3. In the Figure below, ?1 = 16 V, ?2 = 5.0 V, ?3 = 7.0 V,...

3. In the Figure below, ?1 = 16 V, ?2 = 5.0 V, ?3 = 7.0 V, R1 = 2.0 ?, R2 = 3.0 ?, and R3 = 5.0 ?. Find electric currents i1, i2 , and i3. (Hint: use the junction a and the two loops to find the currents). (30 points) Use the following equations: Junction rule: i_in = i_out Loop rule: ? Voltages = 0

Transcript

00:01 About this question we are given a electric circuit and we need to find the current and world current i1 i2 and i3 so the values of all the all the batteries are given so this is 16 volt this is 5 volt and this is 7 volt r1 is 2 oms r2 is 3 oms r2 is 3 oms and r 3 is 5 oms so let's apply the the loop rule and the junction rules.

00:33 So as for the junction rule over here, i1 is going to be equal to i2 plus i3.

00:39 Let's call this equation 1.

00:42 In the first loop, in the first loop, that's going to be equal to 16 minus i1 r1 minus i2 r2 minus e2, which is 5 is equal to 0.

00:57 So that's going to be 16 minus 2 i1 we're going to just substitute the value of i1 and i2 and uh sorry r1 and r2 and this is going to be three times i 2 minus 5 is equal to 0 which means that uh 11 is equal to 2 i 1 plus 3 i 2 and i 1 is nothing but i 2 plus i 3 from 1 so we're going to replace that as well so we're going to rewrite this as i 2 plus i 3 i 2 and this is going to be equal to 11 is equal to 2 i2 plus 3 i 2 is going to be 5 i 2 and this is going to be 2 i 3 let's call this equation 2 likewise we are going to apply a loop rule over here as well so that's going to look like 3 i 2 minus 5 i 3 plus 7 plus 5 is going to be equal to 0 which means that 3 i2 minus 5 i 3 is equal to minus 12 let's call this equation 3 so we have two equations in terms of i 1 i 2 and i 3 and we need to eliminate one of them definitely to get the value of both so let's say from from equation 3 we can get the value of i 2 as let me just rewrite this you get the value value of i2 as 5 i3 minus 12 over over 3 so this is going to this is something which we're going to substitute an equation to so if we do that we get 11 is equal to 5 times 5 i 3 minus 12 over 3 plus 2 i 3 so this is going to be 11 is equal to 20 5 times 5 is 25 we're just opening up the parentheses 3 times 5 is 6 plus 6 i3 over 3 so let's multiply three both sides we have 33 is equal to 25 i 325 plus 6 is going to be 31 31 i 3 minus 60 adding 60 both sides that's going to be 93 is equal to 31 i i 3 so i 3 is going to come as to 93 over 31 ampere and now we can easily get the value of i 2 by substituted duting i3 over here.

03:38 So we have five times 93 over 31 minus 12 over three...

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