Electromagnetic induction > AC electrical circuits > Problem 11.4.12

Statement

$11.4.12.$ a. In the vector diagram, the current $I = I_0cosωt$ is defined as the projection onto the $x$-axis of the vector $I_0$, which rotates around the point $O$ with angular velocity $\omega$. How are the voltage drop vectors located on the same diagram when this current flows through the resistance $R$, the inductor $L$ and the capacitance capacitor $C$? What are the amplitudes of the voltage vectors?

b. Using a vector diagram, determine the voltage drop in the center of a series-connected inductor $L$, resistance $R$, and capacitance capacitor $C$ and the phase shift between current and voltage in the circuit if the current in the circuit varies cosinusoidally: $I = I_0coswt$.

Solution

a) The voltage drop across the resistor is just Ohm's law:

$$V_R = I_0 R cos \omega t$$

There is a charge on the capacitor:

$$q = V_C C= \int_{0}^{t} I(t) dt$$

$$V_C C = \frac{I_0}{\omega} sin(\omega t)$$

$$V_C = \frac{I_0}{\omega C} cos(\omega t - \frac{\pi}{2})$$

Finally, we write down the equation for the voltage across the coil:

$$V_L = L \frac{dI}{dt} = -LI_0 \omega sin(\omega t)$$

$$V_L = \frac{dI}{dt} = LI_0 \omega cos(\omega t + \frac{\pi}{2})$$

The phase shift as seen in the coil $+\frac{\pi}{2}$ depends on the current, in the capacitor $-\frac{\pi}{2}$, the resistance has no shift, the amplitudes are equal:

$$V_R = I_0 R$$
$$V_C = \frac{I_0}{\omega C}$$
$$V_L = LI_0 \omega$$

b)
To find the amplitude, it is enough to add up our voltages vector based on the diagram obtained in a:

$$V = I_0 \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$$

Since we have a right triangle formed from vectors, the phase shift is easy to express in terms of a tangent.:

$$\varphi = arctg \frac{\omega L - \frac{1}{\omega C}}{R}$$

Answer

a)
$$V_R = I_0 R$$
$$V_C = \frac{I_0}{\omega C}$$
$$V_L = LI_0 \omega$$

$$\varphi = arctg \frac{\omega L - \frac{1}{\omega C}}{R}$$