A resistor (R = 9.00 times 10^2 Ohm), a capacitor (C = 0.250 mu F), and an induc

Question

A resistor (R = 9.00 times 10^2 Ohm), a capacitor (C = 0.250 mu F), and an inductor (L = 1.30 H) are connected in series across a 2.40 times 10^2-Hz AC source for which Delta V_max = 1.50 times 10^2 V. (a) Calculate the impedance of the circuit. (b) Calculate the maximum current delivered by the source. (c) Calculate the phase angle between the current and voltage. (d) Is the current leading or lagging behind the voltage? The current leads the voltage. The current lags behind the voltage. There is no phase difference between the current and voltage.

Explanation / Answer

In order to find the impedance, we must first calculate the inductive and capacitive reactance.

XC = 1 / (2pi * F * C)
XL = 2pi * F * L

XC = 2,652 ohms
XL = 1960 ohms

Inductive and capacitive reactances cancel each other out, so the total reactance of the circuit is XL - XC

X = 2652 - 1960 = 692 ohms

Notice that the inductive reactance is larger than the capacitive reactance. This means that the circuit is primarily inductive, and that the current will lag the voltage. If the capacitive reactance was larger, current would lead the voltage.

The impedance, in rectangular form, is:

Z = 900 + J692

The total impedance is the magnitude of the phasor sum of the reactance and resistance.

Z = sqrt( 900^2 + 692^2 )
Z = 1135 ohms

The phase angle is the angle between the circuit phasor and the origin.
Theta = arctan( 692 / 900 )
Theta = 37.56degrees

To find the maximum current, you need Ohm's law, and the voltage expressed in peak, which it already is.

I = 150 / 1135 = 0.13215A = 132.15 mA

So, the answers to your questions, in order:

1) What is the impedance of the circuit?
1135 ohms

2) The maximum current delivered by the source?
132.15 mA

3) The phase angle between the current and voltage?
37.56 degrees

4) Is the current leading or lagging the voltage?
Lagging

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