Worked examples · Sheet 2

Resistors, capacitors and inductors in alternating current

Theory used: Alternating voltage and current Rms value Rms value of a sinusoidal quantity Resistor in AC Capacitor in AC Inductor in AC Reactances and frequency

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

An electric heater, which can be modeled as a resistor, absorbs the average power $P = 1.0 kW$ when connected to the grid, with rms voltage $V_{rms} = 230 V$ and frequency $f = 50 Hz$ . Compute the resistance, the rms current, the current amplitude and the maximum value of the instantaneous power, and write the expression of $i (t)$ taking $v (t) = V_{0} sin (ω t)$ .

P = \frac{V _{rms}^{2}}{R} ⟹ R = \frac{V _{rms}^{2}}{P}, I_{rms} = \frac{V _{rms}}{R}, I_{0} = 2 I_{rms}, p_{m a x} = V_{0} I_{0} = 2 V_{rms} I_{rms} = 2 P

R I_{rms} I_{0} p_{m a x} i (t) = \frac{23 0 ^{2}}{1.0 \cdot 1 0 ^{3}} Ω \approx 52.9 Ω \approx 53 Ω = \frac{230}{52.9} A \approx 4.3 A = 2 \cdot 4.35 A \approx 6.1 A = 2 \cdot 1.0 kW = 2.0 kW \approx 6.1 sin (314 t) A, t in seconds

For a resistor the average power is $R I_{rms}^{2}$ , by the definition of rms value, and since $I_{rms} = \frac{V _{rms}}{R}$ it can also be written as $\frac{V _{rms}^{2}}{R}$ ; from this we obtain $R$ . The current amplitude is $2$ times the rms value, by the theorem on the rms value of a sinusoidal quantity.

In the resistor the current is in phase with the voltage, so the instantaneous power $p = V_{0} I_{0} sin^{2} (ω t)$ oscillates between $0$ and $V_{0} I_{0}$ . The maximum value is $V_{0} I_{0} = 2 V_{rms} \cdot 2 I_{rms} = 2 P$ , that is, twice the average power. The angular frequency is $ω = 2 π \cdot 50 rad/s \approx 314 rad/s$ , and the phase of the current is the same as that of the voltage.

Exercise 2

A capacitor of capacitance $C = 15 μ F$ is connected to the grid, with $V_{rms} = 230 V$ and $f = 50 Hz$ . Compute the capacitive reactance, the rms current and the expression of $i (t)$ taking $v (t) = V_{0} sin (ω t)$ . Then compute the rms current if the frequency doubles, with the same rms voltage.

X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}, I_{rms} = \frac{V _{rms}}{X _{C}} = 2 π f C V_{rms}, i (t) = \frac{V _{0}}{X _{C}} sin (ω t + \frac{π}{2})

X_{C} I_{rms} I_{0} i (t) I_{rms}^{'} = \frac{1}{2 π \cdot 50 \cdot 15 \cdot 1 0 ^{- 6}} Ω \approx 212 Ω = \frac{230}{212} A \approx 1.08 A = 2 \cdot 1.08 A \approx 1.53 A \approx 1.53 sin (314 t + \frac{π}{2}) A = 2 I_{rms} \approx 2.17 A

By the theorem on the capacitor in AC the amplitudes of the voltage and the current are related by $V_{0} = X_{C} I_{0}$ , and dividing by $2$ the same relation holds between the rms values. The current leads the voltage by $\frac{π}{2}$ , so its phase is $+ \frac{π}{2}$ .

The rms current is proportional to the frequency, because $X_{C}$ is inversely proportional to $f$ . Doubling the frequency halves the reactance, to about $106 Ω$ , and the rms current doubles, to about $2.2 A$ . In direct current, in the limit $f \to 0$ , the reactance tends to infinity and in steady state the current is zero.

Exercise 3

A coil of negligible resistance, connected to a source with $V_{rms} = 24 V$ and $f = 50 Hz$ , carries the rms current $I_{rms} = 0.40 A$ . Compute the reactance and the inductance of the coil, write $i (t)$ taking $v (t) = V_{0} sin (ω t)$ , and compute the rms current if the frequency of the source becomes $200 Hz$ with the same rms voltage.

X_{L} = \frac{V _{rms}}{I _{rms}}, L = \frac{X _{L}}{2 π f}, i (t) = 2 I_{rms} sin (ω t - \frac{π}{2}), I_{rms}^{'} = \frac{V _{rms}}{2 π f ^{'} L} = I_{rms} \frac{f}{f ^{'}}

X_{L} L i (t) I_{rms}^{'} = \frac{24}{0.40} Ω = 60 Ω = \frac{60}{2 π \cdot 50} H \approx 0.19 H \approx 0.57 sin (314 t - \frac{π}{2}) A = 0.40 \cdot \frac{50}{200} A = 0.10 A

The coil has negligible resistance, so it is an ideal inductor and by the theorem on the inductor in AC the ratio of the rms values of voltage and current is the inductive reactance $X_{L} = ω L$ . From $X_{L}$ we obtain $L$ by dividing by the angular frequency $2 π f$ . The current lags the voltage by $\frac{π}{2}$ , and its amplitude is $2 \cdot 0.40 A \approx 0.57 A$ .

At four times the frequency the inductive reactance becomes four times larger, $240 Ω$ , because it is proportional to $f$ ; with the same voltage the rms current therefore becomes one quarter of the initial one. The behavior is opposite to that of the capacitor, whose current grows with the frequency.