The voltage applied to a purely inductive coil of self inductance 15.9 m H is given by the equation `V = 100 sin 314 t + 75 sin 942t + 50 sin 1570t`.
Find the equation of the resulting current.

Answer

To find the equation of the resulting current in a purely inductive coil given the voltage equation, we can follow these steps:

### Step 1: Identify the given values
- Self-inductance \( L = 15.9 \, \text{mH} = 15.9 \times 10^{-3} \, \text{H} \)
- Voltage equation \( V(t) = 100 \sin(314t) + 75 \sin(942t) + 50 \sin(1570t) \)

### Step 2: Use the relationship between voltage and current in an inductor
In a purely inductive circuit, the voltage \( V \) across the inductor is related to the current \( I \) through the equation:
\[ V = L \frac{dI}{dt} \]
From this, we can express the current \( I \) as:
\[ I = \frac{1}{L} \int V \, dt \]

### Step 3: Substitute the voltage equation into the integral
We substitute the given voltage equation into the integral:
\[ I(t) = \frac{1}{L} \int (100 \sin(314t) + 75 \sin(942t) + 50 \sin(1570t)) \, dt \]

### Step 4: Integrate each term separately
Now we integrate each term:
1. For \( 100 \sin(314t) \):
\[ \int 100 \sin(314t) \, dt = -\frac{100}{314} \cos(314t) \]
2. For \( 75 \sin(942t) \):
\[ \int 75 \sin(942t) \, dt = -\frac{75}{942} \cos(942t) \]
3. For \( 50 \sin(1570t) \):
\[ \int 50 \sin(1570t) \, dt = -\frac{50}{1570} \cos(1570t) \]

### Step 5: Combine the results of the integrals
Combining these results, we have:
\[ I(t) = \frac{1}{L} \left( -\frac{100}{314} \cos(314t) - \frac{75}{942} \cos(942t) - \frac{50}{1570} \cos(1570t) \right) \]

### Step 6: Substitute the value of \( L \)
Now substitute \( L = 15.9 \times 10^{-3} \):
\[ I(t) = \frac{1}{15.9 \times 10^{-3}} \left( -\frac{100}{314} \cos(314t) - \frac{75}{942} \cos(942t) - \frac{50}{1570} \cos(1570t) \right) \]

### Step 7: Simplify the expression
Calculating the constants:
- \( \frac{100}{314} \approx 0.3185 \)
- \( \frac{75}{942} \approx 0.0795 \)
- \( \frac{50}{1570} \approx 0.0318 \)

Thus, the current equation becomes:
\[ I(t) \approx -20 \cos(314t) - 5 \cos(942t) - 2 \cos(1570t) \]

### Final Result
The equation of the resulting current is:
\[ I(t) = -20 \cos(314t) - 5 \cos(942t) - 2 \cos(1570t) \]