# Voltage to Current Converter with Floating and Grounded Load – 2 MCQ’s

This set of Linear Integrated Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Voltage to Current Converter with Floating and Grounded Load – 2″.

1. Determine the current through the diode, when the switch is in position 1, 2& 3. Assuming op-amps initially nulled.

a) Io (LED) =4.01mA; Io (Zener) =4.01mA; Io (rectifier) =8.33mA
b) Io (LED) =25mA; Io (Zener) =4.01mA; Io (rectifier) =4.01mA
c) Io (LED) =16.67mA; Io (Zener) =16.66mA; Io (rectifier) =4.01mA
d) Io (LED) =8.33mA; Io (Zener) =8.33mA; Io (rectifier) =8.33mA

2. A diode match finder circuit has input voltage of 2.6v and output voltage is 5.78v. Calculate the voltage drop across diode 1N4735
a) 2.22v
b) 8.38v
c) 3.18v
d) 15.02v

3. Determine the full scale range for the input voltage if the resistance in series with meters are 1kΩ, 2kΩ, 47kΩ and full scale meter movement is 1mA in low voltage AC voltmeter?
a) 1.0 to 7.48 Vrms
b) 1.1 to 7.48 Vrms
c) 1.2 to 7.48 Vrms
d) 1.3 to 7.48 Vrms

4. Find the voltage drop across the zener diode in the zener diode tester from the given specifications: IZk=1mA, VZ =6.2v, input voltage= 1.2v, output voltage =3.2v and resistance in series with meter =150Ω.
a) 6.2mA
b) 8mA
c) 21.33mA
d) Cannot be determined

5. The maximum current through the load in all application that uses voltage to current converter with floating is
a) 100mA
b) 75mA
c) 25mA
d) 50mA

6. Find the gain of the voltage to current converter with grounded load?
a) 2
b) 1
c) ∞
d) 0

7. Which among the following is preferred to display device in digital application?
a) Matched zener diode
b) Matched LEDs
c) Matched rectifier diode
d) All of the mentioned

8. For voltage to current converter with grounded load, establish a relation between the non-inverting input terminals and load current
a) V1 = [Vin+Vo-(IL×R)] /2
b) V1 = [Vin-Vo-(IL×R)] /2
c) V1 = [Vin+Vo-IL+R] /2
d) V1 = [Vin+Vo+(IL×R)] /2

9. Find the output voltage and the load current for the circuit given below. Assume that the op-amp is initially nulled V1 =2.5v

a) IL=0.42mA, Vo =10v
b) IL=0.42mA, Vo =3.4v
c) IL=0.42mA, Vo =6.1v
d) IL=0.42mA, Vo =5v

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