*Question G5C12:
What is the capacitance of a 20-microfarad capacitor connected in series with a 50-microfarad capacitor?*  

*Correct Answer is  B: 14.3 microfarads*  

### AI Response: (may contain errors)  

When capacitors are connected in series, the total capacitance is less than the smallest individual capacitor in the series. The formula to calculate the total capacitance ($C_T$) for capacitors in series is:

$$
\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}
$$

For this question, we have two capacitors:

- $C_1 = 20 \, \text{microfarads}$
- $C_2 = 50 \, \text{microfarads}$

Plug these values into the formula:

$$
\frac{1}{C_T} = \frac{1}{20} + \frac{1}{50}
$$

Calculate each term:

- $\frac{1}{20} = 0.05$
- $\frac{1}{50} = 0.02$

Add these values:

$$
\frac{1}{C_T} = 0.05 + 0.02 = 0.07
$$

Now, take the reciprocal to find $C_T$:

$$
C_T = \frac{1}{0.07} \approx 14.3 \, \text{microfarads}
$$

Thus, the correct answer is 14.3 microfarads (Choice B).

**Why the other choices are incorrect:**

- **Choice A: 0.07 microfarads**  
  This is the reciprocal of the total capacitance, not the capacitance itself.

- **Choice C: 70 microfarads**  
  This value is incorrect because it is larger than the smallest capacitor in the series, which is not possible for series capacitance.

- **Choice D: 1,000 microfarads**  
  This is far too large and not feasible for the given series configuration. 

Remember, for capacitors in series, the total capacitance is always less than the smallest individual capacitor.