To determine the other current, we use the vector addition formula for two AC currents that are out of phase:

$I_R = \sqrt{I_1^2 + I_2^2 + 2 I_1 I_2 \cos \theta}$

where:

• $I_R = 70$ A (resultant current)
• $I_1 = 50$ A (one of the currents)
• $I_2 = ?$ (the unknown current)
• $\theta = 60^\circ$ (phase difference)

Step 1: Substitute the values into the equation:

$70^2 = 50^2 + I_2^2 + 2(50)(I_2) \cos 60^\circ$

Since $\cos 60^\circ = 0.5$, the equation simplifies to:

$4900 = 2500 + I_2^2 + 50 I_2$

Step 2: Rearrange into quadratic form:

$I_2^2 + 50 I_2 - 2400 = 0$

Step 3: Solve using the quadratic formula:

$I_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where:

• $a = 1$,
• $b = 50$,
• $c = -2400$

$I_2 = \frac{-50 \pm \sqrt{50^2 - 4(1)(-2400)}}{2(1)}$

$I_2 = \frac{-50 \pm \sqrt{2500 + 9600}}{2}$

$I_2 = \frac{-50 \pm \sqrt{12100}}{2}$

$I_2 = \frac{-50 \pm 110}{2}$

Step 4: Compute the two possible solutions:

$I_2 = \frac{-50 + 110}{2} = \frac{60}{2} = 30$

$I_2 = \frac{-50 - 110}{2} = \frac{-160}{2} = -80$

Since current cannot be negative, we take $I_2 = 30$ A.

Final Answer:

The other current is 30 A. ✅

Would you like a detailed vector diagram to illustrate this?

---

Related Questions:

1. How does the phase difference affect the resultant current in AC circuits?
2. What is the significance of vector addition in AC circuit analysis?
3. Can the resultant current ever be larger than the sum of individual currents?
4. How does the power factor change with phase differences in AC circuits?
5. What happens if the phase difference is 90 degrees instead of 60 degrees?

Tip:

In AC circuit analysis, always use phasor representation for easy vector calculations when dealing with phase differences. 📉