x.Hmidth=\"750\" bgcolor=\"#730018\" border=\"0\" cellpadding=\"0\" cellspacing=

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x.Hmidth="750" bgcolor="#730018" border="0" cellpadding="0" cellspacing="0"> 18.5 Another Application of Kirchhoff's Rules Goal Find the currents in a circuit with three currents and two batteries when some current directions are chosen wrongly. Problem Find I1, I2, and I3 in Figure 18.15a on page 604, where = 13 V. 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. Figure 18.15 (a) (Example 18.5) (b) (Exercise 18.5) Solution Apply Kirchhoff's junction rule to junction c. Because of the chosen current directions, I1 and I2are directed into the junction and I3 is directed out of the junction. (1) I3 = I1 + I2 Apply Kirchhoff's loop rule to the loops abcda andbefcb. 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 I1. (2) Loop abcda: 10 V - (6.0 )I1 - (2.0 )I3 = 0 (3) Loop befcb: -13 V + (6.0 )I1 - 10 V - (4.0 )I2 = 0 Using Equation (1), eliminate I3 from Equation (2) (ignore units for the moment). 10 - 6.0 I1 - 2.0(I1 + I2) = 0 (4) 10 = 8.0 I1 + 2.0 I2 Divide each term in Equation (3) by 2 and rearrange the equation so that the currents are on the right side. (5) -11.5 = -3.0 I1 + 2.0 I2 Subtracting Equation (5) from Equation (4) eliminates I2 and gives I1. 21.5 = 11 I

Explanation / Answer

      -i1 + 3.3i3 = 5         (1)
Loop i1: -5 + i1 + 1(i1 - i3) + 5 = 0          2i1 - i3 = 0          (2)
From (1) & (2):

i1 = (25/28)    = 0.8929A
i2 = (i3 - i1)    = (25/28)    = 0.8929A
i3 = (25/14)    = 1.79 A

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