For the circuit shown in the figure below,we wish to find the currents I1, I2. a
ID: 2040535 • Letter: F
Question
For the circuit shown in the figure below,we wish to find the currents I1, I2. and I3 18.0 V , 5.00 ? 8.00 ? 1.0I+ 1 12'Y 7.00 ? ? -, l+ 5.00 ? -I+ 30.0 V (a) Use Kirchhoff's rules to complete the equation for the upper loop. (Use any variable or symbol stated above as necessary. All currents are given in amperes. Do not enter any units.) (b) Use Kirchhoff's rules to complete the equation for the lower loop. (Use any variable or symbol stated above as necessary. All currents are given in amperes. Do not enter any units.) -24 V- (c) Use Kirchhoff's rules to obtain an equation for the junction on the left side. (Use any variable or symbol stated above as necessary. All currents are given in amperes.) (d) Solve the junction equation for I3. (e) Using the equation found in part (d), eliminate 13 from the equation found in part (b). (Use any variable or symbol stated above as necessary. All currents are given in amperes. Do not enter any units.) 24 V (f) Solve the equations found in part (a) and part (e) simultaneously for the two unknowns for 11 and I2, respectively (9) Substitute the answers found in part (f) into the junction equation found in part (d), solving for 13. (h) What is the significance of the negative answer for I2?Explanation / Answer
a) in upper loop
18 - 8*I1 - 11*I2 + 12 - 7*I2 - 5*I1 = 0
18 + 12 = 13*I1 + 18*I2
30 V = 13*I1 + 18*I2 <<<<<<------------Answer
b) in lower loop
-11*I2 + 12 - 7*I2 - 36 + 5*I3 = 0
12 - 36 = 18*I2 - 5*I3
-24 V = 18*I2 - 5*I3 <<<<<--------------Answer
c) I1 = I2 + I3
d) I3 = I1 - I2
e) -24 V = 18*I2 - 5*I3
-24 = 18*I2 - 5*(I1 - I2)
-24 = 18*I2 - 5*I1 + 5*I2
-24 = 23*I2 - 5*I1
24 = 5*I1 - 23*I2 <<<<<--------------Answer
f) I1 = 2.88 A
I2 = -0.416 A
g) I3 = I1 - I2
= 2.88 - (-0.416)
= 3.30 A <<<<<--------------Answer