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You connect a 250-2 resistor, a 1.20-mH inductor, and a 1.80-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. When you connect a voltmeter across the resistor in this circuit, the meter reads approximately A) 25.7 mV B) 83.6 V C) 120 V D) 14.2 V E) 84.9 V

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Answer:

Explanation:

 reactance of inductor = wL = 2 X 3.14 X 60 X 1.2 X 10⁻³ = .45 ohm.

reactance of capacitor = 1/wC = 1/( 2 X 3.14 X 60 X 1.8 X 10⁻⁶ ) = 1474.4

Impedence of the circuit =[ R² + ( I/wC - wL)  ]¹/²  = [250² + ( 1474.4-.45 )]¹/²

Impedence = 1495 ohm.

RMS Voltage = 120/ 1.414 = 84.86 V

current = 84.86 / 1495 = 0.0576

Potential over resistance = 0.0576 x 250 = 14.2 V.

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