### Discussion :: Op-Amp Applications - General Questions (Q.No.1)

Ram said: (May 31, 2011) | |

vo=v1+v2. |

Chandan Natkar said: (Jul 30, 2011) | |

Vo=(1+(Rf/R1)(Rf2/R2)[V1-V2] |

Bipin S said: (Sep 14, 2011) | |

Vo= (1+(2Rfa/R1a))*(Rfb/R1b)(V2-V1) = (1+(2*5k/500))*(5k/5K)*(2.5-2.25) = (1+20)*1*0.25) = 21*0.25 = 5.25 |

Mallik said: (Mar 27, 2012) | |

@ Bipin S. You did a wonderful work but I would like to make slight modification to your formula used. Instead of being (V2-V1) this should have been (V1-V2). Though you have used it correctly in the following step by saying (2.5 - 2.25). Nothing offensive. Just wanted to keep it clear. |

Mathi said: (Jul 21, 2013) | |

Output Voltage = (Rf/R1)(1+(2R'/R))(v1-v2). |

Jinju said: (Aug 17, 2013) | |

Vo = (v2-v1)(1+2R1/Rgain)(R3/R2), R1=R2=R3=5K, Rgain=500, = (.25)(21)(1). = 5.25V. |

Masta said: (Jan 10, 2014) | |

What is Rf and R1? I don't get it. |

Suji said: (Feb 25, 2014) | |

The formula for finding output of op-amp is = A(V2-V1). |

Raji said: (Apr 20, 2014) | |

Output is = v1+v2 = 4.75 then = 10-4.75 = 5.25. |

Arnesh Sen said: (Jun 30, 2014) | |

v1 = 2.5v and v2 = 2.25v. Then for virtual short v1(-) = 2.5v and v2(-) = 2.25. Thus the voltage difference across 500 ohm is 2.5v-2.25v = 0.25v. Then current through the 500 ohm resistor is (0.25/500) = (5x10^-4 Amp). Now 500 ohm, and two 5k resistors are connected in series. So the current through this total branch is (5x10^-4) amp. That's why the total voltage difference at this branch is (5k+5K+.5k)*(5x10^-04) = 5.25volt. Now the 2nd part is simply a differential circuit whose output is the differences of the outputs of the two op amp which is 5.25volt. If any one have any question please reply me. |

Nisha said: (Sep 26, 2014) | |

Why you have considered 2R1 in the formula? |

Javed said: (Jan 10, 2015) | |

We can calculate the current between two voltage and the find drop across the resistor? |

Akshay said: (Aug 22, 2015) | |

@Arnesh. How is the three resistances are connected in series? |

Wasim said: (Mar 18, 2016) | |

Is it instrumentation amplifier? |

Husain Kapadia said: (Mar 30, 2016) | |

For those, who don't want to remember the formula. Consider the above op-amp to be 2 having o/p Vo2 and below op-amp to be 1 having o/p Vo1. Since the op-amps are ideal we assume no currents flow into the input terminals, so drop across the 500 Ohm resistor is (V1-V2). Current flowing through the 500 Ohm resistor is 0.5 mA. Vo1 = V1 + drop across 5k Ohm = 5V (simple KVL path). Vo2 = - (drop across 5k Ohm - V2) = -0.25 V (simple KVL path). These o/ps Vo1 and Vo2 act as i/p to the third op-amp. Apply superposition, Vo = Vo1' + Vo2' Consider Vo2 to be GND. Vo1' = (1 + Rf/R1) * Vo1 * (5/10) = 5V Vo2' = -Rf/R1*Vo2 = 0.25V Vo = 5.25V |

Prachi Tewari said: (Sep 10, 2016) | |

@Arnesh Sen answer is correct . |

Sachin said: (Nov 26, 2016) | |

v0 =A * (V2 - V1), A = 1 + (2 * 5k)/500. |

Sanjay said: (Dec 26, 2016) | |

Asking a normal q here. When amplifier at v1 and v2 have supply voltage +10 &-10v then how v1 and v2 amplified out can use on the third opamp? |

Harsh Tyagi said: (Nov 28, 2017) | |

First apply the nodal equation and find the output of first two op-amp. The upper one has output -.25 volt. and the lower one has 5 volt. |

Charan said: (Sep 9, 2018) | |

Why they took 2R1 in the formula? Anyone explain me. |

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