Worked examples · Sheet 3

Impedance and phase shift in series circuits

Theory used: Impedance of the series RLC circuit Complex impedance Phasor Inductive circuit and capacitive circuit tensioni sfasate Resistor in AC

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A real coil, connected to a direct voltage of $12 V$ , carries in steady state the current $0.50 A$ . Connected to an alternating voltage of rms value $12 V$ and frequency $50 Hz$ , it carries the rms current $0.30 A$ . Compute the resistance and the inductance of the coil and the phase shift between voltage and current in AC.

R = \frac{V}{I}, Z = \frac{V _{rms}}{I _{rms}} = R^{2} + X_{L}^{2} ⟹ X_{L} = Z^{2} - R^{2}, L = \frac{X _{L}}{2 π f}, tan φ = \frac{X _{L}}{R}

R Z X_{L} L φ = \frac{12}{0.50} Ω = 24 Ω = \frac{12}{0.30} Ω = 40 Ω = 4 0^{2} - 2 4^{2} Ω = 32 Ω = \frac{32}{2 π \cdot 50} H \approx 0.10 H = arctan \frac{32}{24} \approx 5 3^{\circ}

A real coil is modeled as a resistor $R$ , the resistance of the wire, in series with an ideal inductor $L$ . In direct current in steady state the current is constant, so the voltage across the inductor $L \frac{d i}{d t}$ is zero and the coil reduces to the resistor; by Ohm’s law we obtain $R$ .

In AC the theorem on the impedance of the series RLC circuit applies with $X_{C} = 0$ , that is, to an $R L$ circuit. The ratio of the rms values gives the impedance, and the impedance is the hypotenuse of the right triangle with legs $R$ and $X_{L}$ ; therefore $X_{L} = Z^{2} - R^{2}$ , and dividing by the angular frequency we obtain $L$ . The phase shift is positive, so the current lags the voltage by about $5 3^{\circ}$ , as in every inductive circuit.

Exercise 2

A resistor of resistance $R = 100 Ω$ and a capacitor of capacitance $C = 20 μ F$ are connected in series to the grid, with $V_{rms} = 230 V$ and $f = 50 Hz$ . Compute the impedance, the rms current, the phase shift and the rms voltages across the two components.

X_{C} = \frac{1}{2 π f C}, Z = R^{2} + X_{C}^{2}, I_{rms} = \frac{V _{rms}}{Z}, tan φ = - \frac{X _{C}}{R}, V_{R} = R I_{rms}, V_{C} = X_{C} I_{rms}

X_{C} Z I_{rms} φ V_{R} = \frac{1}{2 π \cdot 50 \cdot 20 \cdot 1 0 ^{- 6}} Ω \approx 159 Ω = 10 0^{2} + 15 9^{2} Ω \approx 188 Ω = \frac{230}{188} A \approx 1.22 A = - arctan \frac{159}{100} \approx - 5 8^{\circ} = 100 \cdot 1.22 V \approx 122 V, V_{C} = 159 \cdot 1.22 V \approx 195 V

The circuit is a series $R L C$ circuit without an inductor, so in the theorem on the impedance of the series RLC circuit we set $X_{L} = 0$ . The phase shift $φ$ is negative, therefore the current leads the source voltage by about $5 8^{\circ}$ and the circuit is capacitive.

The sum $V_{R} + V_{C} \approx 317 V$ exceeds the source voltage, and there is no contradiction. The two voltages are out of phase by $\frac{π}{2}$ , because the one across the resistor is in phase with the current and the one across the capacitor lags by $\frac{π}{2}$ ; their phasors are perpendicular and add as vectors. Indeed $12 2^{2} + 19 5^{2} V \approx 230 V$ , which is the rms voltage of the source.

Exercise 3

A series $R L C$ circuit with $R = 30 Ω$ , $L = 0.25 H$ and $C = 60 μ F$ is supplied by the voltage $v (t) = V_{0} sin (ω t)$ , with $V_{rms} = 120 V$ and $f = 50 Hz$ . Compute the impedance, the rms current, the phase shift, the rms voltages across the three components and the expression of $i (t)$ .

Z = R^{2} + (X_{L} - X_{C})^{2}, tan φ = \frac{X _{L} - X _{C}}{R}, i (t) = 2 \frac{V _{rms}}{Z} sin (ω t - φ)

X_{L} X_{C} Z I_{rms} φ V_{R} i (t) = 2 π \cdot 50 \cdot 0.25 Ω \approx 78.5 Ω = \frac{1}{2 π \cdot 50 \cdot 60 \cdot 1 0 ^{- 6}} Ω \approx 53.1 Ω = 3 0^{2} + 25. 5^{2} Ω \approx 39.4 Ω = \frac{120}{39.4} A \approx 3.05 A = arctan \frac{25.5}{30} \approx 0.70 rad \approx 4 0^{\circ} \approx 91 V, V_{L} \approx 239 V, V_{C} \approx 162 V \approx 4.3 sin (314 t - 0.70) A

Phasors of the rms voltages:

V_{R} \approx 91 V

V_{L} \approx 239 V

V_{C} \approx 162 V

; their vector sum is the source voltage,

V = 120 V

We first compute the reactances at the frequency of the source and then apply the theorem on the impedance of the series RLC circuit. The inductive reactance exceeds the capacitive one by $25.5 Ω$ , so the phase shift is positive and the current lags the voltage by about $4 0^{\circ}$ ; the circuit is inductive. The current amplitude is $2 \cdot 3.05 A \approx 4.3 A$ .

The rms voltages across the components are the products of the rms current with $R$ , $X_{L}$ and $X_{C}$ . The voltage across the inductor, about $239 V$ , is twice that of the source. In the phasor diagram the voltages across the inductor and the capacitor are opposite and their sum has magnitude $V_{L} - V_{C} \approx 78 V$ ; together with $V_{R}$ it forms the right triangle whose hypotenuse is $9 1^{2} + 7 8^{2} V \approx 120 V$ , the source voltage.