# PHYS 2212 Module 11.2

## Simple AC Circuits

11.2 Simple AC Circuits

### Learning Objectives

By the end of this section, you will be able to:

• Interpret phasor diagrams and apply them to ac circuits with resistors, capacitors, and inductors
• Define the reactance for a resistor, capacitor, and inductor to help understand how current in the circuit behaves compared to each of these devices

Practice!

If the frequency of this AC source is 60 Hz, then the angular frequency, , is equal to . The angular frequency is equal to times the frequency and is measured in units of radians per second. The units for frequency are Hz, which are equivalent to s-1 or cycles per second. Since there are radians in one cycle, the product of the angular displacement and the frequency gives the angular frequency.

If the frequency of this AC source is 60 Hz, then the period is 1/60 s. The period is always determined by calculating 1/f. The units for frequency are Hz, which are equivalent to s-1 or cycles per second. The inverse of Hz will be seconds per cycle, which is the definition of period.

### Resistors in AC Circuits

Practice!

When a resistor is connected in series with an AC source, the current in the resistor is in phase with the voltage and can be described with the function , where I is the maximum current. The voltage across the resistor is always equal to the voltage of the AC power source since the resistor is directly connected to the AC source. The current through the resistor can be determined using Ohm’s Law: . Since the resistance, R, is constant and does not vary in time, the current through the resistor, iR(t), will be in phase with the voltage across the resistor, vR(t).

There will be some instants in time when the current through the resistor is equal to zero, but this will only occur at the instants in time that the voltage supplied by the AC source is equal to zero. Since the voltage from the AC source is sinusoidal, the voltage is equal to zero for an instant twice during every period of oscillation.

When a resistor is connected to an AC source, the average power supplied by this source is IV/2, where I is the maximum current and V is the maximum voltage. Power is equal to the current multiplied by the voltage but we cannot simply multiply the maximum current and the maximum voltage because the current and voltage are not at their maximum values all the time. They only reach their maximum values for an instant twice during one cycle.

If you multiply the current times the voltage, you get: , where I is the maximum current and V is the maximum voltage. To find the average power, you need to determine the average value of , which is equal to ½. So IV is the maximum power supplied by the source and IV/2 is the average power supplied.

When a resistor is connected to an AC source, the rms current in the resistor is I/√2. The rms current is the “root-mean-square” value of the current and is determined by squaring the current function, calculating the mean (or average value) of that squared function, and then taking the square root of the mean of the squared function. Since the current through the resistor is , this function squared is . The mean of this squared function is since the average value of is equal to ½. Finally the square root of the mean of the squared function is .

The rms current is always a positive value, even though the AC current is negative during half of every cycle. The rms current is the amount of direct current that when passed through a resistor in a certain period of time would dissipate the same amount of energy that an AC current dissipates in the same resistor in the same period of time.

When a resistor is connected to an AC source, the rms voltage across the resistor is V/√2. The rms voltage is the “root-mean-square” value of the voltage and is determined by squaring the voltage function, calculating the mean (or average value) of that squared function, and then taking the square root of the mean of the squared function. Since the voltage across the resistor is , this function squared is . The mean of this squared function is since the average value of is equal to ½. Finally the square root of the mean of the squared function is .

The rms voltage is always a positive value, even though the AC voltage is negative during half of every cycle. The rms voltage is the amount of DC voltage that when provided across a resistor in a certain period of time would dissipate the same amount of energy that an AC voltage dissipates in the same resistor in the same period of time.

Discuss!

The voltage output of an AC source is given by the expression:

∆V = (180 V) sin(ωt).

(a) Find the rms current in the circuit when this source is connected to a 110-Ω resistor.

(b) What if a second resistor of 85 Ω is connected parallel to the existing one, find the rms current in the circuit.

### Capacitors in AC Circuits

Practice!

When a capacitor is connected in series with an AC source, the current in the circuit is out of phase with the voltage and can be described with the function , where I is the maximum current. The current in this circuit is proportional to the rate of change of charge on the capacitor. When the capacitor is fully charged and the voltage across the capacitor is a maximum, the current is zero. As the capacitor discharges and charge is removed from the capacitor plates, the current increases. The current increases more and more as the charge comes off of the capacitor faster and faster. When all the charge comes off the capacitor and the capacitor is fully discharged, the current is flowing at its maximum rate. Then charge begins to collect on the capacitor again and the current reduces.

So when the voltage across the capacitor is a maximum, the current is zero and when the current is a maximum, the voltage is zero. The current and the voltage are completely out of phase.

If the angular frequency of the AC source is increased from to 3, the capacitive reactance, XC, decreases by a factor of 3. The capacitive reactance is equal to . If the angular frequency of the source is increased by a factor of 3, the new capacitive reactance is , which is 1/3 of the initial capacitive reactance.

The capacitive reactance is also equal to maximum voltage across the capacitor divided by the maximum current in the circuit: . This ratio has units of V/A, which are equivalent to units of ohms ().

If a second identical capacitor is wired in series with the first capacitor, the total capacitive reactance of the circuit, XC, increases by a factor of 2. The capacitive reactance is equal to , where C is the equivalent capacitance of the two capacitors in series. The equivalent capacitance of two identical capacitors in series is equal to C/2 because wiring capacitors in series effectively doubles the space between the positive and negative plates of the equivalent capacitor. To calculate the equivalent capacitance of two identical capacitors in series, you can use the relationship: and solving for Ceq gives Ceq = C/2. The capacitive reactance with two identical capacitors in series is , which is 2 times the capacitive reactance with one capacitor.

Discuss!

A 7.10 μF capacitor is connected to the terminals of a 60.0 Hz AC source whose rms voltage is 160 V.

(a) Find the capacitive reactance and the rms current in the circuit.

(b) What if we change the frequency, and the rms current is measured to be 0.75 A? Find the new frequency.

### Inductors in AC Circuits

Practice!

If the angular frequency of the AC source is increased from to 3, the inductive reactance, XL, increases by a factor of 3. The inductive reactance is equal to . If the angular frequency of the source is increased by a factor of 3, the new inductive reactance is , which is 3 times the initial inductive reactance.

The inductive reactance is also equal to maximum voltage across the inductor divided by the maximum current in the circuit: . This ratio has units of V/A, which are equivalent to units of ohms ().

If a second identical inductor is wired in series with the first inductor, the total inductive reactance of the circuit, XL, increases by a factor of 2. The inductive reactance is equal to , where L is the equivalent inductance of the two inductors in series. The equivalent inductance of two identical inductors in series is equal to 2L because wiring inductors in series effectively doubles the length of the equivalent inductor. To calculate the equivalent inductance of two identical inductors in series, you can use the relationship: Leq = L + L = 2L.

The inductive reactance with two identical inductors in series is , which is 2 times the inductive reactance with one inductor.

Discuss!

In a purely inductive AC circuit, L = 29.0 mH and the rms voltage is 280 V.

(a) Calculate the inductive reactance and rms current in the circuit if the frequency is 60.0 Hz.

(b) What if the frequency is now 200 Hz, and the maximum voltage (not rms) is 280 V? Find the rms current in the circuit.