Apply Kirchhoffs junction rule to junction c. Because of the chosen current dire

Question

Apply Kirchhoffs junction rule to junction c. Because of the chosen current directions, I1 and I2 are directed into the junction and I3 is directed out of the junction. I3 = I1 + I2 Apply Kirchhoffs loop rule to the loops abcda and befcb. 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. Loop abcda: 10 V - (6. 0 )I1 - (2. 0 )I3 = 0 Loop befcb: -11 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 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. -10. 5 = -3. 0 I1 + 2. 0 I2 Subtracting Equation (5) from Equation (4) eliminates I2 and gives I1. 20. 5 = 11 t1 Substituting this value of I1 into Equation (5) gives I2. 2. 0 I2 = 3. 0 I1 - 10. 5 Finally, substitute the values found for I1 and I2 into Equation (1) and obtain I3. I3 = The fact that i2 and I3 are both negative indicates that the wrong directions were chosen for these currents. Nonetheless, the magnitudes are correct. Choosing the right directions of the currents at the outset is unimportant because the equations are linear, and wrong choices result only in a minus sign in the answer. Find the three currents in Figure 18. 15b, where R = 2. 9 . (Note that the direction of one current may have been chosen wrongly!) I1 = I2 = I3 =

Explanation / Answer

I1 = 20.5/11 = 1.86364 A

2I2 = 3I1 - 10.5 => I2 = -2.4545 A

I3 = I1 + I2 = -0.59091 A

For 18.4) Figure is not provided. If you can provide figure, I can solve that problem. Thank You.

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