Question

Recommendation: In case not able to finish solving problems, write down equations to get partial credits. Direct writing down answer without showing steps has no credit. 1. Solve for the node voltages $V_1$ and $V_2$ shown in figure below. (10 pt.) $R_{10\Omega}$ $j_{10\Omega}$ $V_1$ $V_2$ $10\Omega$ $i_1 \angle 30^\circ$ $5\angle 120^\circ$ $i_2$ $-j10\Omega$ $i_1 = 5(0)30 = 4.33A$

Name: recommendation in case not able to finish solving problems write down equations to get partial credits direct writing down answer without showing steps has no credit 1 solve for the node vol 67725
Uploaded: 2024-05-25T05:39:35-08:00
Duration: 5 min 17 s
Channel: Adi S
Description: recommendation in case not able to finish solving problems write down equations to get partial credits direct writing down answer without showing steps has no credit 1 solve for the node vol 67725

          Recommendation: In case not able to finish solving problems, write down equations to get
partial credits. Direct writing down answer without showing steps has no credit.
1. Solve for the node voltages $V_1$ and $V_2$ shown in figure below. (10 pt.)
$R_{10\Omega}$
$j_{10\Omega}$
$V_1$
$V_2$
$10\Omega$
$i_1 \angle 30^\circ$
$5\angle 120^\circ$
$i_2$
$-j10\Omega$
$i_1 = 5(0)30 = 4.33A$

recommendation in case not able to finish solving problems write down equations to get partial credits direct writing down answer without showing steps has no credit 1 solve for the node vol 67725

Added by Diana J.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

05/25/2024

Step 1

Step 1: Apply KCL at node $V_1$: $$\frac{V_1}{10} + \frac{V_1 - V_2}{10} - 5\angle 30^\circ = 0$$ $$V_1 + V_1 - V_2 = 50\angle 30^\circ$$ $$2V_1 - V_2 = 50\angle 30^\circ$$ Show more…

Show all steps

Thanks for your feedback!

Recommendation: In case not able to finish solving problems, write down equations to get partial credits. Direct writing down answer without showing steps has no credit. 1. Solve for the node voltages $V_1$ and $V_2$ shown in figure below. (10 pt.) $R_{10\Omega}$ $j_{10\Omega}$ $V_1$ $V_2$ $10\Omega$ $i_1 \angle 30^\circ$ $5\angle 120^\circ$ $i_2$ $-j10\Omega$ $i_1 = 5(0)30 = 4.33A$