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

Question: Prove that the equivalent capacitance of two capacitors connected in series is always less than the smaller of the two capacitances. Answer: By using the formula for equivalent capacitance in series, $C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}}$, we can show that if we assume $C_1 \leq C_2$, then $C_{eq} < C_{1}$. This proves that the equivalent capacitance is always less than the smaller capacitance.

Step-by-step solution

Write the formula for equivalent capacitance in series

The formula for the equivalent capacitance \(C_{eq}\) when two capacitors are connected in series is given by: $$ \frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} $$

Rewrite the expression for equivalent capacitance

To make the comparison with \(C_1\) and \(C_2\) easier, let's rewrite the expression for \(C_{eq}\): $$ C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}} $$

Compare the equivalent capacitance to the smaller capacitance

Now we will compare \(C_{eq}\) to \(C_1\) and \(C_2\) separately. Without loss of generality, let's suppose that \(C_1 \leq C_2\). We will demonstrate that \(C_{eq} < C_1\): $$ C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}} < \frac{C_{1}C_{2}}{C_{1}} = C_{2} $$ Since \(C_1 \leq C_2\), we have \(C_{eq} < C_{1}\). Hence, the equivalent capacitance is always less than the smaller capacitance.