To prove that the instantaneous voltage in an AC circuit is given by $$V_m \sin \omega t$$ using a phasor diagram, let's break it down step-by-step:
Step 1: Phasor representation
A phasor represents the amplitude and phase of a sinusoidal quantity as a rotating vector in the complex plane. The voltage phasor rotates counterclockwise with an angular velocity $$\omega$$, corresponding to the AC source frequency.
Step 2: Voltage components in phasor form
If the peak voltage (amplitude) of the sinusoidal signal is $$V_m$$, then the phasor at time $$t$$ is given by: $$V = V_m \angle (\omega t)$$ The angle $$\omega t$$ indicates the instantaneous phase of the voltage.
Step 3: Relationship between phasor and sinusoidal voltage
To find the instantaneous voltage, take the projection of the phasor onto the vertical axis (y-axis). This projection corresponds to: $$v(t) = V_m \sin \omega t$$
This is because the y-component of the phasor, which oscillates sinusoidally as the vector rotates, represents the instantaneous voltage. The cosine projection along the x-axis would similarly represent a $$\cos \omega t$$ waveform.
Step 4: Visualizing with a phasor diagram
In the phasor diagram:
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The length of the vector is $$V_m$$ (the amplitude).
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As the phasor rotates about the origin, its projection on the y-axis changes sinusoidally with time.
Thus, the phasor diagram provides a geometric proof for the expression $$v(t) = V_m \sin \omega t$$ as the instantaneous voltage in an AC circuit.
Let me know if you'd like a detailed explanation of how this applies to specific circuit components or want more illustrations!
