I-0, and by Kirchhoff\'s loop rule, V = 0, Clockwise around junction closed loop
ID: 1584061 • Letter: I
Question
I-0, and by Kirchhoff's loop rule, V = 0, Clockwise around junction closed loop the left-hand loop containing 11 and 12, we have +11(6.45 )-12(5.00 )-12(1.00 )-4.00 V = 0 This becomes (6.45 )11-(6.00 )12-4.00 V 0 Applying Kirchhoff's loop rule clockwise around the right-hand loop containing I2 and 13 and combining resistors and voltage, gives (6.00 )12 + (5.15 )13-9.50 V = 0 To solve three independent equations in three unknowns, we substitute (I1 + 12) for I3 and reduce the three equations to two equations as follows (6.45 )11-(6.00 )12-4.00 V 0 (5.15 )(11 12) (6.00 )12-9.50 V = 0 Solving the first equation for 12 gives (6.45 )11-4.00 V 6.00 Rearranging the second of the pair of equations gives 11.15 5.15 9.50 V 5.15 We substitute for I12 to obtain one equation in one unknown. Then we solve for the current 11 down the 6.45-2 resistor. We have Your response differs from the correct answer by more than 10%. Double check your calculations. A. Now we can substitute the value we calculated for 11 to find the current down the middle branch A) - (4.00 v) 6.45 6.00 and the current in the right-hand loop 3=11 + 12 =Explanation / Answer
According to the concept of the kirchhoff's second law
i+i2=i3......................(1)
for second law =>6.45 i1- 6i2=4...............(2)
=>6i2+5.15i3=9.5
=>6i2+5.15(i1+i2)=9.5
=>5.15i1+11.15 i2=9.5.............(3)
now we solve eq2 and eq3
(eq2)*5.15=>33.22 i1-30.9i2=20.6
(eq3)*6.45=>33.22i1+71.92i2=61.3
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-102.82 i2=-40.7
current i2=0.396 A
now we current i2 vaue sub in eq 2 we get
6.45i1-(6*0.396)=4
6.45i1=6.376
current i1=0.99 A
current i3=0.99+0.396=1.386 A