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 :

Integrated Publishing, Inc.
6230 Stone Rd, Unit Q
Port Richey, FL 34668
Phone For Parts Inquiries: (727) 493-0744 Google + |