Example

Question:

A resistor of \(200\,\Omega\) and a capacitor of \(15.0\,\mu\text{F}\) are connected in series to a \(220\,\text{V},\ 50\,\text{Hz}\) AC source.
(a) Calculate the current in the circuit.
(b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.

Solution:

Given: \(R = 200\,\Omega,\ C = 15.0\,\mu\text{F} = 15.0 \times 10^{-6}\text{F},\ V = 220\,\text{V},\ \nu = 50\,\text{Hz}\)
(a) Impedance: \[ Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + (2\pi \nu C)^{-2}} \] \[ Z = \sqrt{(200)^2 + (212.3)^2} = 291.67\,\Omega \] Current: \[ I = \frac{V}{Z} = \frac{220}{291.67} = 0.755\,\text{A} \] (b) Voltage across resistor: \[ V_R = I R = 0.755 \times 200 = 151\,\text{V} \] Voltage across capacitor: \[ V_C = I X_C = 0.755 \times 212.3 = 160.3\,\text{V} \] Algebraic sum \(V_R + V_C = 311.3\,\text{V}\), which is more than the source voltage (\(220\,\text{V}\)).
Resolution: The voltages are not in the same phase; they must be added via Pythagoras: \[ V_{R+C} = \sqrt{V_R^2 + V_C^2} = 220\,\text{V} \] Thus, the total matches the source voltage when phase is accounted for.