Peoples Physics Book Version-2

16.3. Examples http://www.ck12.org

16.3 Examples

Example 1

Question: Consider the op-amp circuit diagram shown here. Note the fixed-voltage leads are omitted for clarity.
(This is typical.)

Let’s begin with an input voltage at point A of 0.5V. a) If the op-amp is “doing its job,” what is the electric

potential at point B? b) What current is flowing through the 10Ωresistor? c) What current must be flowing

through the 100Ωresistor? d) Whatmustthe output voltage be? Now let’s adjust the input voltage at point A

to 0.75V. e) What is the output voltage now? f) By what factor is the op-amp amplifying the input voltage?

Answer:

a) The op-amp is supposed to make the two input voltages as close to equal as possible, or in other words,

VA−VB= 0

Therefore if the input voltage at pointAis.5V, then the input voltage at pointBshould also be.5V.

b) We will use Ohm’s Law to find the current going through the 10Ωresistor.

V=IR⇒

V

R

=I⇒I=

.5V

10 Ω

=.05A

c) Recall that no current ever flowsintoan op-amp. Therefore, the current must be the same as the current running
through the 10Ωresistor, which is.05A.

d) We will again use Ohm’s law. First we must find the total resistance and then we can plug in the known values to
solve for the voltage.

Rtotal=R 1 +R 2 = 10 Ω+ 100 Ω= 110 Ω

V=IR=.05A× 110 Ω= 5 .5V