Section 2.3: Calculating AC Circuits #

The impedance triangle has given us one number so far: the magnitude. Its angle completes the description, telling how far voltage and current are out of step. One more calculator function finds it, and a little care with signs keeps it pointing the right way once an inductor and a capacitor share a circuit.

An Angle from the Triangle #

In the impedance triangle, the vertical side divided by the horizontal side, $X/R$ , is the tangent of the angle. A calculator’s inverse tangent function works backward: give it that ratio and it finds the angle. Written as a formula for the phase angle $\phi$ (phi),

Look for $\tan^{-1}$ or atan on your calculator, often through a shift or second-function key. Here the $-1$ means inverse tangent, not “1 divided by tangent.” Set the calculator to degrees, the unit used for these angles. As a check, $\tan^{-1}(1)$ should give 45°; a result near 0.785 means it’s set to radians, a classic exam-day trap.

The previous section’s $300+j400\ \Omega$ triangle gives $\tan^{-1}(400/300)\approx53.1^\circ$ . With the 500 Ω magnitude we already found, its polar form is 500 Ω at +53.1°.

This angle describes voltage relative to current. A positive angle means voltage leads; a negative angle means voltage lags. The +53.1° load is inductive, and its voltage leads, just as an inductor’s does.

Net Reactance Sets the Angle #

Put an inductor and a capacitor in series, and their reactances play tug-of-war with the impedance point, one pulling up and one pulling down. Find where that leaves the point before calculating its angle; adding both reactance magnitudes would put it in the wrong place.

A series RLC circuit contains resistance $R$ , inductance $L$ , and capacitance $C$ . Its net reactance is

Positive $X$ means the circuit is inductive overall; negative $X$ means it is capacitive. That net value becomes the vertical side of the triangle.

Take $X_C=500\ \Omega$ , $X_L=250\ \Omega$ , and $R=1\ \mathrm{k}\Omega$ , or 1000 Ω:

1. Subtract the reactances: $250-500=-250\ \Omega$
2. Divide by resistance: $-250/1000=-0.25$
3. Find the angle: $\tan^{-1}(-0.25)\approx-14.0^\circ$

The figure below draws this example on the impedance graph.

The capacitor has the larger reactance, so we expected a capacitive result: voltage lags, as the minus sign says. You can check the angle’s size too. Ignoring signs, a net reactance smaller than the resistance gives an angle under 45°, and a larger one gives an angle over 45°. Both checks catch calculator slips.

Two more circuits use the same steps. With $X_L=100\ \Omega$ and $X_C=300\ \Omega$ , net reactance is −200 Ω. Dividing by the 100 Ω resistance gives −2; inverse tangent gives about −63.4 degrees.

With $X_L=75\ \Omega$ and $X_C=25\ \Omega$ , net reactance is +50 Ω. The resistance is again 100 Ω, so the ratio is +0.5 and the angle is about +26.6 degrees.

Key Information: In a series RLC circuit:

• With $X_C=500\ \Omega$ , $R=1\ \mathrm{k}\Omega$ , and $X_L=250\ \Omega$ , the voltage lags the current by 14.0 degrees.
• With $X_C=300\ \Omega$ , $R=100\ \Omega$ , and $X_L=100\ \Omega$ , the voltage lags the current by 63 degrees.
• With $X_C=25\ \Omega$ , $R=100\ \Omega$ , and $X_L=75\ \Omega$ , the voltage leads the current by 27 degrees.

Turning the Ratio Over #

Every part of a series circuit carries the same current. Parallel branches share the same voltage instead, and their currents add. Working with current per volt makes those branches easier to combine: we can add their admittances. Impedance is voltage over current, $Z=E/I$ . Admittance flips that ratio to current over voltage, $Y=I/E$ , and is measured in siemens, abbreviated S. For a 50 Ω resistor, divide 1 by 50 to get 0.02 S. Apply 1 V and it passes 0.02 A.

Like impedance, admittance has two perpendicular components:

The real part, $G$ , is called conductance.

Key Information:

• Admittance is the inverse of impedance: $Y=1/Z$ .
• Susceptance is the imaginary part of admittance and is commonly represented by the letter $B$ .

In “inverse of impedance,” inverse means the reciprocal, 1 divided by the impedance, not the inverse-tangent sense used earlier. An impedance given in polar form has both a magnitude and an angle to convert. For an impedance of $100\ \Omega$ at $+30^\circ$ :

• Divide 1 by 100: the admittance magnitude is 0.01 S.
• Reverse the sign of +30°: the admittance angle is −30°.

Both changes come from flipping the ratio: the size relationship reverses, and so does which quantity leads.

A pure inductive reactance of $+j100\ \Omega$ has an admittance of $-j0.01\ \mathrm{S}$ . The magnitude becomes $1/100$ , and the angle changes from $+90^\circ$ to $-90^\circ$ . With no resistance, all of that admittance is susceptance. The magnitude is always positive; the sign belongs to the phase angle.

Key Information:

• Converting the magnitude of pure reactance to susceptance replaces it with its reciprocal.
• To convert polar impedance to admittance, take the reciprocal of the magnitude and change the sign of the angle.