Example 21.8 A Multiloop Circuit Find the currents I ,I, and ); in the circuit shown in the figure where / = 15 V. 14.0 V + SOLVE IT 24.00 Conceptualize We cannot simplify the circuit by the rules associated with combining resistances in series and in parallel. (If b C the 10.0 V battery were not present, we could reduce the remaining circuit with series and parallel combinations.) 10.0 V 6.0 0 Categorize Because the circuit is not a simple series and parallel combination of resistances, this problem is one in which we must -M use Kirchhoff's rules. 2.00 A circuit containing different branches. Analyze We arbitrarily choose the directions of the currents as labeled in the figure. Apply Kirchhoff's junction rule to junction (1) I + 2 - 13=0 We now have one equation with three abcda: (2) 10.0 V - (6.0 0)1, - (2.0 0); = 0 unknowns: J J,, and I?. There are three loops in the circuit: aboda, befeb, and aefda. befeb: -(4.0 0)1, + (-15.0 V) + (6.0 0), - 10.0 V = 0 We need only two loop equations to (3) -25.0 V + (6.0 0)1, - (4.0 0) 12 = 0 determine the unknown currents. (The third loop equation would give no new information.) Let's choose to traverse these loops in the clockwise direction. Apply Kirchhoff's loop rule to loops aboda and befcb: Solve Equation (1) for /} and substitute 10.0 V - (6.0 0), - (2.0 0)(1, + 12) = 0 into Equation (2): (4) 10.0 V - (8.0 0) 1 - (2.0 0)1, = 0 Multiply each term in Equation (3) by 4 ) -100.0 V + (24.0 0)1, - (16.0 0)1, = 0 and each term in Equation (4) by 3: (6) 30.0 V - (24.0 0)1, - (6.0 0)12 = 0 Add Equation (6) to Equation (5) to -70.0 V - (22.0 0), = 0 eliminate I, and find I,: 12 = -3.18 A Use this value of I, in Equation (3) to find -25.0 V + (6.0 0)1, - (4.0 0)(-3.182 A) = 0 I : -25.0 V + (6.0 0)1, + 12.728 V = 0 I = 2.05 A Use Equation (1) to find Is: I3 = 1, + 12 = (2.045 A) + (-3.182 A) = -1.137 A Finalize Because our values for J, and I, are negative, the directions of these currents are opposite those indicated in the figure. The numerical values for the currents are correct. Despite the incorrect direction, we must continue to use these negative values in subsequent calculations because our equations were established with our original choice of direction. . .... MASTER IT HINTS: GETTING STARTED | I'M STUCK! Using Kirchhoff's rules, find the currents Jury, and ; in the circuit shown where R, = 1.1 n, R, = 2.3 0, and R. = 6.4 0. A A 10 V R