximum current.

**Hint:** The phase difference between voltage and current in an inductor is crucial.

### Step 3: Calculate the Instantaneous Power
The instantaneous power \(P(t)\) in the circuit is given by the product of the instantaneous voltage and current:
\[ P(t) = E(t) \cdot I(t) = E_0 \sin(\omega t) \cdot I_0 \sin\left(\omega t - \frac{\pi}{2}\right) \]

**Hint:** The power is the product of voltage and current.

### Step 4: Use the Trigonometric Identity
Using the trigonometric identity for the product of sine functions:
\[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]
we can substitute \(A = \omega t\) and \(B = \left(\omega t - \frac{\pi}{2}\right)\):
\[ P(t) = E_0 I_0 \cdot \frac{1}{2} \left[\cos\left(\omega t - \left(\omega t - \frac{\pi}{2}\right)\right) - \cos\left(\omega t + \left(\omega t - \frac{\pi}{2}\right)\right)\right] \]
This simplifies to:
\[ P(t) = \frac{E_0 I_0}{2} \left[\cos\left(\frac{\pi}{2}\right) - \cos(2\omega t - \frac{\pi}{2})\right] \]

**Hint:** Remember to simplify the cosine terms carefully.

### Step 5: Simplify the Expression
Since \(\cos\left(\frac{\pi}{2}\right) = 0\), we have:
\[ P(t) = -\frac{E_0 I_0}{2} \cos(2\omega t - \frac{\pi}{2}) \]
This shows that the instantaneous power oscillates with an angular frequency of \(2\omega\).

**Hint:** The negative sign indicates the phase shift but does not affect the frequency.

### Conclusion
The angular frequency of the instantaneous power developed in the circuit is:
\[ \text{Angular frequency of } P(t) = 2\omega \]

Thus, the answer is option **D**.