8. You have a resistor of resistance 200 and a 6.00-uF capacitor. Suppose you ta
ID: 1786163 • Letter: 8
Question
8. You have a resistor of resistance 200 and a 6.00-uF capacitor. Suppose you take the resistor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 250 rad/s. a) What is the impedance of the circuit? b) What is the current amplitude? c) What is the voltage amplitude across the resistor? d) What is the phase angle of the source voltage with respect to the current? e) Does the source voltage lag or lead the current? f) What is the power being delivered to this circuit by the voltage source? g) Construct the phasor diagram.Explanation / Answer
The real part of the impedance (Z_r) is:
Z_r = 200
The imaginary part of the impedance (Z_i: is
Z_i = 1/C
where = 250 rad/s and C = 6*10^(-6) F
Z_i = 1/(250)(0.000006) = 666.67
The magnitude of the impedance is the square root of the sum of the squares of the real and imaginary parts:
|Z| = sqrt{(200)² + (666.67)²} 696
The magnitude of the current (I) is
|I| = 30.0 V/696 4.31*10(-2) A
The magnitude (amplitude) of the voltage across the resistor is:
|V| = |I|Z_r = (0.043)(200) = 8.62 V
The magnitude (amplitude) of the voltage across the inductor is:
|V| = |I|Z_i = 0.043(666.67) = 28.7 V
The phase angle () of the impedance is
= tan^-1(Z_i/Z_r) = tan^-1(666.67/200) 73.3°
To make the equation
V = I(Z)
come out that the voltage is at a zero net phase angle the current must be equal but opposite in sign with the impedance:
This makes the phase angle of the current become -73.3°
e). The voltage lags the current.
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