x=2V sinft-) Consider the above series AC circuit, determine the series current I, in its phasor format OI, = 1.414mAZ-113.13° I, 1mA/113.13 I, 1.414mA/113.13 R = 640 X₂=UND X = 160 voo OI, = 1mA/36.87
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- A clipper is a device that removes either the positive half (top half) or negative half (bottom half), or both positive and negative halves of the input AC signal. In other words, a clipper is a device that limits the positive amplitude or negative amplitude or both positive and negative amplitudes of the input AC signal. In some cases, a clipper removes a small portion of the positive half cycle or negative half cycle or both positive and negative half cycles. In the below circuit diagram, the positive half cycles are removed by using the series positive clipper. QUESTION: In your own opinion. Why do we need to clip a certain amount of voltage in positive or negative or on both sides? What is the benefit of that in our devices or circuits in doing such thing?Q6. Calculate the total current in the circuit below if the circuit is powered by the sine wave voltage source of 150 Vm and frequency 60 Hz. Draw the current phasor diagram and determine if this circuit fulfils the criteria of resonance. Vm Sine 60 Hz C L1 0.7 H L2 0.5 H relee cele Figure Q6 C1 5.1 mF C2 0.12 mF R1 120 22 R2 330 ΩC. Draw a general sine wave V vs t and I vs t plot for a purely resistive and a purely inductive circuit and show leading lagging parameters and clearly mark the phase difference.