Electrical Engineering - Alternating Current and Voltage - Discussion
Discussion Forum : 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
Discussion:
11 comments Page 1 of 2.
Muneeb said:
7 years ago
I(peak) = 1.414 * Vrms,
= 1.414 * 14,
= 19.796.
I(P-P). = 19.796 * 2 = 39.59.
= 1.414 * 14,
= 19.796.
I(P-P). = 19.796 * 2 = 39.59.
(1)
Puja said:
9 years ago
Please tell me, How the Irms = Ip/sqrt 2?
(1)
Satya said:
9 years ago
@Rafiqul.
How its possible 14*0.707=9.898?
Solve it once again.
How its possible 14*0.707=9.898?
Solve it once again.
AKSHAY D. KAMATHE said:
1 decade ago
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.
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.
Vijay Malavalli said:
1 decade ago
14*2root2 = 39.59 mA.
Rafiqul Islam said:
1 decade ago
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.
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.
(1)
Nitishkumar said:
1 decade ago
Irms_ = ip/1.414(under root2) So, Ip in this ques is 14*1.414 = 19.796 now,
Peak to peak is 2*Ip.
Peak to peak is 2*Ip.
(1)
Yasmin said:
1 decade ago
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.
= Irms*1.414.
Since it is a sine wave, +14 and -14.
Hence Irms=28.
Ipeak= 28*1.414
= 39.59
=39.6mA.
Kamran Alam said:
1 decade ago
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.
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.
(1)
Roshith said:
1 decade ago
peak to peak=2*peak
so peak=14ma*sqrt2=19.79ma
ie peak to peak=2*19=39.6
so peak=14ma*sqrt2=19.79ma
ie peak to peak=2*19=39.6
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