Capacitors in Series

When two capacitors are in series we can replace them with a single equivalent capacitor whose capacitance is given by: \(C_{eq} = \frac{C_1 C_2}{C_1 + C_2}\) OR \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\) This are the same equations as for resistors in parallel.

Capacitors in series always have a lower capacitance then any of the individual capacitors.

If three or more capacitors are in series we reduce them two at a time just like with resistors.

When capacitors are in series they all have the same charge across them. This charge is given by \(Q = C_{eq}\times V\) where \(C_{eq}\) is the capacitance of the single, equivalent capacitor and \(V\) is the voltage across the equivalent capacitor.

When several capacitors are in series the voltage across each one is different depending on its capacitance since they all have the same charge \(Q\) across them as given above. To find the voltage across each capacitor calculate the charge across each capacitor then apply the formula \(Q = C V\) for that individual capacitor.

Find the voltage across each of the capacitors in the circuit below.

First we combine the capacitors into an equivalent capacitor, then we find the charge across it - this will be the charge across both the original capacitors. Then we'll use our formula to find the voltage across each capacitor since we'll have its capacitance and the charge on it. So we find that \(C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{50}{15} = 3.33\)F From the circuit we can see that the voltage across \(C_{eq}\) is 40V so the charge across it is: \(Q = C_{eq} V = 3.33 \times 40 = 133.2\) Coulombs. Since \(C_1\) and \(C_2\) are in series \(Q_{C_1} = Q_{C_2} = Q\). Now the voltage across \(C_1\) is given by: \(V_{C_1} = \frac{Q}{C_1} = 13.32\)V and the voltage across \(C_2\) is given by: \(V_{C_2} = \frac{Q}{C_2} = 26.64\)V