# Multiplier and Divider Using Op-Amp MCQ [Free PDF] – Objective Question Answer for Multiplier and Divider Using Op-Amp Quiz

1. Determine output voltage of analog multiplier provided with two input signals Vx and Vy.

A. Vo = (Vx ×Vx) / Vy
B. Vo = (Vx ×Vy / Vref
C. Vo = (Vy ×Vy) / Vx
D. Vo = (Vx ×Vy) / Vref2

The output is the product of two inputs divided by a reference voltage in the analog multiplier. Thus, the output voltage is a scaled version of x and y inputs.

=> Vo =Vx ×Vy / Vref.

2. Match the List-I with list-II

 List-I List-II 1. One quadrant multiplier i. Input 1- Positive, Input 2- Either positive or negative 2. Two quadrant multiplier ii. Input 1- Positive, Input 2 – Positive 3. Four quadrant multiplier iii. Input 1- Either positive or negative, Input 2- Either positive or negative

A. 1-ii, 2-i, 3-iii
B. 1-ii, 2-ii, 3-ii
C. 1-iii, 2-I, 3-ii
D. 1-I, 2-iii, 3-i

If both inputs are positive, the IC is said to be a one quadrant multiplier. A two-quadrant multiplier function properly, if one input is held positive and the other is allowed to swing. Similarly, for a four-quadrant multiplier both the inputs are allowed to swing.

3. What is the disadvantage of the log-antilog multiplier?

B. Provides one quadrant multiplication only
C. Provides two and four-quadrant multiplication only
D. Provides one, two, and four-quadrant multiplication only

Log amplifier requires the input and reference voltage to be of the same polarity. This restricts the log-antilog multiplier to one quadrant operation.

4. An input of Vsinωt is applied to an ideal frequency doubler. Compute its output voltage?

A. Vo = [(Vx×Vy) /Vref2] × [1-cos2ωt/2].
B. Vo = [(Vx2×Vy2) /Vref] × [1-cos2ωt/2].
C. Vo = [(Vx×Vy)2 /Vref] × [1-cos2ωt/2].
D. Vo = [(Vx×Vy) /( Vref] × [1-cos2ωt/2].

In an ideal frequency doubler, same frequency is applied to both inputs.

∴ Vx = Vxsinωt and Vy = Vysinωt

=> Vo = (Vx×Vy × sin2ωt) / Vref

= [(Vx×Vy) / Vref] × [1-cos2ωt/2].

5. Find the output voltage for the squarer circuit given below, choose input frequency as 10kHz and Vref =10v

A. Vo = 5.0-(5.0×cos4π×104t)
B. Vo = 2.75-(2.75×cos4π×104t)
C. Vo = 1.25-(1.25×cos4π×104t)
D. None of the mentioned

Output voltage of frequency Vo =Vi2 / Vref

=> Vi = 5sinωt = 5sin2π×104t

Vo = [5×(sin2π×104t)2 ]/10

= 2.5×[1/2-(1/2cos2π ×2×104t)] = 1.25-

(1.25×cos4π×104t).

6. Calculate the phase difference between two input signals applied to a multiplier, if the input signals are Vx= 2sinωt and Vy= 4sin(ωt+θ). (Take Vref= 12v).

A. θ = 1.019
B. θ = 30.626
C. θ = 13.87
D. θ = 45.667

Vo= [Vmx×Vmy /(2×Vref)] ×cosθ

=> (Vo×2×Vref)/ (Vmx × Vmy) = cosθ

=> cosθ = (10×2×12)/(2×4) = 30.

=> θ = cos-130 =1.019.

7. Express the output voltage equation of the divider circuit

A. Vo= -(Vref/2)×(Vz/Vx)
B. Vo= -(2×Vref)×(Vz/Vx)
C. Vo= -(Vref)×(Vz/Vx)
D. Vo= -Vref2×(Vz/Vx)

The output voltage of the divider,

Vo= -Vref×(Vz/Vx).

Where Vz –> dividend and Vx –> divisor.

8. Find the input current for the circuit given below.

A. IZ = 0.5372mA
B. IZ = 1.581mA
C. IZ = 2.436mA
D. IZ =9.347mA

Input current, IZ = -(Vx×Vo)/(Vref×R)

= −(4.79v×16.5v)/(10×5kΩ)

= 1.581mA.

9. Find the condition at which the output will not saturate?

A. Vx > 10v ; Vy > 10v
B. Vx < 10v ; Vy > 10v
C. Vx < 10v ; Vy < 10v
D. Vx > 10v ; Vy < 10v

In an analog multiplier, Vref is internally set to 10v. As long as Vx < Vref and Vy < Vref, the output of the multiplier will not saturate.

10. Determine the relationship between log-antilog method.

A. lnVx×lnVy = ln(Vx+Vy)
B. lnVx / lnVy = ln(Vx-Vy)
C. lnVx -lnVy = ln(Vx/Vy)
D. lnVx+ lnVy = ln(Vx×Vy)