Question

Part A Consider the circuit in (Figure 1). Assume $i = 11$ A Is the interconnection in (Figure 1) valid? Yes, the interconnection is valid. No, the interconnection is not valid. Figure 10 V 50 V 5 A 40 V 1 of 1 ? Correct Previous Answers Part B Find the total power developed by the current sources. Express your answer to three significant figures and include the appropriate units. P = Value Units Submit Previous Answers Request Answer Incorrect; Try Again; 8 attempts remaining Provide Feedback 

          Part A
Consider the circuit in (Figure 1). Assume $i = 11$ A
Is the interconnection in (Figure 1) valid?
Yes, the interconnection is valid.
No, the interconnection is not valid.
Figure
10 V
50 V
5 A
40 V
1 of 1
? Correct
Previous Answers
Part B
Find the total power developed by the current sources.
Express your answer to three significant figures and include the appropriate units.
P = Value Units
Submit Previous Answers Request Answer
Incorrect; Try Again; 8 attempts remaining
Provide Feedback
Show more…
Part A Consider the circuit in (Figure 1). Assume i = 11 A Is the interconnection in (Figure 1) valid? Yes, the interconnection is valid. No, the interconnection is not valid. Figure 10 V 50 V 5 A 40 V 1 of 1 ? Correct Previous Answers Part B Find the total power developed by the current sources. Express your answer to three significant figures and include the appropriate units. P = Value Units Submit Previous Answers Request Answer Incorrect; Try Again; 8 attempts remaining Provide Feedback

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Problem 2.4 Consider the circuit in (Figure 1). Assume i=11A Part A Is the interconnection in (Figure 1) valid? Yes, the interconnection is valid. No, the interconnection is not valid. Previous Answers Correct Part B Find the total power developed by the current sources. Express your answer to three significant figures and include the appropriate units. X Incorrect; Try Again; 8 attempts remaining Provide Feedback Consider the circuit in (Figure 1). Assume i=11A Is the interconnection in (Figure 1) valid? Yes, the interconnection is valid. No, the interconnection is not valid. Previous Answers Correct Part B Find the total power developed by the current sources. Express your answer to three significant figures and include the appropriate units. Previous Answers Request Answer Incorrect; Try Again; 8 attempts remaining Provide Feedback Figure Part A i=11A Correct Part B 10V 5A
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Transcript

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00:01 Hello, the question is taken from electricity and magnetism and in this question we have to find the we have to find the current flowing through the circuit.
00:12 Okay, so the value of each battery is already given to us and the internal resistance is also given to us let me first draw the diagram and then these are a b c d is f and the value of each batteries first is even even is equal to 16 .5 e2 is equal to 3 .5 is in volt e3 and then e4 the value of e3 and e4 are 8 .5 and 24 .4 in volts and even e2 e3 and this is e4 and the value of each resistance is 0 .5 om, 20, current from here is i1 and this resistance is 6 ome, that is r2, and next, current flowing through here is i3, i2, this is smaller internal resistance, this is r1, then r4, 15 om, then r3 8 om then small r3 0 .5m there is a resistance here also r2 0 .25 om r3 0 .5 om next r4 0 .75 om okay so these are the required value which are given to us and we have to find the all current that is passing in this circuit okay so let us solve it applying the kvl loop here and here okay so uh first one is i'm using the kcel so out outgoing current is equal to incoming current so i1 is equal to i2 plus i3 so first a loop in first this is one loop and this is second so for first minus e1 plus i1 into small r1 and then i1 into capital r1 plus this i 3 into r2 minus e2 i 3 into small r2 and current that is going here this the current coming here is i 2 and this is i 3 so this is i 1 so i 1 into r 4 that is equal to 0 okay so on substituting the value we get this equation e1 is minus 16 .5 small r1 is 0 .5 plus capital r1 20 .5 plus 15 20 .5 plus 15 is 35 .5 i1 and corresponding to i 3 i 3 these two resistances 6 .25 i 3 minus 8 2 is 3 .5 that is equal to 0 so this equation is 35 .5 i 1 6 .25 i3 is equal to the sum of these 2 is 20 next applying kvl in second loop so let us start from e2.
05:25 So that is this value is e2, e2 minus i3 into r2 plus i2 into r3.
05:41 The direction of current is opposite in that we are going minus e3 plus i2 into r3.
06:04 So this is small r3 internal resistance and then i2 into small.
06:14 R4 plus e4 that is equal to zero so from here substituting e2 is equal to 3 .5 let us combine i3 terms only one term once again i forgot this term so that is minus i 3 into small r2 let me write here that is equal to 0 so that is small r2 is 0 .25 0 .25 plus 6 .25 plus 6 6 .25 into i3 and capital r3 r3 plus r4 8 .5 8 .5 plus 0 .75 9 .25 plus 9 .25 into i2 and minus e3 is 8 .5.
07:38 E4 is 24 .5.
07:40 E4 is 24.
07:42 4 that is equal to 0 okay so now 3 .5 minus 8 .5 minus 5 minus 5 plus 24 .4 is 19 .4 so this value is with a negative sign is equal to 9 .25 i2 minus 6 .25 i3 now there are two equations three equations first one is i 1 is equal to i 2 plus i 3 so let us substitute this value in equation 1 so that is 35 .5 i2 plus 35 .5 i 3 plus 6 .25 i 3 is equal to 20 so this value is 35 .5 i2 plus these these values are 41 .75 i3 is equal to 20 okay and second equation is 9 .25 i2 minus 6 .25 i 3 is equal to minus 19 .4.
09:03 Okay so let me evaluate the value of i 2 i2 is equal to this into this so that is 41 .75.
09:12 Let us keep it on the left that is fine 75 into minus 19 .4 plus 6 .25 into 20.
09:26 So no let us use the direct way to evaluate this.
09:37 So let us first evaluate the value of i2 form equation this equation...
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