A resistor, a coil and a capacitor are joined in series across a \(200\,\text{V}\) AC supply. A voltmeter across the resistor reads \(120\,\text{V}\). Across the coil it reads \(320\,\text{V}\), and across the capacitor \(160\,\text{V}\). The three readings add to \(600\,\text{V}\), three times the supply. The meters are not faulty, and Kirchhoff's loop rule still holds at every instant.
Every voltage is a sinusoid of one frequency
In a series circuit the same current \(i = I_0\sin\omega t\) flows through every element. Each element answers with a voltage of the same \(\omega\), but not always at the same instant. The voltage across a resistor is \(iR\), in phase with the current. The voltage across a pure inductor leads the current by a quarter cycle. The voltage across a capacitor lags the current by a quarter cycle. So each voltage is a phasor: its length is the peak or rms value, and its angle is fixed relative to the current.
The quantities \(X_L = \omega L\) and \(X_C = 1/\omega C\) are the reactances, measured in ohms. They play the part that resistance plays for a resistor, but they come with a quarter-turn twist. Take them as given here; the chapter on alternating current derives them. Because \(V_L\) is a quarter turn ahead and \(V_C\) a quarter turn behind, the two point in exactly opposite directions.
The series diagram
The loop rule says that the supply voltage equals \(v_R + v_L + v_C\) at every instant. All four are sinusoids of one \(\omega\), so this sum of instantaneous values is a phasor sum. Draw \(V_R\) along the current. From its tip, draw \(V_L\) up and \(V_C\) down. The supply voltage \(V\) is the arrow from O to the point \(V_R\) across and \(V_L - V_C\) up.
\[ V_R = IR, \quad V_L = IX_L, \quad V_C = IX_C \]
Why. each voltage is the common current times the resistance or the reactance of that element.
\[ V_L - V_C \ \text{lies at } 90° \text{ to } V_R \]
Why. \(V_L\) and \(V_C\) point opposite ways along one line, and that line is a quarter turn from the current.
\[ V^2 = V_R^2 + (V_L - V_C)^2 \]
Why. the phasor rule for two arrows at right angles, by Pythagoras.
\[ V = I\sqrt{R^2 + (X_L - X_C)^2} \]
Why. put step 1 into step 3 and take the common factor \(I\) outside the root.
\[ Z = \dfrac{V}{I} = \sqrt{R^2 + (X_L - X_C)^2}, \qquad \tan\phi = \dfrac{X_L - X_C}{R} \]
Read it. the impedance \(Z\) is the ratio \(V/I\), and the phase angle is the angle of \(V\) from the current.
Now the opening problem is solved. The resistor's \(120\,\text{V}\) lies along the current. The coil's \(320\,\text{V}\) and the capacitor's \(160\,\text{V}\) point opposite ways and leave \(160\,\text{V}\) upright. So \(V = \sqrt{120^2 + 160^2} = 200\,\text{V}\), exactly the supply. The meters read the sizes of the arrows. The arrows reach their peaks at different instants, so their sizes do not add as numbers.
This is the worked example of Alternating Current and the LCR Circuit, lesson 4, done with the tool. A \(30\,\Omega\) resistor, an inductor of reactance \(80\,\Omega\) and a capacitor of reactance \(40\,\Omega\) are in series across \(200\,\text{V}\) rms. Find the impedance, the current, the three voltages and the phase angle.
\(R = 30\,\Omega\), \(X_L - X_C = 80 - 40 = 40\,\Omega\), \(V = 200\,\text{V}\).
\(Z = \sqrt{R^2 + (X_L - X_C)^2}\), \(I = V/Z\), \(\tan\phi = (X_L - X_C)/R\).
\(Z = \sqrt{900 + 1600} = 50\,\Omega\) and \(I = 200/50 = 4\,\text{A}\). \(V_R = 120\,\text{V}\), \(V_L = 320\,\text{V}\), \(V_C = 160\,\text{V}\). \(\tan\phi = 40/30 = 4/3\), so \(\phi = 53°\).
\(Z = 50\,\Omega\), \(I = 4\,\text{A}\), and the voltage leads the current by \(53°\). Check: \(\sqrt{120^2 + 160^2} = 200\,\text{V}\), the supply. These are the three voltmeters of the opening problem.
A capacitor is connected across an AC supply. How is the current in it related to the voltage across it?
A coil of resistance \(40\,\Omega\) and reactance \(30\,\Omega\) is put across a \(100\,\text{V}\) rms supply. What is the rms current, in amperes?
A \(60\,\Omega\) resistor and a capacitor of reactance \(80\,\Omega\) are in series on an AC supply. How is the current related to the supply voltage?
Changing the frequency
The reactances depend on the frequency in opposite ways. \(X_L = \omega L\) grows as the frequency rises, and \(X_C = 1/\omega C\) falls. At a low frequency \(V_C\) is the longer of the two, \(V\) lies behind the current, and the circuit is capacitive. At a high frequency \(V_L\) is longer, \(V\) lies ahead, and the circuit is inductive. At one frequency the two are equal and cancel.
At the frequency where \(X_L = X_C\), the phasors \(V_L\) and \(V_C\) are equal and opposite, so they cancel completely. The supply voltage then lies along the current, \(\phi = 0\), and \(Z = R\), its smallest value. The current \(V/R\) is the largest the supply can drive. This is resonance. The condition \(\omega L = 1/\omega C\) gives \(\omega_0 = 1/\sqrt{LC}\).
In the sim, at \(50\,\text{Hz}\) the current is \(200/40 = 5\,\text{A}\), and \(V_L = V_C = 5 \times 60 = 300\,\text{V}\). Each is larger than the \(200\,\text{V}\) supply. A voltmeter across the coil alone would read \(300\,\text{V}\). The two large voltages are opposite at every instant, so they never add to anything across the pair.
A series circuit of \(0.1\,\text{H}\), \(10\,\mu\text{F}\) and \(10\,\Omega\) is at resonance on a \(20\,\text{V}\) supply. The voltage across the inductor is then \(200\,\text{V}\). How can this be?
The power factor, read from the diagram
Split the current phasor into two parts: one along the supply voltage, \(I\cos\phi\), and one at right angles to it, \(I\sin\phi\). The part along \(V\) carries energy to the circuit, so the average power is \(P = VI\cos\phi\), with rms values. The part at right angles gives power out for one quarter cycle and takes it back the next. The factor \(\cos\phi\) is the power factor, and the diagram gives it at once as \(\cos\phi = V_R/V = R/Z\).
The series circuit of the example has \(R = 30\,\Omega\), \(X_L = 80\,\Omega\) and \(X_C = 40\,\Omega\) on \(200\,\text{V}\) rms, with a current of \(4\,\text{A}\). What is the average power taken, in watts?
In an LC circuit the charge on the capacitor is \(q = q_0\cos\omega t\) and the current is \(i = -q_0\omega\sin\omega t\). Where is the phasor of the current relative to the phasor of the charge?
The phasor diagram describes the steady state of a sinusoidal supply. For a short while after a switch is closed, the current is not yet a pure sinusoid. A square-wave supply is not a sinusoid at all. Keep all the arrows as rms values or all as peak values; never mix the two in one diagram. A real coil also has some resistance, so the voltage across it leads the current by a little less than \(90°\).
Alternating Current and the LCR Circuit uses this diagram for the phasor itself and for the resistor. It shows the lagging current of an inductor and the leading current of a capacitor with it. It uses it again for the series LCR circuit, its impedance and phase angle, and for inductive and capacitive circuits. The same chapter explains resonance, the voltages at resonance and the wattless current with it. Transformers, LC Oscillations and Electromagnetic Waves uses the phasors of charge and current in the LC circuit.
- Adding the three voltmeter readings to get the supply voltage.The readings are sizes of phasors that peak at different instants. Add them as arrows: \(V = \sqrt{V_R^2 + (V_L - V_C)^2}\).
- Adding \(R\), \(X_L\) and \(X_C\) as plain numbers.\(X_L\) and \(X_C\) point opposite ways and at right angles to \(R\): \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).
- Mixing up which lags in an inductor and in a capacitor.In an inductor the current lags the voltage; in a capacitor the current leads it. Draw \(V_L\) up and \(V_C\) down from the current.