Op Amp Q1-4 PUBLIC

1. The circuit is currently in series, so by using 4-volt total drop and the equivalent resistance of 40 kΩ (10k + 20k + 10k), the current is uniform throughout with a value of 100 µA, including across each resistor. VN1 = 4 V because it is first, VR4¬ = (100u)(10k) = 1 V drop, making VN2 = 3 V, and VR5 = (100u)(20k) = 2 V drop, leaving VN3 = 1 V. IR4 = IR5 = IR6 = 100 uA.

2. For their difference to be zero, both inputs must be equal, thus the Voltage at Node B is 2 V. The Voltage drop across R1 changes to meet this demand, shifting from 1 V to 2 V. The resistance is static, so to compensate for the change in Voltage, the current changed from 100uA to 200uA. R2 will similarly have a current of 200uA. VR2 = (200u)(20k) = 4 V. Dropping 4 Volts across R2 leaves Node C at -2 V.

3. The Voltages do not change when reversed to use the non-inverting terminal.

4. The output is reversed, and increases in voltage. Whereas the regular circuit changes from -8 to 8 V, the inverted one goes from -12 to 12 V.

Not provided.