I have to find \$V_o\$ (voltage over \$R_3\$), \$I_o\$ (current through \$R_3\$) and the power absorbed by the 1 Ohm resistor in this circuit:


simulate this circuit – Schematic created using CircuitLab

I used \$V=IR\$ to find the current through \$R_3\$ and \$R_1\$ to be 1A. Then I multiplied it by 3 Ohms to find that the voltage is 3V. After this, I found the voltage over \$R_2\$ to be 1V the same way. Then I subtracted the V over \$R_3\$ from V over \$R_2\$ to get 2V, which I answered for A.

I also said \$I_o = 1A\$. After this, I used \$P=I^2*R\$ to find the power absorbed by \$R_3 = 1W\$.

When I entered the values I got, I got parts B and C wrong, but I'm confused on how I got A right if I used Io to find \$V_o\$.

  • \$\begingroup\$ Assume Vin+= 5V and same current flows from input R1 to Vin- and feedback current must be the same as we ignore (for now) all current going into Op Amp . Also dont use Ohms, use KOhms or MOhms. Op Amps cant drive that much current (20mA typ) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 9 at 2:12
  • \$\begingroup\$ 5 v across r1, so 1amp. thus 3volts across r3, vo = v2 - 3v \$\endgroup\$ – analogsystemsrf Apr 9 at 2:43
  • \$\begingroup\$ Always assume for linear operation ( not saturated output to rail) Vin+=Vin- \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 9 at 2:48

Step by step:

Let's assume the hypothetical super-power TL081 can drive the currents as per this task. We have V+, V- and Vout in this circuit.

By OA theory, the voltage between V- and V+ will be always driven to zero with this negative feedback setup and under ideal conditions. Therefore the V- must be at +5V in the steady state. To make this happen the current over R1 must be -1A, for a drop of 5 V from the 10V source V1.

To keep V- at +5V the R3 must sink all this 1A current since an ideal OA have infinite impedance and the inputs don't consume any current. All current must be provided over R3.

Since R3 is 3 Ω, the voltage drop will be 3V. To get 3V across R3 the Vout must be at +2V, and this potential will be maintained by internal drivers of the OA. Therefore the Vout will be kept at +2V.

The load R2 is 1 Ω and therefore the dissipated power (with 2 V across) will be 2^2/1 = 4W.

The current over R1 is therefore 2 A; 1 A will come from R3, and another 1 A will be supplied by OA.


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