Question 5 · Adding two alternating voltages

Sum of two sinusoids with the same frequency using phasors; amplitude and phase shift; period and frequency; reversed source

Maximum mark · 16 marks

Two alternating current sources with the same frequency produce the voltages, in volts,

V_1 = 120\sin\left(100\pi t + \dfrac{\pi}{6}\right), \qquad V_2 = 90\sin\left(100\pi t + \dfrac{2\pi}{3}\right),

where $t$ is the time in seconds. The two sources are connected so that the resulting voltage is $V = V_1 + V_2$.

Justify that $V = \operatorname{Im}\left(e^{100\pi t\,i}\left(120e^{\frac{\pi}{6}i} + 90e^{\frac{2\pi}{3}i}\right)\right)$. 3 p
Write $V$ in the form $R\sin(100\pi t + \alpha)$, with $R > 0$ and $\alpha$ in radians. 5 p
Find the period and the frequency of the resulting voltage. 3 p
The second source is connected the other way round, which is equivalent to adding $\pi$ to its phase shift. Find the amplitude and phase shift of the new resulting voltage and explain why the amplitude has not changed. 5 p

IB AI HL TANS 1.13

Step-by-step solution

a) Why the imaginary part works

For each source,

\operatorname{Im}\left(e^{100\pi t\,i}\,A e^{\varphi i}\right) = \operatorname{Im}\left(A e^{(100\pi t + \varphi)i}\right) = A\sin(100\pi t + \varphi)

Since the imaginary part of a sum is the sum of the imaginary parts, and both sources have the same frequency, $e^{100\pi t\,i}$ can be taken out as a common factor and the two complex numbers added.

Each source as an imaginary part 2 marks, common factor 1 mark.

b) Amplitude and phase shift

120e^{\frac{\pi}{6}i} + 90e^{\frac{2\pi}{3}i} \approx 58.9 + 137.9\,i

The two terms are at an angle of $\frac{2\pi}{3} - \frac{\pi}{6} = \frac{\pi}{2}$, so the modulus follows from Pythagoras:

R = \sqrt{120^2 + 90^2} = 150, \qquad \alpha = \arg(58.9 + 137.9\,i) \approx 1.17\ \text{rad}\ (\approx 66.9^\circ)

$V = 150\sin(100\pi t + 1.17)$

Sum of the complex numbers 2 marks, modulus 1 mark, argument 1 mark, final expression 1 mark.

c) Period and frequency

b = 100\pi \;\Longrightarrow\; T = \dfrac{2\pi}{100\pi} = 0.02\ \text{s}, \qquad f = \dfrac{1}{T} = 50\ \text{Hz}

Period 2 marks, frequency 1 mark.

d) The reversed source

Now $90e^{\left(\frac{2\pi}{3} + \pi\right)i} = -90e^{\frac{2\pi}{3}i}$, so

120e^{\frac{\pi}{6}i} - 90e^{\frac{2\pi}{3}i} \approx 148.9 - 17.9\,i

with modulus $150$ and argument $-0.120$ rad ($\approx -6.87^\circ$):

$V = 150\sin(100\pi t - 0.120)$

The amplitude is unchanged because the phase difference between the two sources is still $90^\circ$ in absolute value: $\frac{2\pi}{3} - \frac{\pi}{6} = \frac{\pi}{2}$ and $\frac{2\pi}{3} + \pi - \frac{\pi}{6} = \frac{3\pi}{2}$. The two vectors in the complex plane are still perpendicular, so by Pythagoras $R = \sqrt{120^2 + 90^2} = 150$ in both cases; only the direction of the sum, that is the phase shift, changes.

New sum 2 marks, modulus and argument 1 mark, expression 1 mark, explanation 1 mark.