7-96. With the same inputs that were used for the subtractor (E1 = +3 V; E2 = +12 V)

the output of the difference amplifier should be five times the output of the subtractor

(EOUT = +45 V). Following the same steps used for the subtractor, first compute the value

of R2 plus R4:

R2 + R4 = 1 kΩ + 5 kΩ

R2 + R4 = 6 kΩ

With this value, compute the current through R2 (IR2):

E2

I R2 =

R2 = R4

+ 12V

=

I R2

6kΩ

= + 2 mA

I R2

Next, compute the voltage drop across R2 (ER2):

ER2 = (R2) x (IR2)

ER2 = (1 k Ω) x (+2 mA)

ER2 = +2 V

Then, compute the voltage at point B:

Voltage at point B = E2 - ER2

Voltage at point B = (+12 V) - (+2 V)

Voltage at point B = +10 V

Since point A and point B will be the same potential in an operational amplifier:

Voltage at point A = +10 V

Now compute the voltage developed by the R1 (ER1):

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