Solution:
Method 1
The sensitivity of the meter movement, Sdc=1/Ifs = 1/1mA = 1kΩ/V
Rs= Sdc X Rangedc – Rm
=1k X (0.45Erms)/1 – 300
= 1k X 0.45(10) – 300
= 4.2kΩ
Method 2
The ac sensitivity for half wave rectifier, Sac = 0.45Sdc = 0.45(1k) = 450W/V
Rs= Sac X Rangeac – Rm
= 450 X 10 –300
= 4.2kΩ
Method 3
Rs=(0.45Erms)/Ifs – Rm
=(0.45X10)/1m –300
= 4.2kΩ
2.Each diode in the full wave rectifier circuit in figure has an average forward resistance of 50Ω and is assumed to have an infinite resistance in the reverse direction:
Calculates:
a) 𝑅𝑠 b) ac sensitivity c) dc sensitivity
Solution:
a) Let analyze in +ve half cycle
b)
c)
3) Figure 1 shows a simple series circuit of R1 and R2 connected to a 100 Vdc source. If the voltage across R2 is to be measured by voltmeters having:
i) a sensitivity of 1000 Ω/V
ii) a sensitivity of 20000 Ω/V
Identify the voltage reading and percentage of error measured by using the both voltmeters. Find which voltmeter will read the accurate value of voltage across R2. Both meters are used on the 100 V range. Why have error and how to reduce the error of measurement because the voltmeters (Loading Effect).
Figure 4
4) The series ohmmeter in Figure 4 is made up of a 1.5 V battery, a 100 μA meter, and R1 which makes (R1 + Rm) = 15 kΩ.
i) Determine the instrument indication when Rx =0
ii) Determine how the resistance scale should be marked at FSD.
iii) Determine how the resistance scale should be marked at 0.75 FSD.
iv) Determine how the resistance scale should be marked at 0.5 FSD.
v) Determine how the resistance scale should be marked at 0.25 FSD.