Test: Electrical & Electronic Measurements- 6 - Electrical Engineering (EE) MCQ

V(t) = 1 + 2 sin 50 πt
Dual slope integrated DVM measures the average value of input.
V_avg = 1 V
Hence the meter reading will be 1.000 V

Frequency = 400 kHz.
f_clk = 400kHz
t = 20m sec
⇒ Clock pulses, n = t f_clk = 20m sec × 400 kHz = 8000

Resolution on 10 V scale 
N = no. of full bits

Reading of 0.6973 = 0.697 volts

Given that, V_ref = 100 mV
T₁ = 100 msec
V­_in = 100 mV
We know that,
V_in T₁ = V_ref T₂
⇒ (100) (100) = (100) (T₂)
⇒ T₂ = 100 msec
Total conversion time
T = T₁ + T₂
T = 100 + 100 = 200 ms

The meter uses a full wave rectifier circuit it indicates value of 2.22 V the form factor for full wave rectified sinusoidal waveform is 1.11 
Average value of voltage,
V_avg = 2 V
For triangular wave shape, peak value of voltage V_m = 2 V_avg = 4 V

The output voltage of dual slope integrating type AD converter,

= 80 t
Given that, output voltage cannot exceed 10 V
⇒ 80 t = 10 ⇒ 125 msec

In the absence of D₂, leakage current would have flowed in D₁. To bypass this reverse leakage current of D₁ during the negative half cycle of the input. The absence of D₂ reduces the average value for complete cycle than it would have been in half wave rectification.

The actual voltage of R₂ is:

The voltage after inserting the voltmeter will then be

Thus, error percentage

Meter A:
Total resistance = 2 + 22 = 24 kΩ

Meter B:
Total resistance = 2 + 324 = 326 kΩ

Meter A is more sensitive