*Question G5C10:
What is the inductance of three 10-millihenry inductors connected in parallel?*  

*Correct Answer is  C: 3.3 millihenries*  

### AI Response: (may contain errors)  

When inductors are connected in parallel, the total inductance is less than the smallest individual inductance. The formula for calculating the total inductance ($L_T$) of inductors in parallel is similar to that for resistors in parallel:

$$
\frac{1}{L_T} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}
$$

For three 10-millihenry inductors in parallel:

- $L_1 = 10 \, \text{mH}$
- $L_2 = 10 \, \text{mH}$
- $L_3 = 10 \, \text{mH}$

Substitute these values into the formula:

$$
\frac{1}{L_T} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10}
$$

$$
\frac{1}{L_T} = \frac{3}{10}
$$

To find $L_T$, take the reciprocal:

$$
L_T = \frac{10}{3} \, \text{mH} \approx 3.3 \, \text{mH}
$$

Thus, the correct answer is 3.3 millihenries (Choice C).

### Why the Other Choices are Incorrect:
- **Choice A: 0.30 henries**  
  This is equivalent to 300 millihenries, which is much larger than any of the individual inductors. Inductance in parallel should be less than the smallest inductor.

- **Choice B: 3.3 henries**  
  This is equivalent to 3300 millihenries, which is incorrect for the same reason as Choice A.

- **Choice D: 30 millihenries**  
  This is simply the sum of the inductors if they were in series, not parallel. In parallel, the total inductance is always less than the smallest individual inductance.

Remember, for inductors in parallel, the total inductance is always less than the smallest individual inductance, and you use the reciprocal formula to calculate it.