The voltage  =  ( + ) is applied to a   capacitor, calculate the current through the capacitor, and Draw phasor diagram?

To calculate the current through the capacitor when a voltage is applied, we will follow these steps:

1. Break Down the Problem

We need to find the current $I$ through the capacitor when a sinusoidal voltage $V = V_m \sin(\omega t + \phi)$ is applied across it.

2. Relevant Concepts

The current $I$ through a capacitor can be described by the following formula: $I = C \frac{dV}{dt}$ where:

• $I$ = current through the capacitor (in Amperes)
• $C$ = capacitance of the capacitor (in Farads)
• $V$ = voltage across the capacitor (in Volts)

In phasor form, the relationship between voltage and current for a capacitor is: $I = j\omega C V$ where $j$ is the imaginary unit.

3. Analysis and Detail

Let’s assume:

• $V = V_m \sin(\omega t + \phi)$
• $\omega$ is the angular frequency in radians per second.
• $V_m$ is the peak voltage.

The derivative of the voltage with respect to time will be: $\frac{dV}{dt} = V_m \omega \cos(\omega t + \phi)$

Now substituting it into the current formula: $I = C \frac{dV}{dt} = C V_m \omega \cos(\omega t + \phi)$

In phasor form: $I = j \omega C V_m e^{j(\phi + \frac{\pi}{2})}$

4. Verify and Summarize

If we need to summarize:

1. The current through the capacitor in the time domain is: $I(t) = C V_m \omega \cos(\omega t + \phi)$

2. For phasor representation, we can denote the current as: $I = j \omega C V_m e^{j\phi}$

Final Answer

The current through the capacitor is given by: $I(t) = C V_m \omega \cos(\omega t + \phi) \quad \text{(in time domain)}$ In phasor form: $I = j \omega C V_m e^{j \phi}$

Phasor Diagram

• The voltage phasor $V$ rotates counter-clockwise in the complex plane.
• The current phasor $I$ will lead the voltage phasor by $90^\circ$.

Note: The drawing of the phasor diagram is not possible in a text-based format. However, you should visualize the voltage vector along the real axis and the current vector $90^\circ$ ahead (pointing upwards in the imaginary direction).