asked by @anal363 •
6 months ago •
Physics
• 5 pts

Find the current in the resistor R3 of the figure

The junction rule tells us that the currecnt flowing though R3 is the sum of I1 and I2, because current out equals current in (and vice versa). This gives rise to a 3rd current, I3 (I1+I2=I3). Loop 1 using Kirchoff's rule would be 6-2I1-I3=0. The second loop would be 9-4I2-I3. Substitute the junction equation for I3 in both loop equations, then use algebra to eliminate either I1 or I2. Repeat for the second current, and then solve for I3 which is what the question is asking.

If you are still lost, let me know and I will post a solution.

answered by @juanv2 •
5 months ago

A ball is thrown at an original speed of 8.0 m/s at an angle of 35e above the horizontal. What is the speed of the ball when it returns to the same horizontal level

Can someone please explain this?

answered by @paulettef1 •
3 months ago

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