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3. Consider the AC circuit shown in the figure below, consisting of an alternating voltage source—of voltage V (t) = V0 cos (ωt)—a capacitor (of capacitance C), an inductor (of inductance L), and two resistors (of resistances R1 and R2). Also, note the highlighted points a and b in the circuit. (a) While explaining your reasoning, determine the necessary condition that must be satisfied between the circuit elements such that the potential difference between points a and b is zero for all outputted frequencies from the alternating voltage source. [18] (b) Now, suppose, instead, that the frequency of the voltage source is a very high frequency. While discussing your rationale, determine the amplitude and phase (relative to the source voltage’s phase) of the outputted current. [11] (c) Finally, suppose, instead, that the frequency of the voltage source is a very low frequency. While discussing your rationale, determine the amplitude and phase (relative to the source voltage’s phase) of the outputted current. [11]

3. Consider the AC circuit shown in the figure below, consisting of an alternating voltage source-of voltage V (t) = V cos (w

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Answer #1

31 (a). The impedance of the R-c circuit is Voit meja [w-angular frequency of voltage so voltage at point a is - Zz No coswtAccording to the question- V = Va =) No R, Cosut VolWLY R2 HWL)² cosot | Riºt Cabec) RI => ابيا Z ✓ Ritle) Vrit (w.)? 2 2) 1Ne I e O=ton an- Xe * R Vc tant CUOR V-IZ Now for high frequency ! VIR 2. Oz tan- (o) T VR In R-L circuit for high frequency(C) for low frequency line >>R) in R-c circuit I VO T. 2 Vc VRita) vf Vowe and Oz tanel female word) [vce >>1] ~ n 2 in R-L cIn part (b) for too high frequency the amplitude of current through inductor may be assumed to be zero(as it proportional to the inverse of the frequency). So no current will pass through the inductor for too high frequency.

Similarly in part (c) for too low frequency the amplitude of current through capacitor may be assumed to be zero(as it is proportional to the frequency). So no current will pass through the capacitor for too low frequency.

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