One current, three voltages
Connect a resistor \(R\), an inductor \(L\) and a capacitor \(C\) in series with an alternating source. Charge cannot pile up anywhere along a series path. So the same current \(i = i_0\sin\omega t\) flows through all three at every instant. The current is the one thing they share, so its phasor is the natural reference. We draw it along the horizontal.
Each element puts its own voltage across itself, with its own phase relative to that current. The voltage across the resistor, \(V_R = IR\), is in phase with the current. The current in the inductor lags its voltage, so the voltage \(V_L = IX_L\) is \(90°\) ahead of the current. The current in the capacitor leads, so the voltage \(V_C = IX_C\) is \(90°\) behind it. \(V_L\) points straight up and \(V_C\) straight down. They fight each other.
Kirchhoff's loop rule, from the chapter Circuits: Kirchhoff's Laws, Networks and RC Transients, holds at every instant: \(v = v_R + v_L + v_C\). The three are sinusoids of one frequency. So, as in lesson 2, the source voltage is the vector sum of the three phasors. First combine the two that point opposite ways, into \(V_L - V_C\). Then add \(V_R\) at right angles to it.
Take the first chip at \(50\,\text{Hz}\) in the figure above. \(V_R = 120\,\text{V}\) lies along \(I\). \(V_L = 320\,\text{V}\) points up and \(V_C = 160\,\text{V}\) down, leaving \(160\,\text{V}\) up. The brown source phasor joins the origin to the point \((120, 160)\). Its length is \(\sqrt{120^2 + 160^2} = 200\,\text{V}\), the supply voltage. Now lower the frequency towards \(35\,\text{Hz}\). \(V_L\) shrinks and \(V_C\) grows until they cancel. Then \(V\) lies along \(I\).
In a series LCR circuit, which quantity is the same for the resistor, the inductor and the capacitor at every instant?
\[ V^2 = V_R^2 + (V_L - V_C)^2 \]
Why. \(V_L - V_C\) is at right angles to \(V_R\), so the source phasor is the hypotenuse of a right triangle.
\[ V^2 = I^2R^2 + I^2(X_L - X_C)^2 \]
Why. substitute \(V_R = IR\), \(V_L = IX_L\) and \(V_C = IX_C\), with one current \(I\).
\[ I = \dfrac{V}{\sqrt{R^2 + (X_L - X_C)^2}} \]
Why. take the root and solve for the current.
\[ Z = \dfrac{V}{I} = \sqrt{R^2 + (X_L - X_C)^2} \]
Read it. the ratio of the rms voltage to the rms current is the impedance.
The impedance \(Z\) of a circuit is the ratio of its rms voltage to its rms current, \(Z = V/I\), in ohms. It plays the part of resistance for the whole circuit. The source phasor makes an angle \(\phi\) with the current phasor, where \(\tan\phi\) \(= (V_L - V_C)/V_R\) \(= (X_L - X_C)/R\). This phase angle is the angle by which the supply voltage leads the current.
Divide every side of the voltage triangle by the current \(I\). The result is the impedance triangle: \(R\) along the base, \(X_L - X_C\) as the height and \(Z\) as the hypotenuse. The angle at its corner is \(\phi\). So \(\cos\phi = R/Z\) and \(\sin\phi = (X_L - X_C)/Z\).
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 \(53°\), with \(\cos\phi = 0.6\). Check: \(\sqrt{120^2 + 160^2} = 200\,\text{V}\). Values used: \(\tan 53° = 4/3\) and \(\tan 37° = 0.75\), as the sheets take them.
A series circuit has \(R = 16\,\Omega\), \(X_L = 24\,\Omega\) and \(X_C = 12\,\Omega\). Find its impedance, in ohms.
The circuit of the last check carries \(5\,\text{A}\) rms. Find the rms supply voltage, in volts.
Lagging or leading
The sign of \(X_L - X_C\) decides which way the source phasor tilts. If \(X_L > X_C\), the circuit is inductive: \(\phi\) is positive, the voltage leads, and the current lags. If \(X_C > X_L\), the circuit is capacitive: \(\phi\) is negative and the current leads the voltage. A higher frequency raises \(X_L\) and lowers \(X_C\). So the same circuit is capacitive at low frequency and inductive at high frequency.
A circuit with only a resistor and an inductor, or a real coil, is the same diagram with \(V_C\) missing. So \(Z = \sqrt{R^2 + X_L^2}\) and the current lags. A resistor and a capacitor give \(Z = \sqrt{R^2 + X_C^2}\), with the current leading. Try the second and third chips of the figure above.
(a) A coil of resistance \(40\,\Omega\) and reactance \(30\,\Omega\) is put across \(100\,\text{V}\). Find the impedance, the current, the two voltages and the phase. (b) A \(60\,\Omega\) resistor and a capacitor of reactance \(80\,\Omega\) are in series. Find the impedance and the phase. (c) A coil takes \(1\,\text{A}\) from \(100\,\text{V}\) DC and \(0.5\,\text{A}\) from \(100\,\text{V}\) at \(50\,\text{Hz}\). Find its resistance and inductance.
(a) \(Z\) \(= \sqrt{40^2 + 30^2}\) \(= 50\,\Omega\), \(I = 100/50 = 2\,\text{A}\). \(V_R = 80\,\text{V}\) and \(V_L = 60\,\text{V}\); check \(\sqrt{80^2 + 60^2} = 100\,\text{V}\). \(\tan\phi = 30/40 = 0.75\): the current lags by \(37°\).
(b) \(Z\) \(= \sqrt{60^2 + 80^2}\) \(= 100\,\Omega\). \(\tan\phi = 80/60 = 4/3\): the current leads by \(53°\).
(c) On DC the coil has no reactance, so \(R = 100/1 = 100\,\Omega\). On AC, \(Z = 100/0.5 = 200\,\Omega\), so \(X_L\) \(= \sqrt{200^2 - 100^2}\) \(= 173\,\Omega\) and \(L\) \(= 173/314\) \(= 0.55\,\text{H}\).
A \(60\,\Omega\) resistor and a capacitor of reactance \(80\,\Omega\) are in series on an alternating supply. How is the current related to the supply voltage?
A series circuit has \(R = 40\,\Omega\), \(X_L = 50\,\Omega\) and \(X_C = 80\,\Omega\) at the supply frequency. What is the circuit?
Why the meters do not add up
The voltmeters of Fig 4.1 read \(120 + 320 + 160 = 600\,\text{V}\) across a \(200\,\text{V}\) supply. This looks like a failure of the loop rule. It is not. A voltmeter reads the rms size of the voltage and nothing about its timing. The three voltages reach their peaks at different instants. \(V_L\) and \(V_C\) are exactly opposite, so at every instant one partly cancels the other. \(V_R\) is a quarter cycle from both.
At every instant the loop rule holds for the instantaneous values. Only the phasor sum of the rms values equals the supply: \(\sqrt{120^2 + (320 - 160)^2} = 200\,\text{V}\). A single voltage in the circuit may even be larger than the supply, as \(V_L = 320\,\text{V}\) is here. Lesson 5 shows how far that can go.
In a series LCR circuit the voltmeters read \(80\,\text{V}\) across \(R\), \(120\,\text{V}\) across \(L\) and \(60\,\text{V}\) across \(C\). Find the supply voltage, in volts.
A student measures \(120\,\text{V}\), \(320\,\text{V}\) and \(160\,\text{V}\) across the resistor, inductor and capacitor of a series circuit on a \(200\,\text{V}\) supply. Explain why this does not break the loop rule.
The loop rule holds for instantaneous voltages, not for rms readings. Each meter reads the rms size of a voltage but not its phase. \(V_L\) is \(90°\) ahead of the current and \(V_C\) \(90°\) behind, so they partly cancel, and \(V_R\) is at right angles to both. The readings add as phasors: \(\sqrt{120^2 + (320 - 160)^2} = 200\,\text{V}\).
Always subtract the two reactances before squaring. \(Z = \sqrt{R^2 + X_L^2 + X_C^2}\) is a common slip, and it gives the wrong impedance. The inductor's and the capacitor's voltages point opposite ways, so their reactances partly cancel.
In a school laboratory, a student wires a resistor, a coil and a capacitor in series across a step-down transformer. A multimeter across each part gives three readings whose sum is larger than the supply. The first thought is a faulty meter. The phasor diagram says the readings are right. The same check can be done on the bench: the square root of \(V_R^2 + (V_L - V_C)^2\) should equal the supply reading. A real coil has some resistance of its own, so its reading slightly spoils this check.
- Adding \(V_R\), \(V_L\) and \(V_C\) as numbers.They are a quarter cycle apart. Add them as phasors: \(V = \sqrt{V_R^2 + (V_L - V_C)^2}\).
- Writing \(Z = R + X_L + X_C\) or \(Z = \sqrt{R^2 + X_L^2 + X_C^2}\).The reactances point opposite ways: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).
- Saying the current leads whenever an inductor is present.The current lags when \(X_L > X_C\) and leads when \(X_C > X_L\). The larger reactance decides.
- Using the resistance alone to find the current.The current is \(V/Z\), with the whole impedance.