TC 9-62
With this value, compute the current through R2 (IR2) (indicating current flow from left to
right):
E2
IR2 =
R2 + R4
+12 V
IR2 =
2 kΩ
IR2 = + 6 mA
Next, compute the voltage drop across R2 (ER2):
ER2 = R2 x IR2
ER2 = 1 kΩ x (+6 mA)
ER2 = +6 V
Then compute the voltage at point B:
Voltage at point B = E2 - ER2
Voltage at point B = (+12 V) - (+6 V)
Voltage at point B = +6 V
Since point B and point A will be at the same potential in an operational amplifier:
Voltage at point A = +6 V
Now compute the voltage developed by R1 (ER1):
ER1 = (voltage at point A) - (E1)
ER1 = (+6 V) - (+3 V)
ER1= +3 V
Compute the current through R1 (IR1):
ER1
IR1 =
R1
+3V
IR1 =
1kΩ
IR1 = + 3 mA
IR1 = IR3
Since :
IR3 = + 3 mA
Then :
7-32
TC 9-62
23 June 2005
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