A Two – phase load draws the following phase currents: $i_1\left(t\right)=I_m\sin \left(\omega t-{\varphi }_1\right),i_2\left(t\right)=I_m\cos \left(\omega t-{\varphi }_2\right)$ These currents are balanced if ${\varphi }_1$ is equal to

Understanding Two-Phase Loads and Balanced Currents

A two-phase system is an electrical system that uses two alternating currents (AC) that are out of phase with each other. These currents are typically separated by a phase difference of 90 degrees (or $\frac{\pi }{2}$ radians). Such systems are common in AC power distribution.

For the currents in a two-phase system to be considered balanced, two main conditions must be met:

- Equal Magnitude: The amplitudes (peak values) of the currents in both phases must be the same.
- Phase Displacement: The phase difference between the currents of the two phases must be exactly 90 degrees ($\frac{\pi }{2}$ radians). This means one current should lead or lag the other by 90 degrees.

Analyzing the Given Phase Currents

The problem provides the expressions for the two phase currents:

- Current in phase 1: $i_1\left(t\right)=I_m\sin \left(\omega t-{\varphi }_1\right)$
- Current in phase 2: $i_2\left(t\right)=I_m\cos \left(\omega t-{\varphi }_2\right)$

Here, $I_m$ is the amplitude, $\omega$ is the angular frequency, $t$ is time, and ${\varphi }_1$ and ${\varphi }_2$ are phase angles.

To easily compare the phase difference, it's helpful to express both currents using the same trigonometric function (either sine or cosine). Let's convert $i_2(t)$ into a sine function. We know the trigonometric identity $\cos (x)=\sin (x+\frac{\pi }{2})$.

Applying this identity to $i_2(t)$:

$i_2\left(t\right)=I_m\cos \left(\omega t-{\varphi }_2\right)=I_m\sin \left(\left(\omega t-{\varphi }_2\right)+\frac{\pi }{2}\right)$ $i_2\left(t\right)=I_m\sin \left(\omega t-{\varphi }_2+\frac{\pi }{2}\right)$

Determining the Condition for Balanced Currents

Now, we have the currents in the same sine form:

- $i_1\left(t\right)=I_m\sin \left(\omega t-{\varphi }_1\right)$
- $i_2\left(t\right)=I_m\sin \left(\omega t-{\varphi }_2+\frac{\pi }{2}\right)$

The amplitudes $I_m$ are already equal. For the currents to be balanced, the phase difference between them must be $\pm \frac{\pi }{2}$.

The phase angle of $i_1(t)$ is $P_1=\omega t-{\varphi }_1$.

The phase angle of $i_2(t)$ is $P_2=\omega t-{\varphi }_2+\frac{\pi }{2}$.

The phase difference is $\Delta P=P_2-P_1$:

$\Delta P=\left(\omega t-{\varphi }_2+\frac{\pi }{2}\right)-\left(\omega t-{\varphi }_1\right)$ $\Delta P=\omega t-{\varphi }_2+\frac{\pi }{2}-\omega t+{\varphi }_1$ $\Delta P={\varphi }_1-{\varphi }_2+\frac{\pi }{2}$

For balanced currents, this phase difference $\Delta P$ must be equal to $\frac{\pi }{2}$ or $-\frac{\pi }{2}$.

**Case 1: $\Delta P=\frac{\pi }{2}$ **

${\varphi }_1-{\varphi }_2+\frac{\pi }{2}=\frac{\pi }{2}$ ${\varphi }_1-{\varphi }_2=0$ ${\varphi }_1={\varphi }_2$

**Case 2: $\Delta P=-\frac{\pi }{2}$ **

${\varphi }_1-{\varphi }_2+\frac{\pi }{2}=-\frac{\pi }{2}$ ${\varphi }_1-{\varphi }_2=-\frac{\pi }{2}-\frac{\pi }{2}$ ${\varphi }_1-{\varphi }_2=-\pi$ ${\varphi }_1={\varphi }_2-\pi$

The condition ${\varphi }_1={\varphi }_2$ ensures a phase difference of $\frac{\pi }{2}$ between the two currents, satisfying the requirement for balanced currents in a two-phase system.

Comparing with Provided Options

We found that the currents are balanced if ${\varphi }_1={\varphi }_2$. Let's compare this with the given options:

1. ${\varphi }_1=-{\varphi }_2$
2. ${\varphi }_1={\varphi }_2$
3. ${\varphi }_1=\left(\frac{\pi }{2}-{\varphi }_2\right)$
4. ${\varphi }_1=\left(\frac{\pi }{2}+{\varphi }_2\right)$

Our derived condition ${\varphi }_1={\varphi }_2$ directly matches Option 2.