Electronics and Communication Engineering - Measurements and Instrumentation - Discussion
Discussion Forum : Measurements and Instrumentation - Section 1 (Q.No. 3)
3.
The coil of a moving iron instrument has a resistance of 500 Ω and an inductance of 1 H. It reads 250 V when a 250 V dc is applied. If series resistance is 2000 Ω, its reading when fed by 250 V, 50 Hz ac will be
Answer: Option
Explanation:
Discussion:
30 comments Page 2 of 3.
Jitedra Kumar Gupta said:
8 years ago
I have a doubt.
As this is an moving iron type instrument so, T (torque)must be proportional to (current)^2.
So, if I1 and I2 are currents dor AC and DC.
THEN,
The answer would come as 246.
Can someone please explain?
As this is an moving iron type instrument so, T (torque)must be proportional to (current)^2.
So, if I1 and I2 are currents dor AC and DC.
THEN,
The answer would come as 246.
Can someone please explain?
Lou said:
9 years ago
It should be 250/(2500+j314) to get 0.09922.
Sidhu said:
9 years ago
xl=j314.16=2πfl.
=2 * 3.14 * 50 * 1 = 314.
250/(2500+314) = 0.08884.
But all of you said 0.09922.
Can anyone explain?
=2 * 3.14 * 50 * 1 = 314.
250/(2500+314) = 0.08884.
But all of you said 0.09922.
Can anyone explain?
Pipo said:
9 years ago
Find first fall Idc.
Find Iac.
Reading = iac/idc * 250.
Find Iac.
Reading = iac/idc * 250.
Ester said:
9 years ago
I've calculated the 2 * pi * f * L and I got 0.8883.
I didn't get the 0.09922. why?
I didn't get the 0.09922. why?
Suma said:
1 decade ago
I can't understood the measuring of ac current. Why you put j314.16?
Surya said:
9 years ago
Ac impedance is Z = R + jXl, so here j314.16 = 2 * pi * 50 * 1 as given inductance is 1H.
Deepshikha said:
9 years ago
Why not considered the 2000 ohm resistance for dc current?
Shikha said:
9 years ago
Inductance value is added in dc it's right but why resistance value is different?
Rahul(power engineer) said:
9 years ago
In AC we have to consider the effect of both resistance and reactance (i.e Xl = 2 pi f l) f = 50, l = 1.
To calculate current I through meter I = voltag/ impedance.
Impedance in AC only = R + j Xl ohms.
To calculate current I through meter I = voltag/ impedance.
Impedance in AC only = R + j Xl ohms.
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