Chapter 05-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

This is solution to problems related Electrical Circuit Analysis course. It was given by Prof. Gurnam Kanth at Punjab Engineering College. Its main points are: Equivalent, Model, Input, Resistance, Loop, Gain, Terminal, Op-amp, Voltage, KCL

Typology: Exercises

2011/2012

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Chapter 5, Problem 1.
The equivalent model of a certain op amp is shown in Fig. 5.43. Determine:
(a) the input resistance.
(b) the output resistance.
(c) the voltage gain in dB.
Figure 5.43 for Prob. 5.1
8x104vd
Chapter 5, Solution 1.
(a) Rin = 1.5 MΩ
(b) R
out = 60 Ω
(c) A = 8x104
Therefore AdB = 20 log 8x104 = 98.0 dB
Chapter 5, Problem 2
The open-loop gain of an op amp is 100,000. Calculate the output voltage when there are
inputs of +10 µV on the inverting terminal and + 20 µV on the noninverting terminal.
Chapter 5, Solution 2.
v
0 = Avd = A(v2 - v1)
= 105 (20-10) x 10-6 = 1V
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Chapter 5, Problem 1.

The equivalent model of a certain op amp is shown in Fig. 5.43. Determine:

(a) the input resistance. (b) the output resistance. (c) the voltage gain in dB.

Figure 5.43 for Prob. 5.

8x10^4 vd

Chapter 5, Solution 1.

(a) Rin = 1.5 M Ω (b) Rout = 60 Ω (c) A = 8x

4

Therefore AdB = 20 log 8x

4 = 98.0 dB

Chapter 5, Problem 2

The open-loop gain of an op amp is 100,000. Calculate the output voltage when there are

inputs of +10 μ V on the inverting terminal and + 20 μ V on the noninverting terminal.

Chapter 5, Solution 2.

v 0 = Avd = A(v 2 - v 1 )

5 (20-10) x 10

  • = 1V

Chapter 5, Problem 3

Determine the output voltage when .20 μ V is applied to the inverting terminal of an op

amp and +30 μ V to its noninverting terminal. Assume that the op amp has an open-loop

gain of 200,000.

Chapter 5, Solution 3.

v 0 = Avd = A(v 2 - v 1 ) = 2 x 10

5 (30 + 20) x 10

  • = 10V

Chapter 5, Problem 4

The output voltage of an op amp is .4 V when the noninverting input is 1 mV. If the

open-loop gain of the op amp is 2 × 10 6 , what is the inverting input?

Chapter 5, Solution 4.

v 0 = Avd = A(v 2 - v 1 )

v 2 - v 1 = 2 V 2 x 10

A

v 6

0 =−μ

v 2 - v 1 = -2 μV = –0.002 mV 1 mV - v 1 = -0.002 mV v 1 = 1.002 mV

Chapter 5, Solution 5.

Av (^) d

v (^) i +

I

Rin

R 0

v (^) d

v (^0)

-

-vi + Avd + (Ri + R 0 ) I = 0 (1)

But vd = RiI,

-vi + (Ri + R 0 + RiA) I = 0

I =

0 i

i R ( 1 A)R

v

-Avd - R 0 I + v 0 = 0

v 0 = Avd + R 0 I = (R 0 + RiA)I = 0 i

0 i i R ( 1 A)R

(R RA)v

4 5

4 5

0 i

0 i

i

0 10 100 ( 1 10 )

100 10 x 10

R ( 1 A)R

R RA

v

v ⋅

4 5

9 10 1 10

Chapter 5, Problem 6

Using the same parameters for the 741 op amp in Example 5.1, find vo in the op amp

circuit of Fig. 5.45.

Figure 5.45 for Prob. 5.

Example 5.

A 741 op amp has an open-loop voltage gain of 2× 10

5 , input resistance of 2 MΩ, and

output resistance of 50Ω. The op amp is used in the circuit of Fig. 5.6(a). Find the closed-

loop gain vo/vs. Determine current i when vs = 2 V.

Chapter 5, Problem 7

The op amp in Fig. 5.46 has Ri = 100 kΩ, Ro = 100 Ω , A = 100 , 000. Find the differential

voltage vd and the output voltage vo.

+

-

Figure 5.46 for Prob. 5.