Question

A series ac circuit consists of an inductor having a reactance of 80 Ω and an inductance of 190 mH, a 40-Ω resistor, a capacitor whose reactance is 100 Ω, and an ac source. The rms current in the circuit is measured to be 2.2 A. What is the rms voltage of the source?

Answer 1

V = IZ

V = rms voltage, I = rms current, Z = total circuit impedance

The impedance Z of a device is sometimes expressed as the sum of a real and imaginary component:

Z = R + jX

R = resistance, j = $\sqrt{-1}$, X = reactance

The impedance of a capacitor is -j/(ωC)

ω = ac source frequency, C = capacitance

We can see that the impedance of a capacitor has no real component and has a negative imaginary component, that is to say:

Z = -jX, where we're given X = 100Ω, so Z = -j100Ω

The impedance of an inductor is jωL, ω = source frequency, L = inductance

We can see that the impedance of an inductor has no real component and has a positive imaginary component, that is to say:

Z = jX, where we're given X = 80Ω, so Z = j80Ω

The impedance of a resistor is simply R, R = resistance

WE can see that the impedance of a resistor has a positive real component and no imaginary component, that is to say:

Z = R, where we're given R = 40Ω, so Z = 40Ω

Add up the impedances to get the total impedance:

Z = 40Ω + j80Ω - j100Ω = (40-j80)Ω

To get the rms source voltage, we multiply the rms current I and the impedance Z. These two quantities will be complex numbers, and the math behind multiplying two complex numbers involves multiplying their magnitudes. We already have the magnitude of the rms current 2.2A, so let's calculate the magnitude of the impedance:

|Z| = √(40²+(-80)²)Ω = 89.44Ω

Now let's calculate the rms source voltage V = IZ:

V = 2.2A(89.44Ω)

V = 197V