Test: Electrical & Electronic Measurements- 2 - Electrical Engineering (EE) MCQ

In positive half cycle, the diode gets forward biased. Hence diode is in on state so, no current will flows through ammeter.
In negative half cycle, the diode gets reverse biased. Hence diode is in off state. So, the current flows through ammeter.

The current waveform will be,

PMMC instrument measures average value.

T_d = K_dI²
for spring control T_c = K_cθ

Given that,
Full scale deflection = 50 mA
Internal resistance of instrument = 10 Ω
We can increase the range of voltage by using series resistance.

750 = (10 + R_S) × 50 × 10^-3
⇒ R_S + 10 = 15,000
⇒ R_S = 14990 Ω
Hence the resistance should be connected in series.

Electrodynamometer voltmeter reads the rms value.
Given that,
e = 200 sin ωt + 40 sin (3ωt + 30°) + 30 cos (5 ωt + 21.5°) V

From the figure (a)
Internal resistance of galvanometer = 50 mV/1 mA
= 50 Ω
Now the circuit becomes,

Total internal resistance of voltmeter for 500 V scale,
R_m = 300 + [(199.5 + 0.05)||600]
= 300 + [200||600]
= 300 + 150 = 450 kΩ
Now, the voltmeter with internal resistance of 450 kΩ and 500 V scale is used to measure voltage in the circuit shown in fig (b).
Now, the circuit becomes

Because of voltmeter resistance, the load resistance becomes = 360||450 = 200 kΩ

By voltage division,

We know that,

Given that,

V = 3000 V
K = 7.06 × 10^-6 Nm/rad

Sensitivity of meter A, S_A = 10 kΩ/V
Full scale reading of meter A, V_FSD(A) = 200 V
Internal resistance of meter A,
R_V(A) = 10 kΩ/V × 200 V
= 2000 KΩ
Guaranteed accuracy of Meter A = 98% at FSD.
Error of meter 
Sensitivity of meter B, S_B = 4KΩ/V
Full Scale reading of meter B, V_FSD = 100V
Internal resistance of meter B,
R_V(B) = 4 KΩ/V × 100 = 400 KΩ
Guaranteed accuracy of meter B = 98.5% at FSD
Error of meter 
True reading across 22 KΩ resister 
If we use meter A, then the reading will be the effective will be, The effective resistance across 22kΩ
= 22k ||2000k = 21.76 kΩ
Now the reading 
But the meter has an error of 4V
Hence the range of reading = [49.71, 57.71]
If we use meter B, then the reading will be
The effective resistance across 22kΩ,
= 22k||400k = 20.85kΩ
Now, the reading 
But the meter has an error of 1.5V
Hence, the range of reading = [50.75, 53.75]
By observing both the readings we can conclude that meter B provides accurate results.

Full scale deflection of instrument (V_FSD) = 100V
Resistance (R_m) = 2 kΩ
Inductance (L_m) = 0.5 H
X_L = 2πfL = 2π × 50 × 0.5 = 157.08 Ω

Current through meter when 100 V at 50 Hz supply is applied

Voltage reading of meter 
 

PMMC measures average value.
The average value

T_c = 200 × 10^-6 N – m
T_d = B I N A
= 1.2 × I × 120 × 50 × 40 × 10^-6
= 288 × 10^-3 I n – m
At equilibrium,
T_c = T_d
⇒ 200 × 10^-6 = 288 × 10^-3 I
⇒ I = 0.694 mA.
Let the resistance of the voltmeter circuit be R voltage across the instrument = 0.694 × 10^-3 R
This produces a deflection of 100 divisions.
Volts per division 
This value should be equal to 1 in order to get 1 volt per division.