Consider the circuit shown in the figure A Use Kirchoffs rul
Consider the circuit shown in the figure:
A) Use Kirchoff\'s rules to determine the equations needed to solve for the currents in each branch.
B) Solve for the currents in each branch.
3.0 v 560 1 Loop 1:-3.0-560 1 +120 121.50 Loop 2:-1.5-12012-3913 = 0 Junction: 11 + 12\" 13 = 0 Loop 1 2 wn- Junction: 2-13 0 1.5 V Loop 2 120 39 Figure 4.3-1: Three branch circuit with batteries in two branches Solution
A)
In the direction of current, resistor and battery show voltage drop
Loop 1:
-3.0 - 560 ohms * I1 + 120 ohms * I2 + 1.5 = 0
Loop 2:
-1.5 - 120 ohm * I2 - 39 * I3 = 0
Junction (current Law):
I1 + I2 = I3
B) From Loop 1, -3 - 560 I1 + 120 I2 + 1.5 = 0
I1 = (120 I2 - 1.5)/560
From loop 2, -1.5 - 120 I2 - 39 I3 = 0
I3 = (-1.5 - 120 I2)/39
From Junction: I1 + I2 = I3
(120 I2 - 1.5)/560 + I2 = (-1.5 - 120 I2)/39
0.214 I2 - 0.00267 + I2 = -58.5 - 3.0769 I2
=> I2 = -13.632 A
=> I1 = -2.9 A
=> I3 = -16.532 A
