Question

In the following circuit, the op-amp is ideal (Vp=Vm, Ip=0A, Im=0A). Find the following:

(a) Vout,
(b) V2,
(c) Iout.

3k
|
3k32 --|--
_|_ V3
/ \
To \_/ Vout
|||
1mA
|
Ima
_|_
/ \
2kN \_/ ka
|
v2
||
Inl
||
I
3mA
|
(1)
4k1
|
V4

Answer 1

In this circuit with an ideal op-amp, the voltage at the positive terminal is equal to the voltage at the negative terminal. Vout can be found by considering voltage division across the resistors. Iout can be found using Kirchhoff's Current Law at the node connected to the 3kΩ resistor.

Step-by-step explanation:

In the given circuit, since the op-amp is ideal, the voltage at the positive terminal (Vp) is equal to the voltage at the negative terminal (Vm), which means Vp = Vm.

To find Vout, we need to consider the voltage division across the resistors connected to the op-amp. In this case, the voltage across the 2kΩ resistor (V2) is given as 2.00 V.

Finally, to find Iout, we can use Kirchhoff's Current Law at the node connected to the 3kΩ resistor. The sum of currents entering the node is equal to the sum of currents leaving the node, which means Iout = 1mA + 3mA = 4mA.