In an L - R - C series circuit, the source has a voltage amplitude of 120 V, R =
ID: 2255941 • Letter: I
Question
In an L-R-C series circuit, the source has a voltage amplitude of 120 V, R = 35.0 ? and the reactance of the capacitor is 440 ?. The voltage amplitude across the capacitor is 360 V. (a) What is the current amplitude in the circuit?1 A
(b) What is the impedance?
2 ?
(c) What two values can the reactance of the inductor have? (Enter your answers from smallest to largest.) 3 ? 4 ?
(d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency?
5 ?
Explain.
6. (a) What is the current amplitude in the circuit?
1 A
(b) What is the impedance?
2 ?
(c) What two values can the reactance of the inductor have? (Enter your answers from smallest to largest.) 3 ? 4 ?
(d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency?
5 ?
Explain.
6. 3 ? 4 ?
Explanation / Answer
a) Voltage = impedance * current
V= IZ
I = V/Z
I = 360/440
I =.818
(b)
Z = source voltage/ circuit current
Z =120/.818
Z = 146 ohms
(c)
Z = sqrt( R^2 + ( Xc - Xl )^2)
146 = sqrt( 35^2 + ( 440 - Xl )^2)
146^2 = ( 35^2 + ( 440 - Xl )^2)
146^2 - 35^2 = 440^2 - 880Xl +Xl^2
rewriting the equation we get
+Xl^2 - 880Xl + 173509 = 0 use quadratic equation to get values of Xl
-165.75 and 1045
d)
Resonance frequency = 1/2pisqrt(LC)
for 1045
because higher inductance means lower resonance frequency as seen from the equation
f) The resonance frequency of the circuit is that value when the complex part of impedance becomes 0
i.e.,
w*L = 1/w C
w = 1/sqrt(LC)
2*pi*f = 1/sqrt(LC)
f=1/(sqr(LC)*2pi
At this frequency the current has a higher value and impedance Z has the least value
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