### Discussion :: Measurements and Instrumentation - Section 1 (Q.No.3)

- «« Measurements and Instrumentation - Section 1
- Measurements and Instrumentation - Section 2
- Measurements and Instrumentation - Section 3
- Measurements and Instrumentation - Section 4
- Measurements and Instrumentation - Section 5
- Measurements and Instrumentation - Section 6
- Measurements and Instrumentation - Section 7
- Measurements and Instrumentation - Section 8
- Measurements and Instrumentation - Section 9
- «« Measurements and Instrumentation - Section 10

Suma said: (May 14, 2014) | |

I can't understood the measuring of ac current. Why you put j314.16? |

Annu Kumari said: (Jun 21, 2014) | |

From where you put the value of j314.16? |

Cheetha said: (Aug 25, 2014) | |

Inductive reactance = 2*pi*f*L. |

Khasim Chotu said: (Dec 14, 2014) | |

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. |

Ritika said: (Jul 30, 2015) | |

Measuring of alternate current sine wave form so imaginary axis will also take as i-axis. Current = Voltage/Resistance. |

Ravinder said: (Nov 6, 2015) | |

Still not understood because if 2pifl = what its not j314.6 then how? |

Zameer said: (Apr 3, 2016) | |

Why didn't you consider the 500 ohms of moving iron at the calculation of ac? |

Shyam Babu Yadav said: (May 20, 2016) | |

I can't understand the measuring of AC current. Why you take j314.16? Please explain that. |

Karthik said: (Aug 1, 2016) | |

I didn't get the value of 2pifl, here f = 50hz, l = 1H, and how to find pi values can you elaborate this please. |

Prabhas said: (Sep 12, 2016) | |

The value of φ = (22/7). |

Prabhas said: (Sep 12, 2016) | |

At last why they have divided AC value and DC value? Please help me. |

Rahul(Power Engineer) said: (Oct 3, 2016) | |

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. |

Deepshikha said: (Dec 12, 2016) | |

Why not considered the 2000 ohm resistance for dc current? |

Shikha said: (Dec 12, 2016) | |

Inductance value is added in dc it's right but why resistance value is different? |

Surya said: (Dec 17, 2016) | |

Ac impedance is Z = R + jXl, so here j314.16 = 2 * pi * 50 * 1 as given inductance is 1H. |

Bijendra said: (Jan 1, 2017) | |

I have doubt in Meter current of AC? Please enplane more about this part. |

Ester said: (Jan 4, 2017) | |

I've calculated the 2 * pi * f * L and I got 0.8883. I didn't get the 0.09922. why? |

Pipo said: (Jan 6, 2017) | |

Find first fall Idc. Find Iac. Reading = iac/idc * 250. |

Sidhu said: (Jan 15, 2017) | |

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? |

Lou said: (Jan 19, 2017) | |

It should be 250/(2500+j314) to get 0.09922. |

Jitedra Kumar Gupta said: (Mar 7, 2017) | |

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? |

Skm said: (Apr 3, 2017) | |

How we calculate i taking j? Please help me. |

Praveen said: (Apr 20, 2017) | |

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). |

Sneha.V.K said: (Nov 14, 2017) | |

V/(R+jXl). 250/(2500+j (2πfl)). 250/(2500+j 314.159). |

Deepika said: (Aug 5, 2018) | |

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. |

Avinash said: (Nov 23, 2018) | |

Here, We need to find the Impedance. X=πr^2+XL^2. |

Mahendra Prajapati said: (Feb 20, 2019) | |

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. |

Lavanya said: (Sep 25, 2019) | |

I am not understanding this, Please explain it clearly to get it. |

Ykj said: (Nov 16, 2019) | |

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 |

Rakesh said: (May 28, 2020) | |

Clearly explained. |

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