Electrical Engineering: Principles & Applications (7th Edition)
7th Edition
ISBN: 9780134484143
Author: Allan R. Hambley
Publisher: PEARSON
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CH1
CH1
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- Shown in the figure below is an "RLC" circuit driven by an AC power source. The AC power source has an RMS voltage of Vps(RMS) = 11.80 Volts and is running at a frequency of f = 147000 Hz. The resistor has a resistance of R = 3910 Ω, the capacitor has a capacitance of C = 4.01e-10 Farads, and the inductor has an inductance of L = 5.97e-03 Henries.Write the FORMULA for the total impedance of the circuit Ztot = Write the FORMULA for the phase of the total impedance of the circuit ?Ztot = Determine the numerical value of Ztot = ΩDetermine the numerical value of ?Ztot = degreesDetermine the current through the circuit: I(PEAK) = Amps I(RMS) = Amps Determine the voltage across the inductor: VL(PEAK) = Volts VL(RMS) = Volts Determine the voltage across the capacitor: VC(PEAK) = Volts VC(RMS) = Volts If a second circuit were connected in parallel with the resistor, this circuit would be considered as: a high-pass filtera flux capacitor a radio tunera phase…arrow_forwardConvert the following to: rectangular form, polar formarrow_forwardWhat impedance vector (0- j15) Ohms represents:A. A pure resistance. C. A pure capacitance.B. A pure inductance. D. An inductance combined with a capacitance.arrow_forward
- In a circuit, there is a series connection of an ideal resistor and an ideal capacitor. The conduction current (in Amperes) through the resistor is 2sin(t+ n/2). The displacement current (in Amperes) through the capacitor isarrow_forwardThe given figure is to be designed to pass a wave having a frequency of 3500 cps and block a 2500 cps wave. The series inductor has an inductance of 0.0023 henry. Calculate the capacitance of the series capacitor and the parallel inductor.arrow_forwardLet Zeq be the equivalent impedance of a parallel connection of an inductor with inductance L and a capacitor with capacitance C. At w=1/√√LC, the equivalent impedance is given by O a. Zeq = 0 b. Zeq = 1 O c. Zeq = 00 O d. Zeq = 0.5arrow_forward
- Shown in the figure below is an "RLC" circuit driven by an AC power source. The AC power source has an RMS voltage of Vps(RMS) = 9.35 Volts and is running at a frequency of f = 113000 Hz. The resistor has a resistance of R = 1980 , the capacitor has a capacitance of C = 1.08e-09 Farads, and the inductor has an inductance of L = 3.71e-03 Henries. mm Vps Write the FORMULA for the total impedance of the circuit Ztot = Write the FORMULA for the phase of the total impedance of the circuit øztot Determine the numerical value of Ztot = 2384 Determine the numerical value of @Z tot Determine the current through the circuit: ● R www I(PEAK) = I(RMS) = ● ● Determine the voltage across the inductor: • VL(PEAK) = Volts Volts VL(RMS) = Amps Amps Determine the voltage across the capacitor: Vc(PEAK) = Volts Volts Vc(RMS) = = 33.85 a capacitive switcher Ω degrees C If a second circuit were connected in parallel with the resistor, this circuit would be considered as: O a high-pass filter a radio tuner O…arrow_forwardCapacitance by AC- Bridge: The impedance of the capacitor of capacitance C is: Select one: a. Directly proportional to 1/C b. Directly proportional to 1/C2 C. Directly proportional to C2 d. Inversely proportional to 1/Carrow_forwardExpress the following in its simplist form: sin(2πft)+sin(2πft-π). Hint: Thinkabout what each seperate waveform looks like.arrow_forward
- The current through a 50 mH inductor is given below. Use Ohm’s Law to find the voltage. Convert to polar form, find the impedance and complete the math, then convert back to the waveform at i1 = 7.5sin(2500t – 20°) mA and i2 = 5.9sin(200t + 40°) mAarrow_forward4. Solve for the equivalent capacitance and voltage drop across C6.arrow_forwardA series circuit consists of a 20-ohm resistance, a 150 mH inductance and an unknown capacitance. The circuit is supplied with a voltage v = 100 sin 314t. Find the value of capacitance at resonancearrow_forward
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