Question

In the Op-Amp circuit, what is the load current \( I_L \)?

• A. \( \frac{v_2}{R_2} \)
• B. \( \frac{v_3}{R_L} \)
• C. \( \frac{v_2}{R_L} \)
• D. \( \frac{v_3}{R_2} \)

Hint:

In Op-Amps, use virtual ground and Ohm’s law to compute currents via resistors.

Answer 1

The Correct Option is A

In a differential amplifier configuration, the Op-Amp will output voltage that causes the inverting and non-inverting terminals to be equal.
Given that \( v_2 \) is at the inverting input through \( R_2 \), the current into that branch is: \[ I_L = \frac{v_2}{R_2} \] This current is the same as the load current due to high input impedance and virtual ground concept.