The instantaneous values, i1 and i2 of two alternating currents are represented by the two sides of a…

Question

The instantaneous values, i1 and i2 of two alternating currents are represented by the two sides of a triangle as shown below. The resultant iR is represented by the third side. Calculate the magnitude of iR and the angle θ between the current i1 and iR

ScanSolve AI Tutor · Accepted Answer

Here's how to calculate the magnitude of $i_R$ and the angle $\theta$ :

The problem describes a vector addition scenario where $i_1$ and $i_2$ are two vectors, and $i_R$ is their resultant. The diagram shows a triangle where $i_1$ and $i_2$ are placed head-to-tail, and $i_R$ closes the triangle.

Given values:
$i_1 = 10$
$i_2 = 15$

The angle shown as $60^\circ$ is the exterior angle at vertex C. Therefore, the interior angle of the triangle at C (the angle between $i_1$ and $i_2$ when placed head-to-tail) is $180^\circ - 60^\circ = 120^\circ$ . This is the angle opposite the resultant $i_R$ .

1) Calculate the magnitude of $i_R$

Use the Law of Cosines to find the magnitude of $i_R$ .
The Law of Cosines states $c^2 = a^2 + b^2 - 2ab \cos C$ , where $C$ is the angle opposite side $c$ .
$i_R^2 = i_1^2 + i_2^2 - 2 i_1 i_2 \cos(120^\circ)$
Substitute the given values and $\cos(120^\circ) = -\frac{1}{2}$ .
$i_R^2 = (10)^2 + (15)^2 - 2(10)(15)\left(-\frac{1}{2}\right)$ $i_R^2 = 100 + 225 - (300)\left(-\frac{1}{2}\right)$ $i_R^2 = 325 + 150$ $i_R^2 = 475$
Solve for $i_R$ .
$i_R = \sqrt{475}$ $i_R = \sqrt{25 \times 19}$ $i_R = \boxed{5\sqrt{19}}$
2) Calculate the angle $\theta$ between $i_1$ and $i_R$
Use the Law of Sines to find the angle $\theta$ .
The angle $\theta$ is opposite the side $i_2$ .
$\frac{\sin \theta}{i_2} = \frac{\sin(120^\circ)}{i_R}$
Substitute the known values.
We know $i_R = 5\sqrt{19}$ and $\sin(120^\circ) = \frac{\sqrt{3}}{2}$ .
$\frac{\sin \theta}{15} = \frac{\frac{\sqrt{3}}{2}}{5\sqrt{19}}$ $\sin \theta = \frac{15 \times \frac{\sqrt{3}}{2}}{5\sqrt{19}}$ $\sin \theta = \frac{3\sqrt{3}}{2\sqrt{19}}$
Rationalize the denominator and calculate $\theta$ .
$\sin \theta = \frac{3\sqrt{3}}{2\sqrt{19}} \times \frac{\sqrt{19}}{\sqrt{19}} = \frac{3\sqrt{57}}{2 \times 19} = \frac{3\sqrt{57}}{38}$ $\theta = \arcsin\left(\frac{3\sqrt{57}}{38}\right)$ $\theta \approx \arcsin(0.5960)$ $\theta \approx \boxed{36.58^\circ}$

The instantaneous values, i1 and i2 of two alternating currents are represented by the two sides of a triangle as shown below. The resultant iR is represented by the third side. Calculate the magnitude of iR and the angle θ between the current i1 and iR

Answer

Step-by-step solution

Use the Law of Cosines to find the magnitude of $i_R$ .

Substitute the given values and $\cos(120^\circ) = -\frac{1}{2}$ .

Solve for $i_R$ .

Use the Law of Sines to find the angle $\theta$ .

Substitute the known values.

Rationalize the denominator and calculate $\theta$ .

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More Physics Questions

The instantaneous values, i1 and i2 of two alternating currents are represented by the two sides of a triangle as shown below. The resultant iR is represented by the third side. Calculate the magnitude of iR and the angle θ between the current i1 and iR

Answer

Step-by-step solution

Use the Law of Cosines to find the magnitude of iRi_RiR​.

Substitute the given values and cos⁡(120∘)=−12\cos(120^\circ) = -\frac{1}{2}cos(120∘)=−21​.

Solve for iRi_RiR​.

Use the Law of Sines to find the angle θ\thetaθ.

Substitute the known values.

Rationalize the denominator and calculate θ\thetaθ.

Try one yourself

Still stuck on this one?

More Physics Questions

Use the Law of Cosines to find the magnitude of $i_R$ .

Substitute the given values and $\cos(120^\circ) = -\frac{1}{2}$ .

Solve for $i_R$ .

Use the Law of Sines to find the angle $\theta$ .

Rationalize the denominator and calculate $\theta$ .