Sounds like a nonsense for me. If ideal opamp has infinite input resistance, you, in fact, have voltage divider which consists of 1k resisotor and infinite input impedance resistor and due to voltage divider formula voltage drop across 1k resistor should be small, not Vin. All this opamp circuits seem to be completely magic.
I wish someone had responded to your question earlier. I just saw it. Anyway, I understand your viewpoint, so maybe I can help you understand why the author is correct. Regarding the inverting amplifier, yes, the 1K resistor IS in series with the op-amp's nearly infinite input resistor. BUT, that nearly-infinite input resistance is paralleled with the 10K resistor that is connected between the inverting input and Vo. All the current flowing through the 1K resistor continues through the 10K resistor on it's way to Vo - the only place that current CAN go. It cannot flow into the inverting input, with it's near-infinite impedance. The op-amp will drive Vo low enough to make the voltage at the inverting input the same as the voltage at the non-inverting input, 0V. So, the entire input voltage is applied across the 1K resistor: the left sideof the 1K is connected to the input, and the right side of the 1K is virtually connected to ground, 0V. So, "looking into" the input, the input signal "sees" only the 1K resistance. The input current will be as if the circuit was simply a 1K resistor connected between the input and ground. Hope that helps.
@@OsoMagna Yeah, I remember spotting this flaw right after i wrote down my comment. Anyway my understanding of opamp circuits improved tremendously since then. Thank you anyway for this long.
Short and concise, thankyou!
Hope it helped! Feel free to send me a message if there is a particular topic you'd like a video on!
Bro saved my life
What about the output impedance?
Thanks sir
I hope it was helpful! As I am building up the channel content, send me a message if there are any topics you would find helpful to have a video on!
Sounds like a nonsense for me. If ideal opamp has infinite input resistance, you, in fact, have voltage divider which consists of 1k resisotor and infinite input impedance resistor and due to voltage divider formula voltage drop across 1k resistor should be small, not Vin. All this opamp circuits seem to be completely magic.
I wish someone had responded to your question earlier. I just saw it. Anyway, I understand your viewpoint, so maybe I can help you understand why the author is correct.
Regarding the inverting amplifier, yes, the 1K resistor IS in series with the op-amp's nearly infinite input resistor. BUT, that nearly-infinite input resistance is paralleled with the 10K resistor that is connected between the inverting input and Vo.
All the current flowing through the 1K resistor continues through the 10K resistor on it's way to Vo - the only place that current CAN go. It cannot flow into the inverting input, with it's near-infinite impedance.
The op-amp will drive Vo low enough to make the voltage at the inverting input the same as the voltage at the non-inverting input, 0V. So, the entire input voltage is applied across the 1K resistor: the left sideof the 1K is connected to the input, and the right side of the 1K is virtually connected to ground, 0V.
So, "looking into" the input, the input signal "sees" only the 1K resistance. The input current will be as if the circuit was simply a 1K resistor connected between the input and ground.
Hope that helps.
@@OsoMagna Yeah, I remember spotting this flaw right after i wrote down my comment. Anyway my understanding of opamp circuits improved tremendously since then. Thank you anyway for this long.