EXAMPLE 18.5 Another Application of Kirchhoff\'s Rules 14V GOAL Find the current
ID: 1585336 • Letter: E
Question
EXAMPLE 18.5 Another Application of Kirchhoff's Rules 14V GOAL Find the currents in a circuit with three currents and two batteries when some current directions are chosen wrongly PROBLEM Find 1, 12, and 13 in figure a. 1.0 l0v 60 STRATEGY Use Kirchhoff's two rules, the junction rule once and the loop rule twice, to develop three equations for the three unknown currents. Solve the equations simultaneously 2.0 (al SOLUTION (1) 13=11 +12 Apply Kirchhoff's junction rule to junction c. Because of the chosen current directions, 11 and 12 are directed into the junction and 13 is directed out of the junction. Loop aboda: 10 V-(6.0 )/1-(2.0 )/3 = 0 Loop bereb:-(4.0 )12-14 V + (6.0 )11-10 V = (2) Apply Kirchhoff's loop rule to the loops abcda and befeb. (Loop aefda gives no new information.) In loop befcb, a positive sign is obtained when the 6.0 resistor is traversed because the direction of the path is opposite the direction of the current 1 (3) Using Equation (1), eliminate 13 from Equation (2) (ignore units for the moment) 10-61 2.0 (2)-o (4) 10=8.0/1 + 2.012 (5) -12--3.02.02 Divide each term in Equation (3) by 2 and rearrange the equation so that the currents are on the right side Subtracting Equation (5) from Equation (4) eliminates 12 and gives 1 2.012 = 3.0/1-12 = 3.0(2.0) -12 =-6.0 A 12 =-3.0 A 13 = 11 +12 = 2.0 A-3.0 A =-1.0 A Substituting this value of 1 into Equation (5) gives 12 Finally, substitute the values found for 1 and 12 into Equation (1) to obtain 13Explanation / Answer
Solving Pratice problem,
First applying KVL in Loop abcd.
10-6I3-2I1=0
6I3+2I1=10.................(1)
And in loop aefd
-4I2-14-2I1=0
-4I2-2I1=14......................(2)
Also, I3+I2=I1..................(3)
Now solving equation 1,2 and 3
I1=-1A
I2=-3A
I3=2A
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