# 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

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.