Question

I'm getting conflicting information on how to solve this. How do you determine which direction current would be flowing, and how do you determine V in each loop; add or subtract? The IR is positive if you're tracing with the flow and negative if you're tracing against it, correct?

4 2 T- 5 V 2 2 4 V 6 3 2 12 3 V Figure 28.16 Use Kirchhoff's solve for the curre laws to

Answer 1

Given network is suppose I_1 & I_2 are currently in respective loops as shown. rightarrow Assume some arbitrary direction for currents. If our assumption is wrong the respective currents comes out to be negative after solving problem. Sign conversion: rightarrow while moving from '-ve' terminal of battery to '+ve' terminal potential increases. rightarrow while crossing resistor potential decreases in the direction of current. So we use '-ve' sign for IR. Apply K.V.L 1) rightarrow 4 - 3I_1 - 3 - I_1 - 6(I_1 - I_2) = 0 rightarrow 10 I_1 - 6 I_2 = 1 ---(1) (2) rightarrow 5 - 4 I_2 - 2 I_2 - 4 - 6(I_2 - I_1) = 0 rightarrow 6 I_1 - 12 I_2 = -1 ---(2) (1) times 3 rightarrow 30 I_1 - 18 I_2 = 3 (2) times 5 rightarrow 30 I_1 - 60 I_2 = -5 0 + 42 I_2 = 8 rightarrow I_2 = 8/42 = 0.19 A (1) rightarrow 10 I_1 - 6(0.19) = 1 rightarrow I_1 = 1 + 6(0.19)/10 = 0.21 A