The output voltage of an AC generator is given by Av = (1.06 x 10 V)sin(32xt). The generator is connected across a 0.500 H inductor. Find the following. (a) frequency of the generator Hz (b) rms voltage across the inductor (c) inductive reactance
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- The output voltage of an AC generator is given by Δ v = (170 V) sin (60πt). The generator is connected across a 20.0-Ω resistor. By inspection, what are the (a) maximum voltage and (b) frequency? Find the (c) rms voltage across the resistor, (d) rms current in the resistor, (e) maximum current in the resistor, (f) power delivered to the resistor, and (g) current when t = 0.005 0 s. (h) Should the argument of the sine function be in degrees or radians? Step 1 Dear student, "Since you have posted a question with multiple sub-parts, we will solve first three sub parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved." Thank you! GIVEN: Output voltage, Δv = (170 V) sin (60πt) Resistance, R = 20.0-Ω arrow_forward Step 2 Part (a) Part (a) Standard equation can be represented as, Δv = Vmsin(ωt) Where, Vm = maximum voltage. ω = angular frequency. Compare the given equation with standard equation. We get, Vm…The output voltage of an AC generator is given by Δ v = (170 V) sin (60πt). The generator is connected across a 20.0-Ω resistor. By inspection, what are the (a) maximum voltage and (b) frequency? Find the (c) rms voltage across the resistor, (d) rms current in the resistor, (e) maximum current in the resistor, (f) power delivered to the resistor, and (g) current when t = 0.005 0 s. (h) Should the argument of the sine function be in degrees or radians? GIVEN: Output voltage, Δv = (170 V) sin (60πt) Resistance, R = 20.0-Ω arrow_forward Step 2 Part (a) Part (a) Standard equation can be represented as, Δv = Vmsin(ωt) Where, Vm = maximum voltage. ω = angular frequency. Compare the given equation with standard equation. We get, Vm = 170 V Answer arrow_forward Step 3 Part (b) Part (b) Angular frequency, ω = 60π ... (1) Also, ω = 2πf .... (2) Equate both, 2πf = 60π f = 60π/2π f = 30 Hz Answer arrow_forward Step 4 Part (c) Part (c) RMS voltage is…A 200 Ω resistor, 1.6 H inductor, and 3.0 μF capacitor are connected in series across a 60 Hz, 120 V power supply. Calculate (a) the inductive reactance of the circuit, (b) its capacitive reactance, (c) its impedance, (d) the current through the circuit, and (e) the phase angle.
- A resistor (R = 9.00 x 102 Ω), a capacitor (C 5 0.250 µF),and an inductor (L = 2.50 H) are connected in series across a2.40 x 102 - Hz AC source for which ΔVmax = 1.40 x 102 V.Calculate (a) the impedance of the circuit, (b) the maximumcurrent delivered by the source, and (c) the phase anglebetween the current and voltage. (d) Is the current leadingor lagging the voltage?An inductor is connected to an AC source. If the inductance of the inductor is 0.448 H and the output voltage of the source is given by Δv = (120 V)sin[(27.0π s−1)t], determine the following. (d) the rms current in the inductor (in A) (e) the maximum current in the inductor (in A)An LC circuit consists of initially fully charged capacitor with a capacitance of 525 x 10-6 F and an inductor. The maximum energy stored by the capacitor is 0.65 J and the resonant frequency of the circuit is 95.5 rad/s. What is the maximum current through the inductor?