From Kirchhoff\'s junction rule 1 = 0, and by Kirchhoff\'s loop rule, V = 0, Clo

Question

From Kirchhoff's junction rule 1 = 0, and by Kirchhoff's loop rule, V = 0, Clockwise around junction closed loop the left-hand loop containing I1 and 12, we have +11(6.15 )-12(5.00 )-12(1.00 )-4.00 V = 0 This becomes (6.15 )11-(6.00 )12-4.00 V = 0 Applying Kirchhoff's loop rule clockwise around the right-hand loop containing I2 and I3 and combining resistors and voltage, gives (6.00 )12 + (5.25 )13" 9.50 V = 0 To solve three independent equations in three unknowns, we substitute (I1 + I2) for I3 and reduce the three equations to two equations as follows (6.15 )11-(6.00 )12-4.00 V = 0 (5.25 )(11 + 12) + (6.00 )12-9.50 V = 0 Solving the first equation for 12 gives 12 = (6.15 )11-4.00 V 6.00 Rearranging the second of the pair of equations gives 11.25 5.25 9.50 V 5.25 www.silvereaglestorage.com substitute Tor 12 to obtain one equation in one unknown. Then we solve for the current 11 down the 6.15- resistor. We have 11 = 1.788 Your response differs from the correct answer by more than 10%. Double check your calculations. V/ = 1.788 Your response differs from the correct answer by more than 10%. Double check your calculations. A.

Explanation / Answer

6.15I1 - 6I2 = 4 ...(1)

5.25I1 + 11.25 I2 = 9.5 ..(2)

by solving these two equation

I1 = 1.013 A

I2 = 0.372 A

I3 = I1 + I2

I3 = 1.385 A

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