Electrical Engineering - Alternating Current and Voltage - Discussion

Discussion :: Alternating Current and Voltage - General Questions (Q.No.10)

10. 

A sinusoidal current has an rms value of 14 mA. The peak-to-peak value is

[A]. 45.12 mA
[B]. 16 mA
[C]. 39.6 mA
[D]. 22.6 mA

Answer: Option C

Explanation:

No answer description available for this question.

Shital Hadiyal said: (Jul 29, 2011)  
Irms = I X 0.707
I = 28/0.707 = 39.6mA

Roshith said: (Aug 12, 2011)  
peak to peak=2*peak
so peak=14ma*sqrt2=19.79ma
ie peak to peak=2*19=39.6

Kamran Alam said: (Mar 23, 2013)  
Given Irms = 14 mA.

We know that Irms = Ip/0.707.

So Ip = Irms* 0.707.

Ip = 14 ma* 0.707 = 19.79 ma.

But this is simply a value of Ip.

We also know that Peak to peak current = 2Ip.

So Peak to peak current = 2*19.79ma = 39.6 mA.

Yasmin said: (Mar 26, 2013)  
Ipeak = Irms*sqrt(2)
= Irms*1.414.

Since it is a sine wave, +14 and -14.
Hence Irms=28.

Ipeak= 28*1.414
= 39.59
=39.6mA.

Nitishkumar said: (Oct 28, 2013)  
Irms_ = ip/1.414(under root2) So, Ip in this ques is 14*1.414 = 19.796 now,

Peak to peak is 2*Ip.

Rafiqul Islam said: (Sep 30, 2014)  
Given,

I rms =14 mA.

We know Irms = Ip/0.707.
=> IP = Irms * 0.707.
=> Ip = 19.8 mA.

Now.
Ipeak to I peak = 2* Ip
= 2*19.8
= 39.597.

So Ip to Ip = 39.6 mA.

Vijay Malavalli said: (May 29, 2015)  
14*2root2 = 39.59 mA.

Akshay D. Kamathe said: (Jul 14, 2015)  
GIVEN:

Irms = 14 m, Irms (p-p) = 14*2 = 28 m.

We have:

I(p-p) = Irms (p-p)*squreroot of 2 or 1.41.
I(p-p) = 28 m*1.41.
I(p-p) = 39.6.

Satya said: (Jan 2, 2017)  
@Rafiqul.

How its possible 14*0.707=9.898?

Solve it once again.

Puja said: (Jan 14, 2017)  
Please tell me, How the Irms = Ip/sqrt 2?

Muneeb said: (Oct 5, 2018)  
I(peak) = 1.414 * Vrms,
= 1.414 * 14,
= 19.796.
I(P-P). = 19.796 * 2 = 39.59.

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