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## Q. 12.19

Express $i(t)=I_1 \cos (\omega t)+ I_2 \sin (\omega t)$, where $I_1=4 mA$ and $I_2=3 mA$, in standard form.

## Verified Solution

From (12.32), the phasor for the sum is the sum of the phasors for the individual terms, which are

$\sum_{n=0}^{N-1} X_n \angle \theta_n=-X_N \angle \theta_N.$      (12.32)

$\tilde{I}_1=I_1 \angle 0, \quad \tilde{I}_2=I_2 \angle-\pi / 2.$

Because we intend to add the phasors, we express the phasors in rectangular form using Euler’s identity:

\begin{aligned} \tilde{I}_1 &=I_1 \angle 0=I_1[\cos (0)+j \sin (0)]=I_1. \\ \tilde{I}_2 &=I_2 \angle-\pi / 2=I_2\left[\cos \left(\frac{\pi}{2}\right)-j \sin \left(\frac{\pi}{2}\right)\right] \\\\ &=-j I_2 . \end{aligned}

Thus, the phasor for the sum is

$\tilde{I}=\tilde{I}_1+\tilde{I}_2=I_1-j I_2 = 4-j 3 mA.$

To obtain the amplitude and phase of the sum, we express the phasor for the sum in polar form $I \angle \theta$

\begin{aligned} &I=|\tilde{I}|=\sqrt{4^2+3^2} mA =5 mA \\ &\theta=∡ \tilde{I}=\tan ^{-1}\left(\frac{-3}{4}\right)=-0.644 \end{aligned}

It follows that

$i(t)=5 \cos (\omega t-0.644) mA.$

Again, such calculations are easy to perform using a pocket calculator.