For the circuit shown in the figure below, we wish to find the currents I_1, I_2
ID: 1530055 • Letter: F
Question
For the circuit shown in the figure below, we wish to find the currents I_1, I_2, and I_3. (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.) (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 I_3. (e) Using the equation found in part (d), eliminate I_3 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.) (f) Solve the equations found in part (a) and part (e) simultaneously for the two unknowns for I_1 and I_2, respectively. (g) Substitute the answers found in part (f) into the junction equation found in part (d), solving for I_3. (h) What is the significance of the negative answer for I_2?Explanation / Answer
(a) Using Kirchoff's voltage rule at upper loop of the circuit
Moving in clockwise
8I1-18+5I1+7I2-12 +11I2=0
13I1+18 I2 =30 V ANS
(b)
Using Kirchoff's voltage rule at lowerr loop of the circuit
Moving in clockwise
-11 I2+12 -7 I2-36 +5 I3=0
-18 I2 +5 I3=-24 V ANS
(c)
At left node of the Junction
I1=I2+I3 ANS
(d)
I3=(I1-I2 ) A ANS
(e) We have to elemenate I3 from equation given below
-18 I2 +5 I3=-24 V
As I3=(I1-I2 )
so
-18 I2 +5 (I1-I2 )=-24
-24=-23 I2 +5 I1 ANS
(f)
Part a contains equation 13I1+18 I2 =30 V
Part e contains eqautaion -23 I2 +5 I1 =-24 V
Solving
I1 +(18/13) I2 =30/13 ----------------------(1)
I1 - (23/5)I2=-24/5 -----------------------(2)
Subtracting equation 2 from eq. 1
I2(18/13 +23/5)=(30/13 +24/5)
I2=1.19 A ANS
putting value of I2 in any of the eq.(1 or 2)
I1=0.663 A ANS
(g )
I3=(0.663 - 1.19 )
=-0.527 A ANS
(h) Negative sign shows the opposite direction of the current.