Question

For the circuit shown in the figure, use Kirchhoff's rules to obtain equations for the following. (Use the following as necessary: I18, I12, and I36)

(a) the upper loop
_______= 30

(b) the lower loop
_____= 24

(c) the node on the left side.
_______= I18

(d) Solve the node equation for I36.
_____= I36

(e) Using the equation found in (d), eliminate I36 from the equation found in part (b).
______= 24


(f) Solve the equations found in part (a) and part (e) simultaneously for the two unknowns for I12 and I18, respectively.
I12 = _____ A
I18 = ______ A

(g) Substitute the answers found in part (f) into the node equation found in part (d), solving for I36.
I36 = ______ A


(h) What is the significance of the negative answer for I12?

Answer 1

Let the current through 18 V battery be i1,thecurrent through the 12 V battery be i2 and the currentthrough the 36 V battery be i3.
Applying Kirchoff's law to the first loop,we get
18 + 8i1 + 11(i1 + i2 -i3) + 12 + 7i2 + 5i2 = 0
or 19i1 + 12i2 - 11i3 = -30------(1)
For the second loop,we get
11(i2 - i3 + i1) + 12 +7i2 - 36 - 5i3 = 0
or 11i1 + 18i2 -16i3 = 24-------(2)
For the entire loop,we get
18 + 8i1 - 5i2 + 5i3 + 36 +5i2 + 5i3 = 0
or 4i1 + 5i3 = -27
or i3 = [(-27 - 4i1)/(5)]--------(3)
Solving equations (1),(2) and (3),we get
i1 = 8.4 A,i2 = -26.9 A andi3 = -12.1 A
The negative sign in the value of current i2 andi3 indicates that the current is flowing in the oppositedirection.