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 1 of 3.
MAHENDRA PRAJAPATI said:
7 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 = voltage/ impedance.
Impedance in AC only = R + j Xl ohms.
Here,
jxL = jwL = J*2*(22/7)*50*1 = J2200/7 = J314.2.
NOW PUT R+JJWL VALUE FOR CURRENT CALCULATION.
To calculate current I through meter I = voltage/ impedance.
Impedance in AC only = R + j Xl ohms.
Here,
jxL = jwL = J*2*(22/7)*50*1 = J2200/7 = J314.2.
NOW PUT R+JJWL VALUE FOR CURRENT CALCULATION.
(4)
Khasim chotu said:
1 decade ago
In dc there will be no term of inductive reactance.
So that meter current reads for ac = (i) = v/r = 250/(2500+j 314.16) = 0.09922 A.
J 314.6 because Inductive reactance = 2*pi*f*L.
NOW voltage(v) = ir = 0.09922*2500 = 248 v.
So that meter current reads for ac = (i) = v/r = 250/(2500+j 314.16) = 0.09922 A.
J 314.6 because Inductive reactance = 2*pi*f*L.
NOW voltage(v) = ir = 0.09922*2500 = 248 v.
Deepika said:
7 years ago
For 250 V DC current is 0.1 A.
But for the same voltage in AC i.e 250 V AC current is 0.9922 A.
Looking from the other side, for 0.1 A reading is 250 V.
for 0.9922 A we need to find the reading!!
so,
0.9922/0.1*250 = 248 V.
But for the same voltage in AC i.e 250 V AC current is 0.9922 A.
Looking from the other side, for 0.1 A reading is 250 V.
for 0.9922 A we need to find the reading!!
so,
0.9922/0.1*250 = 248 V.
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?
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.
Praveen said:
8 years ago
Inductance only makes a contribution in AC circuits.Hence it is neglected in the DC calculation.
@Shikha, @Deepshikha: R value taken in both cases is same (2000+500=2500).
@Shikha, @Deepshikha: R value taken in both cases is same (2000+500=2500).
Ritika said:
1 decade ago
Measuring of alternate current sine wave form so imaginary axis will also take as i-axis.
Current = Voltage/Resistance.
Current = Voltage/Resistance.
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?
YKJ said:
6 years ago
D.C = V/R = 250/(2000+500) = 0.1A.
A.C = V/Z = 250/sqrt(R^2+(XL)^2) = 250/sqrt(2500^2+(2*3.14*50*1)^2) = 0.09922A
A.C = V/Z = 250/sqrt(R^2+(XL)^2) = 250/sqrt(2500^2+(2*3.14*50*1)^2) = 0.09922A
Karthik said:
9 years ago
I didn't get the value of 2pifl, here f = 50hz, l = 1H, and how to find pi values can you elaborate this please.
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