Question
Verify the second equation in Example 21.5 by substituting the values found for the currents I1I_1 and I2I_2.

The equation is −3I2+18−6I1=0-3I_2 + 18 - 6I_1 = 0. I1=4.75 AI_1 = 4.75 \textrm{ A}, I2=−3.50 AI_2 = -3.50 \textrm{ A}.

Question by OpenStax is licensed under CC BY 4.0
Final Answer

Yes, the equation equals zero, as expected.

Solution video

OpenStax College Physics, Chapter 21, Problem 33 (Problems & Exercises)

OpenStax College Physics, Chapter 21, Problem 33 (PE) video thumbnail

In order to watch this solution you need to have a subscription.

Start free trial Log in
vote with a rating of votes with an average rating of .
Chevron left Prev
Select problem
Next Chevron right
Video Transcript
This is College Physics Answers with Shaun Dychko. The equation we are given in the example is negative three times current two plus 18 minus six times current one equals zero. And current one, the example solves 4.75 Amps and current two is negative 3.5 Amps. And so we substitute this number in for I2 and I1 and we end showing that this does indeed equals zero. So we verified the equation.