Download Example- Use circuit reduction techniques to solve for indicated

circuit_reduction.doc
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Example- Use circuit reduction techniques to solve for indicated variables
-
VS
-
V2
-
Req4
Req3
I3
V4
V6
I5 -
R3
R5
Req2
V5
I6 +
I7 +
I8
R6 V 7 R7 V8 R 8
-
+
+
I2 +
R2 V 3
-
+
R4
+
V1
+
+
I4
-
R1
-
IS
I1
Req1
given that IS = 20 mA, R1 = 100, R2 = 500, R3 = 600, R4 =
200, R5 = 300, R6 = 1k, R7 = 1.5k, and R8 = 2.2k.
Step 1
Note resistors R6, R7, & R8 are in parallel. Combine & replace with
equivalent resistance Req1 = R6 // R7 // R8 = [1000-1 + 1500-1 + 2200-1]-1 =
471.4286 and designate the common parallel voltage V678 = V6 = V7 = V8.
Note that currents I6, I7, and I8 are ‘lost’ in the new equivalent circuit.
VS
-
V2
-
Req4
I2 +
I3
R2 V 3 R 3
-
V4
Req3
Req2
R5
V5
+
V678 Req1 = 471.4286 
I5 -
+
+
R4
+
+
V1
I4
-
+
-
R1
-
IS
I1
Step 2
Note resistors R4, Req1, & R5 are in series. Combine & replace with
equivalent resistance Req2 = R4 + Req1 + R5 = 200 + 471.4286 + 300 =
971.4286 and designate the common series current I45 = I4 = I5. Note that
voltages V4, V678, and V5 are ‘lost’ in the new equivalent circuit.
+
V1
+
R1
-
IS
I1
VS
Req4
+
V2
Req3
I2 +
R2 V3
-
I3
R3
I45
Req2 = 971.4286 
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Step 3
Note resistors R2, R3, & Req2 are in parallel. Combine & replace with
equivalent resistance Req3 = R2 // R3 // Req2 = [500-1 + 600-1 + 971.4286-1]-1 =
212.9436 and designate the common parallel voltage V23 = V2 = V3. Note
that currents I2, I3, and I45 are ‘lost’ in the new equivalent circuit.
+
V1
+
R1
VS
-
-
IS
I1
+
V23 Req3 = 212.9436 
-
Req4
Step 4
Note that resistors R1 and Req3 are in series. Combine and replace with
equivalent resistance Req4 = R1 + Req3 = 100 + 212.9436 = 312.9436 with the
common series current I1. Note that voltages V1 and V23 are ‘lost’ in the new
equivalent circuit.
I1
+
IS
VS
-
Req4 = 312.9436 
Now, we can start determining the unknown current I1 and voltage VS for this
circuit. Then, we will ‘re-expand’ or undo or circuit reduction to find the
remaining currents and voltages.
By KCL, I1 = IS = 20 mA
By Ohm’s Law, VS = I1 Req4 = 20*10-3(312.9436)  VS = 6.2589 V
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Step 5
Knowing I1 and VS, go back to the circuit of step 3 and find V1 and V23.
R1 = 100 
+
V1
+
VS = 6.259 V V23 Req3 = 212.9436 
+
IS = 20 mA
-
I1 = 20 mA
Req4 = 312.9436 
There are multiple ways to calculate V1 and V23:
By Ohm’s Law, V1 = I1 R1 = 20*10-3(100) = 2 V,
OR
by voltage division, V1 = Vs (R1/ Req4) = 6.25887(100/312.9) = 2 V.
Next, by Ohm’s Law, V23 = I1 Req3 = 20*10-3(212.9436) = 4.25887 V,
OR
by KVL, -Vs + V1 + V23 = -6.25887 + 2 + V23 = 0  V23 = 4.25887 V,
OR
by voltage division, V23 = Vs (Req3/ Req4) = 6.25887(212.9/312.9) = 4.25887 V
So, V1 = 2 V and V23 = V2 = V3 = 4.2589 V.
Step 6
Knowing I1, V1, V2, V3, & VS, go back to circuit of step 2 and find I2, I3, & I45.
R1
+
V1
Req4
Req3 = 212.9436 
+
VS
-
I2
+
V2 = 4.259 V R2 = 500  V3
-
IS
I1 = 20 mA
+
I3
R3 = 600 
Again, there are multiple ways to calculate I2, I3, & I45:
I45
Req2 = 971.4286 
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By Ohm’s Law, I2 = V2 / R2 = 4.259 / 500 = 8.51774 mA,
OR
by current division, I2 = I1 (Req3/ R2) = 20(212.944 / 500) = 8.51774 mA.
Next, by Ohm’s Law, I3 = V3 / R3 = 4.259 / 600 = 7.09812 mA,
OR
by current division, I3 = I1 (Req3/ R3) = 20(212.944 / 600) = 7.09812 mA.
Last, by Ohm’s Law, I45 = V3 / Req2 = 4.259 / 971.4286 = 4.38413 mA,
OR
by current division, I45 = I1 (Req3/ Req2) = 20(212.94 / 971.43) = 4.3841 mA,
OR
by KCL, I1 = I2 + I3 + I45  I45 = 20 – 8.5177 – 7.0981 = 4.3841 mA.
So, I2 = 8.5177 mA, I3 = 7.0981 mA, and I45 = I4 = I5 = 4.3841 mA.
Step 7
Knowing I1, I2, I3, I4, I5, V1, V2, V3, & VS, go back to circuit of step 1 and find
V4, V5, & V678.
VS
Req4
V2
Req3
I2
I3
V4
+
R2 V3 = 4.259 V R3
R5 = 300 
+
+
V5
Req2 = 971.4286 
+
V678 Req1 = 471.4286 
I5 -
+
+
V1
-
+
I4 = 4.384 mA R4 = 200 
-
R1
-
IS
I1
Again, there are multiple ways to calculate V4, V5, & V678:
By Ohm’s Law, V4 = I4 R4 = 4.38414*10-3(200) = 0.87683 V,
OR
by voltage division, V4 = V3 (R4 / Req2) = 4.25887(200/971.4286) = 0.87683 V.
Next, by Ohm’s Law, V5 = I5 R5 = 4.38414*10-3(300) = 1.31524 V,
OR
by voltage division, V5 = V3(R5 / Req2) = 4.25887(300/971.4286) = 1.31524 V.
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Last, by Ohm’s Law, V678 = I4 Req1 = 4.38414*10-3(471.4286) = 2.06681 V,
OR
by voltage division, V678 = V3(Req1 / Req2) = 4.25887(471.4286/971.4286) =
2.066805 V,
OR
by KVL, V3 = V4 + V678 + V5  V678 = 4.25887 – 0.87683 – 1.31524 = 2.0668 V.
So, V4 = 0.8768 V, V5 = 1.3152 V, and V678 = V6 = V7 = V8 = 2.0668 V.
Step 8
Knowing I1, I2, I3, I4, I5, V1, V2, V3, V4, V5, V6, V7, V8, & VS, go back to
original circuit and find I6, I7, & I8.
Req3
Req2
V4
I6
I7
I8
+
+
+
V6 = 2.067 V R6 = 1 k V7 R7 = 1.5 k V8 R8 = 2.2 k
R5 I5
V5
+
Req4
R4
+
V2
-
+
VS
-
I2 +
I3
R2 V 3 R3
-
+
-
-
+ V1
IS
I4 = 4.384 mA
R1
-
I1
Req1 = 471.4286 
Again, there are multiple ways to calculate I6, I7, & I8:
By Ohm’s Law, I6 = V6 / R6 = 2.06681 / 1000 = 2.06681 mA,
OR
by current division, I6 = I4 (Req1 / R6) = 4.38414(471.4286/1000) = 2.06681 mA.
Next, by Ohm’s Law, I7 = V7 / R7 = 2.06681 / 1500 = 1.37787 mA,
OR
by current division, I7 = I4 (Req1 / R7) = 4.38414(471.4286/1500) = 1.37787 mA.
Last, by Ohm’s Law, I8 = V8 / R8 = 2.06681 / 2200 = 0.93946 mA,
OR
by current division, I8 = I4 (Req1/ R8) = 4.38414(471.4286/2200) = 0.93946 mA,
OR
by KCL, I4 = I6 + I7 + I8  I8 = 4.38414 – 2.06681 – 1.37787 = 0.93946 mA.
So, I6 = 2.0668 mA, I7 = 1.3779 mA, and I8 = 0.9395 mA.