If the current in a primary circuit is `I=I_(0)` Sin `omegat` and the mutual inductance is M, then the value of induced voltage in secondary circuit will be :-

Options

A. `e=MI_(0) omega Cos omega t`
B. `e=MI_(0) omega Cost omega t`
C. `e=[M omega Cos omegat] //I_(0)`
D. `e=(M omega Cos omega t)//I_(0)`

Answer

To find the induced voltage in the secondary circuit when the current in the primary circuit is given by \( I = I_0 \sin(\omega t) \) and the mutual inductance is \( M \), we can follow these steps:

### Step 1: Understand the relationship between current and induced voltage
The induced voltage (or electromotive force, EMF) in the secondary circuit due to a changing current in the primary circuit is given by the formula:
\[ E = M \frac{dI}{dt} \]
where \( E \) is the induced voltage, \( M \) is the mutual inductance, and \( \frac{dI}{dt} \) is the rate of change of current in the primary circuit.

### Step 2: Differentiate the current function
The current in the primary circuit is given by:
\[ I = I_0 \sin(\omega t) \]
To find \( \frac{dI}{dt} \), we differentiate \( I \) with respect to \( t \):
\[ \frac{dI}{dt} = I_0 \frac{d}{dt}(\sin(\omega t)) = I_0 \omega \cos(\omega t) \]
Here, we used the chain rule where the derivative of \( \sin(x) \) is \( \cos(x) \) and multiplied by the derivative of \( \omega t \) which is \( \omega \).

### Step 3: Substitute \( \frac{dI}{dt} \) into the induced voltage formula
Now, we substitute \( \frac{dI}{dt} \) back into the formula for induced voltage:
\[ E = M \frac{dI}{dt} = M \left(I_0 \omega \cos(\omega t)\right) \]
This simplifies to:
\[ E = M I_0 \omega \cos(\omega t) \]

### Step 4: Final expression for the induced voltage
Thus, the induced voltage in the secondary circuit is:
\[ E = M I_0 \omega \cos(\omega t) \]

### Conclusion
The value of the induced voltage in the secondary circuit is \( M I_0 \omega \cos(\omega t) \).

---