An ac generator supplies a peak (not rms) voltage of 208 V at 60.0 Hz. The gener

Question

An ac generator supplies a peak (not rms) voltage of 208 V at 60.0 Hz. The generator is connected in series with a 200-mH inductor, a 45-?F capacitor and a 100-? resistor. Determine the rms current in the circuit.

(a) 1.5 A, V leads I

(b) 1.1 A, I leads V

(c) 1.5 A, I leads V

(d) 2.11 A, V leads I  

* I don't understand the whole V lead, I lead part or how to know it goes from one specific lead to another. Please explain if possible.

PART 2:

At what frequency would the current in the circuit in Problem 10 become maximum?

(a) 60.0 Hz

(b) 53.0 Hz

(c) 85.7 Hz

(d) 98.9 Hz  

Explanation / Answer

(c) 1.5 A, I leads V

explanation

We know that

Irms = Vrms/Z

Vrms = V/1.414 = 208/1.414 = 147 V

Z = sqrt (R^2 + (Xl - Xc)^2)

Z = sqrt[100^2 + (2 x 3.14 x 60 x 200 x 10^-3 - 1/(2 x 3.14 x 60 x 45 x 10^-6))^2] = 101.33

Irms = 147/101.33 = 1.5 A

(c) 1.5 A, I leads V

The current leads voltage in RLC circuit. And phase should be less than 90.

2)For max current

Z = R

this happens when, Xl = Xc

2 pi f L = 1/2 pi f C

f = 1/2 pi sqrt (LC) = 1/2 x 3.14 sqrt (200 x 10^-3 x 45 x 10^-6) = 53.1 Hz

Hence, f = 53.1 Hz

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