### Ohm’s Law for AC Circuits

Ohm’s Law for a dc circuit is a simple relationship among three variables: the current I (in amperes), the voltage E (in volts), and the resistance R (in ohms). Here are the formulas, in case you don’t recall them:
E = IR
I = E/R
R = E/I

In ac circuits containing no reactance, these same formulas apply, as long as you work with rootmean square (rms) voltages and currents. If you need a refresher concerning the meaning of rms, refer to previous topics.

#### Purely Resistive Impedances

When the impedance Z in an ac circuit contains no reactance, so that all of the current and voltage exist through and across a pure resistance R, Ohm’s Law for an ac circuit is expressed as follows:
E = IZ
I = E/Z
Z = E/I

where Z = R, and the values I and E are rms current and voltage.

#### Complex Impedances

When you want to determine the relationship among current, voltage, and resistance in an ac circuit that contains resistance and reactance, things get interesting. Recall the formula for the square of the absolute-value impedance in a series RLC circuit:
Z2 = R2 + X 2
This means that Z is equal to the square root of the quantity R 2 + X 2, as follows:
Z = (R2 + X 2 )1/2
This is the length of the vector R + jX in the complex impedance plane. You learned this in previous topics. This formula applies only for series RLC circuits. The square of the absolute-value impedance for a parallel RLC circuit, in which the resistance is R and the reactance is X, is defined this way:
Z2 = R2X 2/(R 2 + X 2)
This means that the absolute-value impedance, Z, must be calculated using the rather arcane formula:
Z = [R2X 2/(R 2 + X 2)]1/2
The 1⁄ 2 power of a quantity represents the positive square root of that quantity.

#### Problem 1 A series circuit containing resistance and reactance. Illustration for Problems 1 to 4
Suppose a series RX circuit (shown by the generic block diagram of above figure) has a resistance of R = 50.0 Ω and a capacitive reactance of X = −50.0 Ω. Suppose 100-V rms ac is applied to this circuit. What is the current?
First, calculate Z2 = R2 + X 2 = 50.02 + (−50.0)2 = 2500 + 2500 = 5000. Then Z is the square root of 5000, or 70.7. Therefore, I = E/Z = 100/70.7 = 1.41 A rms.

#### Problem 2

What are the rms ac voltages across the resistance and the reactance, respectively, in the circuit described in Problem 16-20?
The Ohm’s Law formulas for dc will work here. Because the current is I = 1.41 A rms, the voltage drop across the resistance is equal to ER = IR = 1.41 × 50.0 = 70.5 V rms. The voltage drop across the reactance is the product of the current and the reactance: EX = IX = 1.41 × (−50.0) = −70.5 V rms. This is an rms ac voltage of equal magnitude to that across the resistance. But the phase is different.

Note that voltages across the resistance and the reactance—a capacitive reactance in this case, because it’s negative—don’t add up to 100 V rms, which is placed across the whole circuit. This is because, in an RX ac circuit, there is always a difference in phase between the voltage across the resistance and the voltage across the reactance. The voltages across the components always add up to the applied voltage vectorially, but not always arithmetically.

#### Problem 3

Suppose a series RX circuit (above figure) has R = 10.0 Ω. and X = 40.0 Ω. The applied voltage is 100- V rms ac. What is the current?
Calculate Z2 = R 2 + X 2 = 100 + 1600 = 1700. This means that Z is the square root of 1700, or 41.2. Therefore, I = E/Z = 100/41.2 = 2.43 A rms.

#### Problem 4

What are the rms ac voltages across the resistance and the reactance, respectively, in the circuit described in Problem 3?
Knowing the current, calculate ER = IR = 2.43 × 10.0 = 24.3 V rms. Also, EX = IX = 2.43 × 40.0 = 97.2 V rms. If you add ER + EX arithmetically, you get 24.3 + 97.2 = 121.5 V as the total across R and X. Again, this differs from the applied voltage! The simple dc rule does not work here, for the same reason it didn’t work in the scenario of Problem 2.

#### Problem 5

Suppose a parallel RX circuit has R = 30.0 Ω and X = −20.0 Ω. The ac supply voltage is 50.0 V rms. What is the total current drawn from the ac supply?
First, find the square of the absolute-value impedance, remembering the formula for parallel circuits: Z2 = R 2X 2/(R2 + X 2) = 360,000/1300 = 277. The absolute-value impedance Z is the square root of 277, or 16.6. The total current is therefore I = E/Z = 50/16.6 = 3.01 A rms.

#### Problem 6

What are the rms currents through the resistance and the reactance, respectively, in the circuit described in Problem 5?
The Ohm’s Law formulas for dc will work here. For the resistance, IR = E/R = 50.0/30.0 = 1.67 A rms. For the reactance, IX = E/X = 50.0/(−20.0) = −2.5 A rms. Note that these currents don’t add up to 3.01 A, the total current. The reason for this is the same as the reason ac voltages don’t add arithmetically in ac circuits that contain reactance. The constituent currents, IR and IX, differ in phase. Vectorially, they add up to 3.01 A rms, but arithmetically, they don’t.