Notes-x D 28Sum2018 Lecture 19.key X and Lecture ure https://learnu credubbcs eb

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Notes-x D 28Sum2018 Lecture 19.key X and Lecture ure https://learnu credubbcs ebdavipid-3242233on-content_id 27aneze treurs espar izlet cem/-. ? IMOb The best tho. ? mtps://www.course orno eunesumoermuesos a ntp:/www.unt.edul P https:/www.master Example Question 19.1 Find the currents 11, 12, I3 in the 2-loop circuit below. we'll use current and energy conservation 1 + 12 -I3 -12(4.0 ?)-14 V + 11(6,0 ?)-10 V :: 0 10 V-11 (6,0 ?)-13 (2.0 ?)-0 14.0 V -12(4.0 ?)-14 V + 11 (6.0 ?)--10 V 0 0 4.02 10 V-11 (6.0 ?)-(h + 12)(2.0 ?) 12(4.0 ?)-14 V + 11 (6.0 ?)-10 V-0 10.0 V 6.0 ? 5 V/?-41,-12 -(5 V/?-41)(4.0 ?)-14 V + 11 (6.0 ?)-10 V : 0 2 2.0 ? initial choice for current directions were wrong for the second and third currents /31 A ees ,a 5 7 8 9

Explanation / Answer

At point c, we can see i1 and i2 add up to give i3,

so i1+i2 =i3

Now the next equation provided in solution is Kirchoffs loop equation for the top loop,

The third equation given is Kirchoffs loop equation for the bottom loop.

Now in next step we put the value of i3 in the kirchoffs loop equations.

Now we are left with two equation and two unknown i1 and i2.

Next, i2 is written in terms of i1 : 5 -4i1 =i2,

puttimg this in above equation we got i1 = 2A,

now i2 = 5 - 4*2 = -3A and i3 = i1+i2 = 2-3 = -1A

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