Question:
In the Op-Amp circuit, what is the load current \( I_L \)?
In the Op-Amp circuit, what is the load current \( I_L \)?
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In Op-Amps, use virtual ground and Ohm’s law to compute currents via resistors.
Updated On: Jun 23, 2025
- \( \frac{v_2}{R_2} \)
- \( \frac{v_3}{R_L} \)
- \( \frac{v_2}{R_L} \)
- \( \frac{v_3}{R_2} \)
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The Correct Option is A
Solution and Explanation
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.
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.
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