Why is the load voltage greater than the supply voltage in an AC circuit?

Question

I was playing with electronics and testing the equation:

$$X = \frac{1}{2\pi f C}$$

where \$X\$ is the resistance of the capacitor, \$f\$ is the frequency, \$C\$ is the capacitance.

The supply is just a 220V/10V transformer without a rectifier, so it produces an AC current. Frequency is 50-60 Hz.

I calculated \$X\$ from the equation and it is equal to 2.8 K .

I wanted to test my calculations so I measured the AC voltage on both \$R\$ and \$C\$, and I would expect that the ratio between voltages will indicate the ratio between resistances as.

I noticed something strange, which is the sum of the voltage across \$R\$ and the voltage across \$C\$ is really greater than the supply voltage: \$V_C + V_R = 14.5\$ volts !! And when I measure the total voltage across \$R\$ and \$C\$ together it is 10.5 V.

Also, the ratio between voltages does not indicates the ratio between resistances.

I made this experiment with different resistors and caps but I got the same issue.

Why is the sum of voltages is greater than the voltage of the supply? Am I missing something?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

The resistor and capacitor voltages are out of phase. Their peak values don't happen at the same time. You can see this in the CircuitLab simulation:

Your equation for the reactance of the capacitor ignores the phase. To include it, you need to use complex numbers:

$$X_C = \frac 1 {j 2 \pi f C}$$

where \$j = \sqrt {-1}\$. For a 1 microfarad capacitor at 55 Hz, this gives:

$$X_C = \frac 1 {j 2 \pi (55\ \mathrm{Hz} \cdot 1\ \mathrm {\mu F})} = -j2.89\ \mathrm{k\Omega}$$

To find the voltage across the capacitor, use your complex reactance in the voltage divider equation:

$$V_C = V_{in} \frac {X_C} {X_C + R} = 10.5\angle 0^\circ\ \mathrm {V} \frac {-j2.89\ \mathrm{k\Omega}}{-j2.89\ \mathrm{k\Omega} + 10\ \mathrm{k\Omega}} = 2.92\angle 74^\circ\ \mathrm V$$

The resistor voltage is the difference between the input voltage and the capacitor voltage:

$$V_R = V_{in} - V_C = 10.5\angle 0^\circ\ \mathrm {V} - 2.92\angle 74^\circ\ \mathrm V = 10.09\angle 16^\circ\ \mathrm V$$

So the peak resistor voltage is about 10 volts, the peak capacitor voltage is about 2.9 volts, and the phase difference between the two voltages is exactly 90 degrees.

The reason for the phase difference is that the capacitor voltage is always 90 degrees out of phase with its current, while the resistor voltage is always in phase with its current. Since the two components share the same current, their voltages must be 90 degrees out of phase with each other.