Explanation Given data: Refer to the reference figure in the question. Write the expression for percent error. % error = [ approximate value true value − 1 ] × 100 (1) The value of resistance, R f = 470 kΩ . The value of resistance, R = 10 kΩ . The value of input voltage, v i n = 15 V . The value of percent error, % error = 1% Calculation: Let v p is the node voltage at the non-inverting input terminal of the op amp, v n is the node voltage at the inverting input terminal of the op amp. Let R 1 = R + Δ R in the circuit. Apply Kirchhoff’s current law at node v p . v p R f + v p R + v p − v i n R 1 = 0 v p ( 1 R f + 1 R + 1 R 1 ) = v i n R 1 v p = R R f v i n R R 1 + R f R 1 + R f R According to the properties of ideal op amp, the voltage at the non-inverting input terminal is equal to the voltage at the inverting input terminal of the op amp. Therefore, v p = v n . v n = R R f v i n R R 1 + R f R 1 + R f R (2) Apply Kirchhoff’s current law at node v n . v n R + v n − v i n R + v n − v o R f = 0 v n ( 1 R + 1 R + 1 R f ) − v o R f = v i n R v o R f = v n ( R + 2 R f R R f ) − v i n R (3) Substitute equation (2) in (3). v o R f = ( R R f v i n R R 1 + R f R 1 + R f R ) ( R + 2 R f R R f ) − v i n R = [ ( R + 2 R f ) v i n R R 1 + R f R 1 + R f R ] − v i n R = ( R + 2 R f R R 1 + R f R 1 + R f R − 1 R ) v i n = ( R 2 + 2 R R f − R R 1 − R f R 1 − R f R R ( R R 1 + R f R 1 + R f R ) ) v i n Simplify the equation as follows v o R f = ( R 2 + 2 R R f − R 1 ( R + R f ) − R f R R ( R 1 ( R + R f ) + R f R ) ) v i n v o = [ R 2 + 2 R R f − R 1 ( R + R f ) − R f R ] R f R [ R 1 ( R + R f ) + R f R ] v i n Substitute R 1 = R + Δ R in above equation

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ISBN: 8220100801792

Author: Riedel

Publisher: YUZU

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Chapter 5, Problem 52P

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Find the largest percent change in resistor R.

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Chapter 5, Problem 51P

Chapter 5, Problem 53P

Chapter 5 Solutions

EBK ELECTRIC CIRCUITS

Ch. 5.2 - Assume that the op amp in the circuit shown is...Ch. 5.3 - The source voltage vs in the circuit in Assessment...Ch. 5.4 - Find vo in the circuit shown if va = 0.1 V and vb...Ch. 5.5 - Assume that the op amp in the circuit shown is...Ch. 5.6 - In the difference amplifier shown, vb = 4.0 V....Ch. 5.7 - The inverting amplifier in the circuit shown has...Ch. 5 - Prob. 1P Ch. 5 - Replace the 2 V source in the circuit in Fig. P5.1...Ch. 5 - Find iL (in milliamperes) in the circuit in Fig....Ch. 5 - The op amp in the circuit in Fig. P5.4 is...

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