Question

Two capacitors with capacitances \(C_{1}\) and \(C_{2}\) are connected in series. Show that, no matter what the values of \(C_{1}\) and \(C_{2}\) are, the equivalent capacitance is always less than the smaller of the two capacitances.

Short answer

Explain. Answer: Yes, the equivalent capacitance in a series connection is always less than the smaller of the two capacitances, no matter what values \(C_{1}\) and \(C_{2}\) have. This is proven by analyzing the relationship between the equivalent capacitance (\(C_{eq}\)) and individual capacitances (\(C_1\) and \(C_2\)) using the formula \(C_{eq} = \frac{C_{1} \cdot C_{2}}{C_{1} + C_{2}}\).

Step-by-step solution

Equivalent capacitance for series connection formula

When capacitors are connected in series, their equivalent capacitance \(C_{eq}\) can be found using the formula: $$ \frac{1}{C_{eq}} = \frac{1}{C_{1}}+\frac{1}{C_{2}} $$

Rearranging the formula to solve for \(C_{eq}\)

We need to solve the formula for \(C_{eq}\) by rearranging the terms: $$ C_{eq} = \frac{C_{1} \cdot C_{2}}{C_{1} + C_{2}} $$

Analyzing the relationship between \(C_{eq}\) and \(C_{1}\), \(C_{2}\)

Now we have the equivalent capacitance formula: $$ C_{eq} = \frac{C_{1} \cdot C_{2}}{C_{1} + C_{2}} $$ Let’s assume, without loss of generality, that \(C_1\) is the smaller capacitance of the two (\(C_1 \le C_2\)). Let's create an inequality to test our assumption: $$ C_{eq} < C_1 $$ If this inequality is true, we will have proven that no matter the values of \(C_{1}\) and \(C_{2}\), the equivalent capacitance will always be less than the smaller of the two capacitances.

Comparing \(C_{eq}\) with \(C_1\)

First, let's substitute \(C_{eq}\) using the equivalent capacitance formula: $$ \frac{C_{1} \cdot C_{2}}{C_{1} + C_{2}} < C_{1} $$ Now, we need to simplify the inequality. We can achieve this by doing some cross-multiplication: $$ C_{1} \cdot C_{2} < C_{1} (C_{1} + C_{2}) $$ To further simplify, we distribute \(C_{1}\): $$ C_{1} \cdot C_{2} < C_{1}^{2} + C_{1} \cdot C_{2} $$ Notice that if we subtract \(C_{1} \cdot C_{2}\) from both sides of the inequality, it becomes: $$ 0 < C_{1}^{2} $$ Since both \(C_1\) and \(C_2\) are capacitances, they must be positive values. This inequality is always true, which supports our claim.

Conclusion

In conclusion, the equivalent capacitance in a series connection is always less than the smaller of the two capacitances, no matter what values \(C_{1}\) and \(C_{2}\) have.